# Although each of these planes is rather large

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Although each of these planes is rather large, from a distance their motion can be
analysed as if each plane were a particle.
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Kinematics of a Particle
CHAPTER OBJECTIVES
• To introduce the concepts of position, displacement, velocity, and
acceleration.
• To study particle motion along a straight line and represent this
motion graphically.
• To investigate particle motion along a curved path using different
coordinate systems.
• To present an analysis of dependent motion of two particles.
• To examine the principles of relative motion of two particles using
translating axes.
12.1 Introduction
Mechanics is a branch of the physical sciences that is concerned with the
state of rest or motion of bodies subjected to the action of forces.
Engineering mechanics is divided into two areas of study, namely, statics
and dynamics. Statics is concerned with the equilibrium of a body that is
either at rest or moves with constant velocity. Here we will consider
dynamics, which deals with the accelerated motion of a body. The subject
of dynamics will be presented in two parts: kinematics, which treats only
the geometric aspects of the motion, and kinetics, which is the analysis of
the forces causing the motion. To develop these principles, the dynamics
of a particle will be discussed first, followed by topics in rigid-body
dynamics in two and then three dimensions.
12
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Historically, the principles of dynamics developed when it was possible
to make an accurate measurement of time. Galileo Galilei (1564–1642)
was one of the first major contributors to this field. His work consisted of
experiments using pendulums and falling bodies. The most significant
contributions in dynamics, however, were made by Isaac Newton
(1642–1727), who is noted for his formulation of the three fundamental
laws of motion and the law of universal gravitational attraction. Shortly
after these laws were postulated, important techniques for their
application were developed by Euler, D’Alembert, Lagrange, and others.
There are many problems in engineering whose solutions require
application of the principles of dynamics. Typically the structural design
of any vehicle, such as an automobile or airplane, requires consideration
of the motion to which it is subjected. This is also true for many
mechanical devices, such as motors, pumps, movable tools, industrial
manipulators, and machinery. Furthermore, predictions of the motions of
artificial satellites, projectiles, and spacecraft are based on the theory of
dynamics. With further advances in technology, there will be an even
greater need for knowing how to apply the principles of this subject.
Problem Solving. Dynamics is considered to be more involved
than statics since both the forces applied to a body and its motion must
be taken into account. Also, many applications require using calculus,
rather than just algebra and trigonometry. In any case, the most effective
way of learning the principles of dynamics is to solve problems. To be
successful at this, it is necessary to present the work in a logical and
orderly manner as suggested by the following sequence of steps:
1. Read the problem carefully and try to correlate the actual physical
situation with the theory you have studied.
2. Draw any necessary diagrams and tabulate the problem data.
3. Establish a coordinate system and apply the relevant principles,
generally in mathematical form.
4. Solve the necessary equations algebraically as far as practical; then,
use a consistent set of units and complete the solution numerically.
Report the answer with no more significant figures than the
accuracy of the given data.
5. Study the answer using technical judgment and common sense to
determine whether or not it seems reasonable.
6. Once the solution has been completed, review the problem. Try to
think of other ways of obtaining the same solution.
In applying this general procedure, do the work as neatly as possible.
Being neat generally stimulates clear and orderly thinking, and vice versa.
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12.2
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5
Rectilinear Kinematics: Continuous
Motion
12
We will begin our study of dynamics by discussing the kinematics of a
particle that moves along a rectilinear or straight line path. Recall that a
particle has a mass but negligible size and shape. Therefore we must limit
application to those objects that have dimensions that are of no
consequence in the analysis of the motion. In most problems, we will be
interested in bodies of finite size, such as rockets, projectiles, or vehicles.
Each of these objects can be considered as a particle, as long as the motion
is characterized by the motion of its mass center and any rotation of the
body is neglected.
Rectilinear Kinematics. The kinematics of a particle is
characterized by specifying, at any given instant, the particle’s position,
velocity, and acceleration.
Position. The straight-line path of a particle will be defined using a
single coordinate axis s, Fig. 12–1a. The origin O on the path is a fixed
point, and from this point the position coordinate s is used to specify the
location of the particle at any given instant. The magnitude of s is the
distance from O to the particle, usually measured in meters (m) or feet
(ft), and the sense of direction is defined by the algebraic sign on s.
Although the choice is arbitrary, in this case s is positive since the
coordinate axis is positive to the right of the origin. Likewise, it is
negative if the particle is located to the left of O. Realize that position is
a vector quantity since it has both magnitude and direction. Here,
however, it is being represented by the algebraic scalar s since the
direction always remains along the coordinate axis.
Displacement. The displacement of the particle is defined as the
change in its position. For example, if the particle moves from one point
to another, Fig. 12–1b, the displacement is
s
O
s
Position
(a)
s
O
s
!s
s¿
¢s = s¿ - s
In this case ¢s is positive since the particle’s final position is to the right
of its initial position, i.e., s¿ 7 s. Likewise, if the final position were to the
left of its initial position, ¢s would be negative.
The displacement of a particle is also a vector quantity, and it should be
distinguished from the distance the particle travels. Specifically, the
distance traveled is a positive scalar that represents the total length of
path over which the particle travels.
Displacement
(b)
Fig. 12–1
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Velocity. If the particle moves through a displacement ¢s during the
12
time interval ¢t, the average velocity of the particle during this time
interval is
¢s
¢t
vavg =
If we take smaller and smaller values of ¢t, the magnitude of ¢s
becomes smaller and smaller. Consequently, the instantaneous velocity is
a vector defined as v = lim 1¢s> ¢t2, or
¢t : 0
+ 2
1:
v
s
O
!s
Velocity
(c)
v =
ds
dt
(12–1)
Since ¢t or dt is always positive, the sign used to define the sense of the
velocity is the same as that of ¢s or ds. For example, if the particle is
moving to the right, Fig. 12–1c, the velocity is positive; whereas if it is
moving to the left, the velocity is negative. (This is emphasized here by
the arrow written at the left of Eq. 12–1.) The magnitude of the velocity
is known as the speed, and it is generally expressed in units of m>s or ft>s.
Occasionally, the term “average speed” is used. The average speed is
always a positive scalar and is defined as the total distance traveled by a
particle, sT , divided by the elapsed time ¢t; i.e.,
1vsp2avg =
sT
¢t
For example, the particle in Fig. 12–1d travels along the path of length sT
in time ¢t, so its average speed is 1vsp2avg = sT>¢t, but its average
velocity is vavg = - ¢s> ¢t.
!s
P¿
P
O
sT
Average velocity and
Average speed
(d)
Fig. 12–1 (cont.)
s
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7
Acceleration. Provided the velocity of the particle is known at two
12
points, the average acceleration of the particle during the time interval ¢t
is defined as
aavg =
¢v
¢t
Here ¢v represents the difference in the velocity during the time
interval ¢t, i.e., ¢v = v¿ - v, Fig. 12–1e.
The instantaneous acceleration at time t is a vector that is found by
taking smaller and smaller values of ¢t and corresponding smaller and
smaller values of ¢v, so that a = lim 1¢v>¢t2, or
a
v
¢t : 0
+ 2
1:
dv
a =
dt
s
O
v¿
Acceleration
(e)
(12–2)
Substituting Eq. 12–1 into this result, we can also write
+ 2
1:
a =
d2s
dt2
Both the average and instantaneous acceleration can be either positive
or negative. In particular, when the particle is slowing down, or its speed
is decreasing, the particle is said to be decelerating. In this case, v¿ in
Fig. 12–1f is less than v, and so ¢v = v¿ - v will be negative.
Consequently, a will also be negative, and therefore it will act to the left,
in the opposite sense to v. Also, note that when the velocity is constant,
the acceleration is zero since ¢v = v - v = 0. Units commonly used to
express the magnitude of acceleration are m>s2 or ft>s2.
Finally, an important differential relation involving the displacement,
velocity, and acceleration along the path may be obtained by eliminating
the time differential dt between Eqs. 12–1 and 12–2, which gives
+ 2
1:
a ds = v dv
a
P
P¿
O
(12–3)
Although we have now produced three important kinematic
equations, realize that the above equation is not independent of
Eqs. 12–1 and 12–2.
v
Deceleration
(f)
v¿
s
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Constant Acceleration, a = ac. When the acceleration is
constant, each of the three kinematic equations ac = dv>dt, v = ds>dt,
and ac ds = v dv can be integrated to obtain formulas that relate a c , v, s,
and t.
Velocity as a Function of Time. Integrate ac = dv>dt, assuming
that initially v = v0 when t = 0.
v
Lv0
+ 2
1:
L0
dv =
t
ac dt
v = v0 + act
Constant Acceleration
(12–4)
Position as a Function of Time. Integrate v = ds>dt = v0 + act,
assuming that initially s = s0 when t = 0.
Ls0
+ 2
1:
s
ds =
L0
t
1v0 + act2 dt
s = s0 + v0t + 12 act2
Constant Acceleration
(12–5)
Velocity as a Function of Position. Either solve for t in Eq. 12–4
and substitute into Eq. 12–5, or integrate v dv = ac ds, assuming that
initially v = v0 at s = s0 .
v
Lv0
+ 2
1:
v dv =
Ls0
s
ac ds
v2 = v20 + 2ac1s - s02
Constant Acceleration
(12–6)
The algebraic signs of s0 , v0 , and ac , used in the above three equations,
are determined from the positive direction of the s axis as indicated by
the arrow written at the left of each equation. Remember that these
equations are useful only when the acceleration is constant and when
t = 0, s = s0 , v = v0 . A typical example of constant accelerated motion
occurs when a body falls freely toward the earth. If air resistance is
neglected and the distance of fall is short, then the downward
acceleration of the body when it is close to the earth is constant and
approximately 9.81 m>s2 or 32.2 ft>s2. The proof of this is given in
Example 13.2.
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9
Important Points
•
•
•
•
•
•
•
•
•
12
Dynamics is concerned with bodies that have accelerated motion.
Kinematics is a study of the geometry of the motion.
Kinetics is a study of the forces that cause the motion.
Rectilinear kinematics refers to straight-line motion.
Speed refers to the magnitude of velocity.
Average speed is the total distance traveled divided by the total
time. This is different from the average velocity, which is the
displacement divided by the time.
A particle that is slowing down is decelerating.
A particle can have an acceleration and yet have zero velocity.
The relationship a ds = v dv is derived from a = dv>dt and
v = ds>dt, by eliminating dt.
s
During the time this rocket undergoes rectilinear
motion, its altitude as a function of time can be
measured and expressed as s = s1t2. Its velocity
can then be found using v = ds>dt, and its
acceleration can be determined from a = dv>dt.
Procedure for Analysis
Coordinate System.
• Establish a position coordinate s along the path and specify its fixed origin and positive direction.
• Since motion is along a straight line, the vector quantities position, velocity, and acceleration can be
represented as algebraic scalars. For analytical work the sense of s, v, and a is then defined by their
algebraic signs.
• The positive sense for each of these scalars can be indicated by an arrow shown alongside each kinematic
equation as it is applied.
Kinematic Equations.
• If a relation is known between any two of the four variables a, v, s and t, then a third variable can be
obtained by using one of the kinematic equations, a = dv>dt, v = ds>dt or a ds = v dv, since each
equation relates all three variables.*
• Whenever integration is performed, it is important that the position and velocity be known at a given
instant in order to evaluate either the constant of integration if an indefinite integral is used, or the limits
of integration if a definite integral is used.
• Remember that Eqs. 12–4 through 12–6 have only limited use. These equations apply only when the
acceleration is constant and the initial conditions are s = s0 and v = v0 when t = 0.
*Some standard differentiation and integration formulas are given in Appendix A.
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EXAMPLE 12.1
The car in Fig. 12–2 moves in a straight line such that for a short time
its velocity is defined by v = 13t2 + 2t2 ft>s, where t is in seconds.
Determine its position and acceleration when t = 3 s. When t = 0,
s = 0.
a, v
s
O
Fig. 12–2
SOLUTION
Coordinate System. The position coordinate extends from the
fixed origin O to the car, positive to the right.
Position. Since v = f1t2, the car’s position can be determined from
v = ds>dt, since this equation relates v, s, and t. Noting that s = 0
when t = 0, we have*
+ 2
1:
v =
L0
s
ds =
ds
= 13t2 + 2t2
dt
L0
s
t
13t2 + 2t2dt
s ` = t3 + t2 `
0
3
s = t + t
When t = 3 s,
t
0
2
s = 1323 + 1322 = 36 ft
Ans.
Acceleration. Since v = f1t2, the acceleration is determined from
a = dv>dt, since this equation relates a, v, and t.
+ 2
1:
When t = 3 s,
dv
d
=
13t2 + 2t2
dt
dt
= 6t + 2
a =
a = 6132 + 2 = 20 ft>s2 :
Ans.
The formulas for constant acceleration cannot be used to
solve this problem, because the acceleration is a function of time.
NOTE:
*The same result can be obtained by evaluating a constant of integration C rather
than using definite limits on the integral. For example, integrating ds = 13t2 + 2t2dt
yields s = t3 + t2 + C. Using the condition that at t = 0, s = 0, then C = 0.
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EXAMPLE 12.2
12
A small projectile is fired vertically downward into a fluid medium with
an initial velocity of 60 m>s. Due to the drag resistance of the fluid the
projectile experiences a deceleration of a = 1-0.4v32 m>s2, where v is in
m>s. Determine the projectile’s velocity and position 4 s after it is fired.
SOLUTION
Coordinate System. Since the motion is downward, the position
coordinate is positive downward, with origin located at O, Fig. 12–3.
Velocity. Here a = f1v2 and so we must determine the velocity as a
function of time using a = dv>dt, since this equation relates v, a, and t.
(Why not use v = v0 + act?) Separating the variables and integrating,
with v0 = 60 m>s when t = 0, yields
dv
1+ T2
a =
= - 0.4v3
dt
v
t
dv
=
dt
3
L60 m>s -0.4v
L0
1
1
1 v
a
b 2` = t - 0
-0.4 -2 v 60
1 1
1
c
d = t
0.8 v2
16022
v = ec
-1>2
1
+ 0.8t d
f m>s
2
1602
Here the positive root is taken, since the projectile will continue to
move downward. When t = 4 s,
v = 0.559 m>sT
Ans.
Position. Knowing v = f1t2, we can obtain the projectile’s position
from v = ds>dt, since this equation relates s, v, and t. Using the initial
condition s = 0, when t = 0, we have
1+ T2
v =
s
-1>2
1
ds
= c
+
0.8t
d
dt
16022
t
-1>2
1
+
0.8t
d
dt
2
L0
L0 1602
1>2 t
1
2
c
+
0.8t
d
s =
`
0.8 16022
0
ds =
s =
When t = 4 s,
11
c
1>2
1
1
1
ec
+ 0.8t d
fm
2
0.4 1602
60
s = 4.43 m
Ans.
O
s
Fig. 12–3
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EXAMPLE 12.3
During a test a rocket travels upward at 75 m>s, and when it is 40 m
from the ground its engine fails. Determine the maximum height sB
reached by the rocket and its speed just before it hits the ground.
While in motion the rocket is subjected to a constant downward
acceleration of 9.81 m>s2 due to gravity. Neglect the effect of air
resistance.
SOLUTION
Coordinate System. The origin O for the position coordinate s is
taken at ground level with positive upward, Fig. 12–4.
Maximum Height.
Since the rocket is traveling upward,
vA = + 75m>s when t = 0. At the maximum height s = sB the velocity
vB = 0. For the entire motion, the acceleration is ac = - 9.81 m>s2
(negative since it acts in the opposite sense to positive velocity or
positive displacement). Since ac is constant the rocket’s position may
be related to its velocity at the two points A and B on the path by using
Eq. 12–6, namely,
vB " 0
B
1+ c 2
v2B = v2A + 2ac1sB - sA2
0 = 175 m>s22 + 21- 9.81 m>s221sB - 40 m2
sB = 327 m
sB
Ans.
Velocity. To obtain the velocity of the rocket just before it hits the
ground, we can apply Eq. 12–6 between points B and C, Fig. 12–4.
vA " 75 m/s
1+ c2
A
sA " 40 m
C
Fig. 12–4
s
v2C = v2B + 2ac1sC - sB2
= 0 + 21-9.81 m>s2210 - 327 m2
vC = - 80.1 m>s = 80.1 m>s T
O
Ans.
The negative root was chosen since the rocket is moving downward.
Similarly, Eq. 12–6 may also be applied between points A and C, i.e.,
1+ c 2
v2C = v2A + 2ac1sC - sA2
= 175 m>s22 + 21- 9.81 m>s2210 - 40 m2
vC = - 80.1 m>s = 80.1 m>s T
Ans.
NOTE: It should be realized that the rocket is subjected to a
deceleration from A to B of 9.81 m>s2, and then from B to C it is
accelerated at this rate. Furthermore, even though the rocket
momentarily comes to rest at B 1vB = 02 the acceleration at B is still
9.81 m>s2 downward!
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13
EXAMPLE 12.4
12
A metallic particle is subjected to the influence of a magnetic field as it
travels downward through a fluid that extends from plate A to plate B,
Fig. 12–5. If the particle is released from rest at the midpoint C,
s = 100 mm, and the acceleration is a = 14s2 m>s2, where s is in
meters, determine the velocity of the particle when it reaches plate B,
s = 200 mm, and the time it takes to travel from C to B.
SOLUTION
Coordinate System. As shown in Fig. 12–5, s is positive downward,
measured from plate A.
Velocity. Since a = f1s2, the velocity as a function of position can
be obtained by using v dv = a ds. Realizing that v = 0 at s = 0.1 m,
we have
1 + T2
L0
A
100 mm
s
C
v
v dv = a ds
s
v dv =
L0.1 m
4s ds
B
1 2 v
4 s
v ` = s2 `
2
2 0.1 m
0
v = 21s2 - 0.0121>2 m>s
(1)
At s = 200 mm = 0.2 m,
vB = 0.346 m>s = 346 mm>s T
Ans.
The positive root is chosen since the particle is traveling downward,
i.e., in the +s direction.
Time. The time for the particle to travel from C to B can be obtained
using v = ds>dt and Eq. 1, where s = 0.1 m when t = 0. From
Appendix A,
1 + T2
ds = v dt
= 21s2 - 0.0121>2dt
s
t
ds
=
2 dt
2
1>2
L0.1 1s - 0.012
L0
ln A 4s2 - 0.01 + s B `
At s = 0.2 m,
s
0.1
= 2t `
t
0
ln A 4s - 0.01 + s B + 2.303 = 2t
2
ln A 410.222 - 0.01 + 0.2 B + 2.303
= 0.658 s
Ans.
2
Note: The formulas for constant acceleration cannot be used here
because the acceleration changes with position, i.e., a = 4s.
t =
200 mm
Fig. 12–5
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EXAMPLE 12.5
A particle moves along a horizontal path with a velocity of
v = 13t2 - 6t2 m>s, where t is the time in seconds. If it is initially
located at the origin O, determine the distance traveled in 3.5 s, and the
particle’s average velocity and average speed during the time interval.
s " #4.0 m
SOLUTION
Coordinate System. Here positive motion is to the right, measured
from the origin O, Fig. 12–6a.
s " 6.125 m
O
t"2s
t"0s
t " 3.5 s
(a)
Distance Traveled. Since v = f1t2, the position as a function of
time may be found by integrating v = ds>dt with t = 0, s = 0.
+ 2
1:
ds = v dt
L0
s
ds =
L0
t
(3t2 - 6t) dt
s = 1t3 - 3t22m
v (m/s)
v " 3t2 # 6t
(0, 0)
= 13t2 - 6t2dt
(2 s, 0)
(1 s, #3 m/s)
(b)
Fig. 12–6
t (s)
(1)
In order to determine the distance traveled in 3.5 s, it is necessary to
investigate the path of motion. If we consider a graph of the velocity
function, Fig. 12–6b, then it reveals that for 0 6 t 6 2 s the velocity is
negative, which means the particle is traveling to the left, and for t 7 2 s
the velocity is positive, and hence the particle is traveling to the right.
Also, note that v = 0 at t = 2 s. The particle’s position when t = 0,
t = 2 s, and t = 3.5 s can now be determined from Eq. 1. This yields
s ƒ t=0 = 0
s ƒ t = 2 s = - 4.0 m
s ƒ t = 3.5 s = 6.125 m
The path is shown in Fig. 12–6a. Hence, the distance traveled in 3.5 s is
sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m
Ans.
Velocity. The displacement from t = 0 to t = 3.5 s is
¢s = s ƒ t = 3.5 s - s ƒ t = 0 = 6.125 m - 0 = 6.125 m
and so the average velocity is
vavg =
¢s
6.125 m
=
= 1.75 m>s :
¢t
3.5 s - 0
Ans.
The average speed is defined in terms of the distance traveled sT . This
positive scalar is
sT
14.125 m
Ans.
=
= 4.04 m>s
1vsp2avg =
¢t
3.5 s - 0
Note: In this problem, the acceleration is a = dv>dt = 16t - 62 m>s2,
which is not constant.
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15
FUNDAMENTAL PROBLEMS
F12–1. Initially, the car travels along a straight road with a
speed of 35 m>s. If the brakes are applied and the speed of
the car is reduced to 10 m>s in 15 s, determine the constant
deceleration of the car.
12
F12–5. The position of the particle is given by
s = (2t2 - 8t + 6) m, where t is in seconds. Determine the
time when the velocity of the particle is zero, and the total
distance traveled by the particle when t = 3 s.
s
F12–1
F12–5
F12–2. A ball is thrown vertically upward with a speed of
15 m>s. Determine the time of flight when it returns to its
original position.
F12–6. A particle travels along a straight line with an
acceleration of a = (10 - 0.2s) m>s2, where s is measured in
meters. Determine the velocity of the particle when s = 10 m
if v = 5 m>s at s = 0.
s
s
s
F12–2
F12–6
F12–3. A particle travels along a straight line with a velocity
of v = (4t - 3t2) m>s, where t is in seconds. Determine the
position of the particle when t = 4 s. s = 0 when t = 0 .
F12–7. A particle moves along a straight line such that its
acceleration is a = (4t2 - 2) m>s2, where t is in seconds.
When t = 0, the particle is located 2 m to the left of the
origin, and when t = 2 s, it is 20 m to the left of the origin.
Determine the position of the particle when t = 4 s.
s
s
F12–3
F12–7
F12–4. A particle travels along a straight line with a speed
v = (0.5t3 - 8t) m>s, where t is in seconds. Determine the
acceleration of the particle when t = 2 s.
F12–8. A particle travels along a straight line with a
velocity of v = (20 - 0.05s2) m>s, where s is in meters.
Determine the acceleration of the particle at s = 15 m.
s
F12–4
s
F12–8
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PROBLEMS
•12–1. A car starts from rest and with constant
acceleration achieves a velocity of 15 m>s when it travels a
distance of 200 m. Determine the acceleration of the car
and the time required.
12–2. A train starts from rest at a station and travels with a
constant acceleration of 1 m>s2. Determine the velocity of the
train when t = 30 s and the distance traveled during this time.
12–3. An elevator descends from rest with an acceleration
of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the
time required and the distance traveled.
*12–4. A car is traveling at 15 m>s, when the traffic light
50 m ahead turns yellow. Determine the required constant
deceleration of the car and the time needed to stop the car
at the light.
•12–5. A particle is moving along a straight line with the
acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds.
Determine the velocity and the position of the particle as a
function of time. When t = 0, v = 0 and s = 15 ft.
12–6. A ball is released from the bottom of an elevator
which is traveling upward with a velocity of 6 ft>s. If the ball
strikes the bottom of the elevator shaft in 3 s, determine the
height of the elevator from the bottom of the shaft at the
instant the ball is released. Also, find the velocity of the ball
when it strikes the bottom of the shaft.
12–7. A car has an initial speed of 25 m>s and a constant
deceleration of 3 m>s2. Determine the velocity of the car
when t = 4 s. What is the displacement of the car during the
4-s time interval? How much time is needed to stop the car?
*12–8. If a particle has an initial velocity of v0 = 12 ft>s to
the right, at s0 = 0, determine its position when t = 10 s, if
a = 2 ft>s2 to the left.
•12–9. The acceleration of a particle traveling along a
straight line is a = k>v, where k is a constant. If s = 0, v = v0
when t = 0, determine the velocity of the particle as a
function of time t.
12–10. Car A starts from rest at t = 0 and travels along a
straight road with a constant acceleration of 6 ft>s2 until it
reaches a speed of 80 ft>s. Afterwards it maintains this
speed. Also, when t = 0, car B located 6000 ft down the
road is traveling towards A at a constant speed of 60 ft>s.
Determine the distance traveled by car A when they pass
each other.
60 ft/s
A
B
6000 ft
Prob. 12–10
12–11. A particle travels along a straight line with a velocity
v = (12 - 3t2) m>s, where t is in seconds. When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from t = 0
to t = 10 s, and the distance the particle travels during this
time period.
*12–12. A sphere is fired downwards into a medium with
an initial speed of 27 m>s. If it experiences a deceleration of
a = ( -6t) m>s2, where t is in seconds, determine the
distance traveled before it stops.
•12–13. A particle travels along a straight line such
that in 2 s it moves from an initial position sA = + 0.5 m to
a position sB = - 1.5 m. Then in another 4 s it moves from
sB to sC = + 2.5 m. Determine the particle’s average
velocity and average speed during the 6-s time interval.
12–14. A particle travels along a straight-line path such
that in 4 s it moves from an initial position sA = -8 m to a
position sB = + 3 m. Then in another 5 s it moves from sB to
sC = -6 m. Determine the particle’s average velocity and
average speed during the 9-s time interval.
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12.2
12–15. Tests reveal that a normal driver takes about 0.75 s
before he or she can react to a situation to avoid a collision. It
takes about 3 s for a driver having 0.1% alcohol in his system
to do the same. If such drivers are traveling on a straight road
at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2 ,
determine the shortest stopping distance d for each from the
moment they see the pedestrians. Moral: If you must drink,
v1 " 44 ft/s
d
RECTILINEAR KINEMATICS: CONTINUOUS MOTION
17
•12–21. Two particles A and B start from rest at the origin
s = 0 and move along a straight line such that 12
aA = (6t - 3) ft>s2 and aB = (12t2 - 8) ft>s2, where t is in
seconds. Determine the distance between them when
t = 4 s and the total distance each has traveled in t = 4 s.
12–22. A particle moving along a straight line is subjected
to a deceleration a = ( -2v3) m>s2, where v is in m>s. If it
has a velocity v = 8 m>s and a position s = 10 m when
t = 0, determine its velocity and position when t = 4 s.
12–23. A particle is moving along a straight line such that
its acceleration is defined as a = ( - 2v) m>s2, where v is in
meters per second. If v = 20 m>s when s = 0 and t = 0,
determine the particle’s position, velocity, and acceleration
as functions of time.
Prob. 12–15
*12–16. As a train accelerates uniformly it passes successive
kilometer marks while traveling at velocities of 2 m>s and
then 10 m>s. Determine the train’s velocity when it passes
the next kilometer mark and the time it takes to travel the
2-km distance.
•12–17. A ball is thrown with an upward velocity of 5 m>s
from the top of a 10-m high building. One second later
another ball is thrown vertically from the ground with a
velocity of 10 m>s. Determine the height from the ground
where the two balls pass each other.
12–18. A car starts from rest and moves with a constant
acceleration of 1.5 m>s2 until it achieves a velocity of 25 m>s .
It then travels with constant velocity for 60 seconds.
Determine the average speed and the total distance
traveled.
12–19. A car is to be hoisted by elevator to the fourth floor
of a parking garage, which is 48 ft above the ground. If the
elevator can accelerate at 0.6 ft>s2, decelerate at 0.3 ft>s2,
and reach a maximum speed of 8 ft>s, determine the shortest
time to make the lift, starting from rest and ending at rest.
*12–20. A particle is moving along a straight line such that
its speed is defined as v = (- 4s2) m>s, where s is in meters.
If s = 2 m when t = 0, determine the velocity and
acceleration as functions of time.
*12–24. A particle starts from rest and travels along a
straight line with an acceleration a = (30 - 0.2v) ft>s2,
where v is in ft>s. Determine the time when the velocity of
the particle is v = 30 ft>s.
•12–25. When a particle is projected vertically upwards
with an initial velocity of v0, it experiences an acceleration
a = -(g + kv2) , where g is the acceleration due to gravity,
k is a constant and v is the velocity of the particle.
Determine the maximum height reached by the particle.
12–26. The acceleration of a particle traveling along a
straight line is a = (0.02et) m>s2, where t is in seconds. If
v = 0, s = 0 when t = 0, determine the velocity and
acceleration of the particle at s = 4 m.
12–27. A particle moves along a straight line with an
acceleration of a = 5>(3s1>3 + s5>2) m>s2, where s is in
meters. Determine the particle’s velocity when s = 2 m, if it
starts from rest when s = 1 m. Use Simpson’s rule to
evaluate the integral.
*12–28. If the effects of atmospheric resistance are
accounted for, a falling body has an acceleration defined by
the equation a = 9.81[1 - v2(10-4)] m>s2, where v is in m>s
and the positive direction is downward. If the body is
released from rest at a very high altitude, determine (a) the
velocity when t = 5 s, and (b) the body’s terminal or
maximum attainable velocity (as t : q ).
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•12–29. The position of a particle along a straight line is
12 given by s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in
seconds. Determine the position of the particle when
t = 6 s and the total distance it travels during the 6-s time
interval. Hint: Plot the path to determine the total distance
traveled.
12–30. The velocity of a particle traveling along a straight
line is v = v0 - ks, where k is constant. If s = 0 when t = 0,
determine the position and acceleration of the particle as a
function of time.
12–31. The acceleration of a particle as it moves along a
straight line is given by a = 12t - 12 m>s2, where t is in
seconds. If s = 1 m and v = 2 m>s when t = 0, determine
the particle’s velocity and position when t = 6 s. Also,
determine the total distance the particle travels during this
time period.
*12–32. Ball A is thrown vertically upward from the top
of a 30-m-high-building with an initial velocity of 5 m>s. At
the same instant another ball B is thrown upward from the
ground with an initial velocity of 20 m>s. Determine the
height from the ground and the time at which they pass.
•12–33. A motorcycle starts from rest at t = 0 and travels
along a straight road with a constant acceleration of 6 ft>s2
until it reaches a speed of 50 ft>s. Afterwards it maintains
this speed. Also, when t = 0, a car located 6000 ft down the
road is traveling toward the motorcycle at a constant speed
of 30 ft>s. Determine the time and the distance traveled by
the motorcycle when they pass each other.
*12–36. The acceleration of a particle traveling along a
straight line is a = (8 - 2s) m>s2, where s is in meters. If
v = 0 at s = 0, determine the velocity of the particle at
s = 2 m, and the position of the particle when the velocity
is maximum.
•12–37. Ball A is thrown vertically upwards with a velocity
of v0. Ball B is thrown upwards from the same point with
the same velocity t seconds later. Determine the elapsed
time t 6 2v0>g from the instant ball A is thrown to when the
balls pass each other, and find the velocity of each ball at
this instant.
12–38. As a body is projected to a high altitude above the
earth’s surface, the variation of the acceleration of gravity
with respect to altitude y must be taken into account.
Neglecting air resistance, this acceleration is determined
from the formula a = -g0[R2>(R + y)2], where g0 is the
constant gravitational acceleration at sea level, R is the
radius of the earth, and the positive direction is measured
upward. If g0 = 9.81 m>s2 and R = 6356 km, determine the
minimum initial velocity (escape velocity) at which a
projectile should be shot vertically from the earth’s surface
so that it does not fall back to the earth. Hint: This requires
that v = 0 as y : q .
12–39. Accounting for the variation of gravitational
acceleration a with respect to altitude y (see Prob. 12–38),
derive an equation that relates the velocity of a freely
falling particle to its altitude. Assume that the particle is
released from rest at an altitude y0 from the earth’s surface.
With what velocity does the particle strike the earth if it is
released from rest at an altitude y0 = 500 km? Use the
numerical data in Prob. 12–38.
12–34. A particle moves along a straight line with a
velocity v = (200s) mm>s, where s is in millimeters.
Determine the acceleration of the particle at s = 2000 mm.
How long does the particle take to reach this position if
s = 500 mm when t = 0?
*12–40. When a particle falls through the air, its initial
acceleration a = g diminishes until it is zero, and
thereafter it falls at a constant or terminal velocity vf . If
this variation of the acceleration can be expressed as
a = 1g>v2f21v2f - v22, determine the time needed for the
velocity to become v = vf>2. Initially the particle falls
from rest.
!12–35. A particle has an initial speed of 27 m>s. If it
experiences a deceleration of a = 1- 6t2 m>s2, where t is in
seconds, determine its velocity, after it has traveled 10 m.
How much time does this take?
•12–41. A particle is moving along a straight line such that
its position from a fixed point is s = (12 - 15t2 + 5t3) m,
where t is in seconds. Determine the total distance traveled
by the particle from t = 1 s to t = 3 s. Also, find the average
speed of the particle during this time interval.
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RECTILINEAR KINEMATICS: ERRATIC MOTION
12.3 Rectilinear Kinematics: Erratic
s
ds
ds
v0 " dt t " 0 v2 " dt t
2
ds
ds
v1 " dt t
v3 " dt t
1
3
Motion
When a particle has erratic or changing motion then its position, velocity,
and acceleration cannot be described by a single continuous mathematical
function along the entire path. Instead, a series of functions will be
required to specify the motion at different intervals. For this reason, it is
convenient to represent the motion as a graph. If a graph of the motion
that relates any two of the variables s,v, a, t can be drawn, then this graph
can be used to construct subsequent graphs relating two other variables
since the variables are related by the differential relationships v = ds>dt,
a = dv>dt, or a ds = v dv. Several situations occur frequently.
19
s2
s1
O
t1
s3
t2
t
t3
(a)
v
The s–t, v–t, and a–t Graphs. To construct the v–t graph given
the s–t graph, Fig. 12–7a, the equation v = ds>dt should be used, since it
relates the variables s and t to v. This equation states that
ds
= v
dt
slope of
= velocity
s–t graph
v0
v2
O
t1
t2
t3
v3
t
(b)
Fig. 12–7
For example, by measuring the slope on the s–t graph when t = t1 , the
velocity is v1 , which is plotted in Fig. 12–7b. The v–t graph can be
constructed by plotting this and other values at each instant.
The a–t graph can be constructed from the v–t graph in a similar
manner, Fig. 12–8, since
v
a " dv
a0 " dv
dt t " 0 2 dt t2
dv
a3 " dt t
a1 " dv
3
dt t1
dv
= a
dt
v3
slope of
= acceleration
v–t graph
Examples of various measurements are shown in Fig. 12–8a and plotted
in Fig. 12–8b.
If the s–t curve for each interval of motion can be expressed by a
mathematical function s = s(t), then the equation of the v–t graph for
the same interval can be obtained by differentiating this function with
respect to time since v = ds/dt. Likewise, the equation of the a–t graph
for the same interval can be determined by differentiating v = v(t) since
a = dv>dt. Since differentiation reduces a polynomial of degree n to
that of degree n – 1, then if the s–t graph is parabolic (a second-degree
curve), the v–t graph will be a sloping line (a first-degree curve), and the
a–t graph will be a constant or a horizontal line (a zero-degree curve).
v1
v2
v0
O
v1
t2
t1
t
t3
(a)
a
a1
O
a0 " 0
t1
a2
t2
(b)
Fig. 12–8
a3
t3
t
12
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If the a–t graph is given, Fig. 12–9a, the v–t graph may be constructed
using a = dv>dt, written as
a
12
Page 20
a0
t1
!v " ! a dt
0
t
t1
(a)
a dt
L
change in
area under
velocity = a–t graph
¢v =
v
!v
v1
v0
t
t1
(b)
Fig. 12–9
Hence, to construct the v–t graph, we begin with the particle’s initial
velocity v0 and then add to this small increments of area 1¢v2
determined from the a–t graph. In this manner successive points,
v1 = v0 + ¢v, etc., for the v–t graph are determined, Fig. 12–9b. Notice
that an algebraic addition of the area increments of the a–t graph is
necessary, since areas lying above the t axis correspond to an increase in
v (“positive” area), whereas those lying below the axis indicate a
decrease in v (“negative” area).
Similarly, if the v–t graph is given, Fig. 12–10a, it is possible to
determine the s–t graph using v = ds>dt, written as
v
v dt
L
area under
displacement = v–t graph
¢s =
v0
t1
!s " ! v dt
0
t
t1
(a)
s
s1
!s
s0
t
t1
(b)
Fig. 12–10
In the same manner as stated above, we begin with the particle’s initial
position s0 and add (algebraically) to this small area increments ¢s
determined from the v–t graph, Fig. 12–10b.
If segments of the a–t graph can be described by a series of equations,
then each of these equations can be integrated to yield equations
describing the corresponding segments of the v–t graph. In a similar
manner, the s–t graph can be obtained by integrating the equations
which describe the segments of the v–t graph. As a result, if the a–t graph
is linear (a first-degree curve), integration will yield a v–t graph that is
parabolic (a second-degree curve) and an s–t graph that is cubic (thirddegree curve).
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RECTILINEAR KINEMATICS: ERRATIC MOTION
The v–s and a–s Graphs. If the a–s graph can be constructed,
then points on the v–s graph can be determined by using v dv = a ds.
Integrating this equation between the limits v = v0 at s = s0 and v = v1
at s = s1 , we have,
21
a
12
s1
a0
!0 a ds " —12 (v12 # v02)
s
s1
(a)
v
1 2
2 1v1
s1
a ds
Ls0
area under
a–s graph
- v202 =
v1
v0
s
s1
(b)
Therefore, if the red area in Fig. 12–11a is determined, and the initial
s
velocity v0 at s0 = 0 is known, then v1 = A 2 1s01a ds + v20 B 1>2, Fig. 12–11b.
Successive points on the v–s graph can be constructed in this manner.
If the v–s graph is known, the acceleration a at any position s can be
determined using a ds = v dv, written as
Fig. 12–11
v
dv
ds
v0
dv
b
ds
velocity times
acceleration = slope of
v–s graph
v
a = va
Thus, at any point (s, v) in Fig. 12–12a, the slope dv>ds of the v–s graph is
measured. Then with v and dv>ds known, the value of a can be
calculated, Fig. 12–12b.
The v–s graph can also be constructed from the a–s graph, or vice versa,
by approximating the known graph in various intervals with mathematical
functions, v = f(s) or a = g(s), and then using a ds = v dv to obtain the
other graph.
s
s
(a)
a
a0
a " v(dv/ds)
s
s
(b)
Fig. 12–12
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EXAMPLE 12.6
A bicycle moves along a straight road such that its position is
described by the graph shown in Fig. 12–13a. Construct the v–t and
a–t graphs for 0 … t … 30 s.
s (ft)
500
s " 20t # 100
100
s " t2
10
30
t (s)
(a)
SOLUTION
v–t Graph. Since v = ds>dt, the v–t graph can be determined by
differentiating the equations defining the s–t graph, Fig. 12–13a.We have
ds
0 … t 6 10 s;
s = (t2) ft
v =
= (2t) ft>s
dt
ds
10 s 6 t … 30 s;
s = (20t - 100) ft
v =
= 20 ft>s
dt
v (ft/s)
v " 2t
v " 20
20
10
30
t (s)
(b)
a (ft/s2)
2
10
30
(c)
Fig. 12–13
t (s)
The results are plotted in Fig. 12–13b. We can also obtain specific
values of v by measuring the slope of the s–t graph at a given instant.
For example, at t = 20 s, the slope of the s–t graph is determined from
the straight line from 10 s to 30 s, i.e.,
¢s
500 ft - 100 ft
t = 20 s;
v =
=
= 20 ft>s
¢t
30 s - 10 s
a–t Graph. Since a = dv>dt, the a–t graph can be determined by
differentiating the equations defining the lines of the v–t graph.
This yields
dv
0 … t 6 10 s;
v = (2t) ft>s
a =
= 2 ft>s2
dt
dv
10 6 t … 30 s;
v = 20 ft>s
a =
= 0
dt
The results are plotted in Fig. 12–13c.
NOTE: Show that a = 2 ft>s2 when t = 5 s by measuring the slope of
the v–t graph.
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RECTILINEAR KINEMATICS: ERRATIC MOTION
23
EXAMPLE 12.7
12
The car in Fig. 12–14a starts from rest and travels along a straight
track such that it accelerates at 10 m>s2 for 10 s, and then decelerates
at 2 m>s2. Draw the v–t and s–t graphs and determine the time t¿
needed to stop the car. How far has the car traveled?
SOLUTION
v–t Graph. Since dv = a dt, the v–t graph is determined by integrating
the straight-line segments of the a–t graph. Using the initial condition
v = 0 when t = 0, we have
v
t
a (m/s2)
10
A1
t¿
#2
10 dt,
v = 10t
L0
L0
When t = 10 s, v = 101102 = 100 m>s. Using this as the initial
condition for the next time period, we have
0 … t 6 10 s;
a = (10) m>s2;
dv =
v
t (s)
A2
10
(a)
t
-2 dt, v = (-2t + 120) m>s
L100 m>s
L10 s
v (m/s)
When t = t¿ we require v = 0. This yields, Fig. 12–14b,
v " 10t
Ans.
t¿ = 60 s
100
A more direct solution for t¿ is possible by realizing that the area
under the a–t graph is equal to the change in the car’s velocity. We
require ¢v = 0 = A 1 + A 2 , Fig. 12–14a. Thus
10 s 6 t … t¿; a = ( -2) m>s2;
dv =
0 = 10 m>s2110 s2 + 1- 2 m>s221t¿ - 10 s2
t¿ = 60 s
Ans.
s–t Graph. Since ds = v dt, integrating the equations of the v–t
graph yields the corresponding equations of the s–t graph. Using the
initial condition s = 0 when t = 0, we have
s
v " #2t \$ 120
10
t¿ " 60
t (s)
(b)
t
10t dt,
s = (5t2) m
L0
L0
When t = 10 s, s = 511022 = 500 m. Using this initial condition,
0 … t … 10 s;
v = (10t) m>s;
ds =
s
10 s … t … 60 s; v = ( -2t + 120) m>s;
2
L500 m
t
ds =
L10 s
2
s - 500 = -t + 120t - [-1102 + 1201102]
1-2t + 1202 dt
s (m)
3000
s = ( -t2 + 120t - 600) m
s " 5t2
When t¿ = 60 s, the position is
s = -16022 + 1201602 - 600 = 3000 m
Ans.
500
The s–t graph is shown in Fig. 12–14c.
A direct solution for s is possible when t¿ = 60 s, since the
triangular area under the v–t graph would yield the displacement
¢s = s - 0 from t = 0 to t¿ = 60 s. Hence,
NOTE:
¢s =
1
2 160
s21100 m>s2 = 3000 m
Ans.
s " #t2 \$ 120t # 600
10
60
(c)
Fig. 12–14
t (s)
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EXAMPLE 12.8
The v–s graph describing the motion of a motorcycle is shown in
Fig. 12–15a. Construct the a–s graph of the motion and determine the
time needed for the motorcycle to reach the position s = 400 ft.
SOLUTION
a–s Graph. Since the equations for segments of the v–s graph are
given, the a–s graph can be determined using a ds = v dv.
v (ft/s)
50
v " 0.2s \$ 10
v " 50
0 … s 6 200 ft;
a = v
10
200
400
s (ft)
(a)
v = (0.2s + 10) ft>s
dv
d
= 10.2s + 102 10.2s + 102 = 0.04s + 2
ds
ds
200 ft 6 s … 400 ft;
v = 50 ft>s
a = v
dv
d
= 1502 1502 = 0
ds
ds
The results are plotted in Fig. 12–15b.
Time. The time can be obtained using the v–s graph and v = ds>dt,
because this equation relates v, s, and t. For the first segment of
motion, s = 0 when t = 0, so
ds
ds
0 … s 6 200 ft;
v = (0.2s + 10) ft>s;
dt =
=
v
0.2s + 10
a (ft/s2)
a " 0.04s \$ 2
10
2
a"0
200
400
(b)
Fig. 12–15
L0
s (ft)
t
s
dt =
ds
0.2s
+ 10
L0
t = (5 ln10.2s + 102 - 5 ln 10) s
At s = 200 ft, t = 5 ln[0.212002 + 10] - 5 ln 10 = 8.05 s. Therefore,
using these initial conditions for the second segment of motion,
200 ft 6 s … 400 ft;
dt =
v = 50 ft>s;
t
ds
ds
=
v
50
s
ds
;
L8.05 s
L200 m 50
s
s
t - 8.05 =
- 4; t = a
+ 4.05b s
50
50
dt =
Therefore, at s = 400 ft,
400
+ 4.05 = 12.0 s
Ans.
50
NOTE: The graphical results can be checked in part by calculating slopes.
For example, at s = 0, a = v1dv>ds2 = 10150 - 102>200 = 2 m>s2.
Also, the results can be checked in part by inspection. The v–s graph
indicates the initial increase in velocity (acceleration) followed by
constant velocity 1a = 02.
t =
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25
FUNDAMENTAL PROBLEMS
F12–9. The particle travels along a straight track such that
its position is described by the s - t graph. Construct the v- t
graph for the same time interval.
s (m)
12
F12–12. The sports car travels along a straight road such
that its position is described by the graph. Construct the v- t
and a -t graphs for the time interval 0 … t … 10 s.
s (m)
225
s ! 108
108
s ! 30t " 75
s ! 0.5 t3
75
s ! 3t2
0
6
8
10
5
t (s)
F12–9
F12–10. A van travels along a straight road with a velocity
described by the graph. Construct the s - t and a - t graphs
during the same period. Take s = 0 when t = 0.
v (ft/s)
t (s)
10
F12–12
F12–13. The dragster starts from rest and has an
acceleration described by the graph. Construct the v- t
graph for the time interval 0 … t … t¿, where t¿ is the time
for the car to come to rest.
a (m/s2)
20
80
v ! "4t # 80
t¿
0
20
t (s)
F12–10
F12–11. A bicycle travels along a straight road where its
velocity is described by the v - s graph. Construct the a-s
graph for the same time interval.
v (m/s)
t (s)
5
"10
F12–13
F12–14. The dragster starts from rest and has a velocity
described by the graph. Construct the s-t graph during the
time interval 0 … t … 15 s. Also, determine the total
distance traveled during this time interval.
v (m/s)
v ! 30 t
10
150
v ! "15 t # 225
v ! 0.25 s
40
F12–11
s (m)
5
15
F12–14
t (s)
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PROBLEMS
12–42. The speed of a train during the first minute has
been recorded as follows:
t (s)
0
20
40
60
v (m>s)
0
16
21
24
Plot the v- t graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.
12–43. A two-stage missile is fired vertically from rest with
the acceleration shown. In 15 s the first stage A burns out
and the second stage B ignites. Plot the v - t and s- t graphs
which describe the two-stage motion of the missile for
0 … t … 20 s.
12–46. A train starts from station A and for the first
kilometer, it travels with a uniform acceleration. Then, for
the next two kilometers, it travels with a uniform speed.
Finally, the train decelerates uniformly for another
kilometer before coming to rest at station B. If the time for
the whole journey is six minutes, draw the v–t graph and
determine the maximum speed of the train.
12–47. The particle travels along a straight line with the
velocity described by the graph. Construct the a- s graph.
v (m/s)
13
10
v!s#7
B
4
A
2
a (m/s )
v ! 2s # 4
s (m)
3
6
Prob. 12–47
25
18
15
20
t (s)
*12–48. The a–s graph for a jeep traveling along a straight
road is given for the first 300 m of its motion. Construct the
v–s graph. At s = 0, v = 0.
Prob. 12–43
a (m/s2)
*12–44. A freight train starts from rest and travels with a
constant acceleration of 0.5 ft>s2. After a time t¿ it
maintains a constant speed so that when t = 160 s it has
traveled 2000 ft. Determine the time t¿ and draw the v–t
graph for the motion.
•12–45. If the position of a particle is defined by
s = [2 sin (p>5)t + 4] m, where t is in seconds, construct
the s- t, v - t, and a- t graphs for 0 … t … 10 s.
2
200
Prob. 12–48
300
s (m)
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12.3
•12–49. A particle travels along a curve defined by the
equation s = (t3 - 3t2 + 2t) m. where t is in seconds. Draw
the s - t, v - t, and a - t graphs for the particle for
0 … t … 3 s.
RECTILINEAR KINEMATICS: ERRATIC MOTION
27
*12–52. A car travels up a hill with the speed shown.
Determine the total distance the car travels until it stops 12
(t = 60 s). Plot the a- t graph.
v (m/s)
12–50. A truck is traveling along the straight line with a
velocity described by the graph. Construct the a-s graph
for 0 … s … 1500 ft.
10
v (ft/s)
30
60
t (s)
Prob. 12–52
v ! 0.6 s3/4
•12–53. The snowmobile moves along a straight course
according to the v–t graph. Construct the s–t and a–t graphs
for the same 50-s time interval. When t = 0, s = 0.
75
s(ft)
625
v (m/s)
1500
Prob. 12–50
12
12–51. A car starts from rest and travels along a straight
road with a velocity described by the graph. Determine the
total distance traveled until the car stops. Construct the s–t
and a–t graphs.
30
50
t (s)
Prob. 12–53
12–54. A motorcyclist at A is traveling at 60 ft>s when he
wishes to pass the truck T which is traveling at a constant
speed of 60 ft>s. To do so the motorcyclist accelerates at
6 ft>s2 until reaching a maximum speed of 85 ft>s. If he then
maintains this speed, determine the time needed for him to
reach a point located 100 ft in front of the truck. Draw the
v-t and s-t graphs for the motorcycle during this time.
v(m/s)
(vm)1 ! 60 ft/s
30
vt ! 60 ft/s
A
v ! "0.5t # 45
v!t
(vm)2 ! 85 ft/s
T
30
60
90
Prob. 12–51
t(s)
40 ft
55 ft
100 ft
Prob. 12–54
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12–55. An airplane traveling at 70 m>s lands on a straight
runway and has a deceleration described by the graph.
Determine the time t¿ and the distance traveled for it to
reach a speed of 5 m>s. Construct the v–t and s -t graphs for
this time interval, 0 … t … t¿ .
•12–57. The dragster starts from rest and travels along a
straight track with an acceleration-deceleration described
by the graph. Construct the v-s graph for 0 … s … s¿, and
determine the distance s¿ traveled before the dragster again
comes to rest.
a(m/s2)
a(m/s2)
5
t¿
t(s)
25
"4
a ! 0.1s # 5
5
"10
s¿
200
s (m)
"15
Prob. 12–55
Prob. 12–57
*12–56. The position of a cyclist traveling along a straight
road is described by the graph. Construct the v–t and a–t
graphs.
12–58. A sports car travels along a straight road with an
acceleration-deceleration described by the graph. If the car
starts from rest, determine the distance s¿ the car travels
until it stops. Construct the v-s graph for 0 … s … s¿ .
s (m)
137.5
a(ft/s2)
s ! "0.625 t2 # 27.5t " 162.5
6
1000
50
s ! 0.05 t3
10
20
Prob. 12–56
t (s)
"4
Prob. 12–58
s¿
s(ft)
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12.3
12–59. A missile starting from rest travels along a straight
track and for 10 s has an acceleration as shown. Draw the
v - t graph that describes the motion and find the distance
traveled in 10 s.
a (m/s2)
RECTILINEAR KINEMATICS: ERRATIC MOTION
29
•12–61. The v- t graph of a car while traveling along a road
12
is shown. Draw the s - t and a -t graphs for the motion.
v (m/s)
40
a ! 2t # 20
30
20
a ! 6t
5
5
20
30
t (s)
t(s)
10
Prob. 12–59
Prob. 12–61
*12–60. A motorcyclist starting from rest travels along a
straight road and for 10 s has an acceleration as shown.
Draw the v -t graph that describes the motion and find the
distance traveled in 10 s.
12–62. The boat travels in a straight line with the
acceleration described by the a -s graph. If it starts from rest,
construct the v-s graph and determine the boat’s maximum
speed. What distance s¿ does it travel before it stops?
a (m/s2)
a(m/s2)
6
6
a ! "0.02s # 6
3
a
1 2
!—
t
6
150
6
Prob. 12–60
10
t (s)
"4
Prob. 12–62
s¿
s(m)
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12–63. The rocket has an acceleration described by the
12 graph. If it starts from rest, construct the v- t and s- t
graphs for the motion for the time interval 0 … t … 14 s.
•12–65. The acceleration of the speed boat starting from
rest is described by the graph. Construct the v-s graph.
a(m/s2)
a(ft/s2)
38
a2 ! 36t
10
a ! 4t " 18
a ! 0.04s # 2
18
2
200
9
500
s(ft)
t(s)
14
Prob. 12–65
Prob. 12–63
*12–64. The jet bike is moving along a straight road with the
speed described by the v - s graph. Construct the a- s graph.
12–66. The boat travels along a straight line with the speed
described by the graph. Construct the s–t and a -s graphs.
Also, determine the time required for the boat to travel a
distance s = 400 m if s = 0 when t = 0.
v(m/s)
v(m/s)
80
1/2
v ! 5s
75
v ! 0.2s
v ! "0.2s # 120
v2 ! 4s
15
225
525
s(m)
20
100
Prob. 12–64
400
Prob. 12–66
s(m)
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12.3
12–67. The s–t graph for a train has been determined
experimentally. From the data, construct the v-t and a–t
graphs for the motion.
s (m)
RECTILINEAR KINEMATICS: ERRATIC MOTION
31
•12–69. The airplane travels along a straight runway with
an acceleration described by the graph. If it starts from rest 12
and requires a velocity of 90 m>s to take off, determine the
minimum length of runway required and the time t¿ for take
off. Construct the v- t and s–t graphs.
a(m/s2)
600
s ! 24t " 360
8
360
a ! 0.8t
s ! 0.4t2
30
10
t (s)
40
t(s)
t¿
Prob. 12–69
Prob. 12–67
*12–68. The airplane lands at 250 ft>s on a straight runway
and has a deceleration described by the graph. Determine
the distance s¿ traveled before its speed is decreased to
25 ft>s. Draw the s -t graph.
a(ft/s2)
12–70. The a–t graph of the bullet train is shown. If the
train starts from rest, determine the elapsed time t¿ before it
again comes to rest. What is the total distance traveled
during this time interval? Construct the v–t and s–t graphs.
a(m/s2)
1750
s¿
s(ft)
a ! 0.1t
3
"7.5
30
1
a ! "( 15 )t # 5
75
"15
Prob. 12–68
Prob. 12–70
t¿
t(s)
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12.4 General Curvilinear Motion
12
Curvilinear motion occurs when a particle moves along a curved path.
Since this path is often described in three dimensions, vector analysis will
be used to formulate the particle’s position, velocity, and acceleration.*
In this section the general aspects of curvilinear motion are discussed, and
in subsequent sections we will consider three types of coordinate systems
often used to analyze this motion.
Position. Consider a particle located at a point on a space curve
s
s
r
O
Position
Path
(a)
\$s
s
\$r
defined by the path function s(t), Fig. 12–16a. The position of the particle,
measured from a fixed point O, will be designated by the position vector
r = r1t2. Notice that both the magnitude and direction of this vector will
change as the particle moves along the curve.
Displacement. Suppose that during a small time interval ¢t the
particle moves a distance ¢s along the curve to a new position, defined
by r¿ = r + ¢r, Fig. 12–16b. The displacement ¢r represents the change
in the particle’s position and is determined by vector subtraction; i.e.,
¢r = r¿ - r.
Velocity. During the time ¢t, the average velocity of the particle is
r¿
vavg =
r
O
Displacement
(b)
The instantaneous velocity is determined from this equation by letting
¢t : 0, and consequently the direction of ¢r approaches the tangent to
the curve. Hence, v = lim 1¢r> ¢t2 or
¢t : 0
v =
v
s
O
r
Velocity
(c)
Fig. 12–16
¢r
¢t
dr
dt
(12–7)
Since dr will be tangent to the curve, the direction of v is also tangent to
the curve, Fig. 12–16c. The magnitude of v, which is called the speed, is
obtained by realizing that the length of the straight line segment ¢r in
Fig. 12–16b approaches the arc length ¢s as ¢t : 0, we have
v = lim 1¢r>¢t2 = lim 1¢s> ¢t2, or
¢t : 0
¢t : 0
v =
ds
dt
(12–8)
Thus, the speed can be obtained by differentiating the path function s
with respect to time.
*A summary of some of the important concepts of vector analysis is given in Appendix B.
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GENERAL CURVILINEAR MOTION
Acceleration. If the particle has a velocity v at time t and a velocity
v¿
33
v
12
v¿ = v + ¢v at t + ¢t, Fig. 12–16d, then the average acceleration of the
particle during the time interval ¢t is
aavg =
¢v
¢t
(d)
where ¢v = v¿ - v. To study this time rate of change, the two velocity
vectors in Fig. 12–16d are plotted in Fig. 12–16e such that their tails are
located at the fixed point O¿ and their arrowheads touch points on a
curve. This curve is called a hodograph, and when constructed, it
describes the locus of points for the arrowhead of the velocity vector in
the same manner as the path s describes the locus of points for the
arrowhead of the position vector, Fig. 12–16a.
To obtain the instantaneous acceleration, let ¢t : 0 in the above
equation. In the limit ¢v will approach the tangent to the hodograph, and
so a = lim 1¢v>¢t2, or
\$v
v
v¿
(e)
¢t : 0
a =
dv
dt
O¿
Hodograph
(12–9)
v
a
O¿
Substituting Eq. 12–7 into this result, we can also write
a =
(f)
d 2r
dt2
By definition of the derivative, a acts tangent to the hodograph,
Fig. 12–16f, and, in general it is not tangent to the path of motion,
Fig. 12–16g. To clarify this point, realize that ¢v and consequently a
must account for the change made in both the magnitude and direction
of the velocity v as the particle moves from one point to the next along
the path, Fig. 12–16d. However, in order for the particle to follow any
curved path, the directional change always “swings” the velocity vector
toward the “inside” or “concave side” of the path, and therefore a
cannot remain tangent to the path. In summary, v is always tangent to
the path and a is always tangent to the hodograph.
a
Acceleration
path
(g)
Fig. 12–16 (cont.)
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12.5 Curvilinear Motion: Rectangular
12
Components
Occasionally the motion of a particle can best be described along a path
that can be expressed in terms of its x, y, z coordinates.
z
Position. If the particle is at point (x, y, z) on the curved path s shown
in Fig. 12–17a, then its location is defined by the position vector
s
k
i
r = xi + yj + zk
z
r ! xi # yj # zk
j
y
x
y
x
Position
(12–10)
When the particle moves, the x, y, z components of r will be functions of
time; i.e., x = x1t2, y = y1t2, z = z1t2, so that r = r1t2.
At any instant the magnitude of r is defined from Eq. C–3 in
Appendix C as
(a)
r = 4x2 + y2 + z2
And the direction of r is specified by the unit vector ur = r>r.
Velocity. The first time derivative of r yields the velocity of the
z
particle. Hence,
v =
s
v ! vxi # vyj # vzk
y
x
Velocity
When taking this derivative, it is necessary to account for changes in both
the magnitude and direction of each of the vector’s components. For
example, the derivative of the i component of r is
d
dx
di
1xi2 =
i + x
dt
dt
dt
(b)
Fig. 12–17
dr
d
d
d
=
1xi2 +
1yj2 +
1zk2
dt
dt
dt
dt
The second term on the right side is zero, provided the x, y, z reference
frame is fixed, and therefore the direction (and the magnitude) of i does
not change with time. Differentiation of the j and k components may be
carried out in a similar manner, which yields the final result,
v =
dr
= vxi + vy j + vzk
dt
(12–11)
where
#
#
#
vx = x vy = y vz = z
(12–12)
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CURVILINEAR MOTION: RECTANGULAR COMPONENTS
35
# # #
The “dot” notation x, y, z represents the first time derivatives of
x = x1t2, y = y1t2, z = z1t2, respectively.
The velocity has a magnitude that is found from
12
v = 4v2x + v2y + v2z
and a direction that is specified by the unit vector uv = v>v. As discussed
in Sec. 12–4, this direction is always tangent to the path, as shown in
Fig. 12–17b.
z
Acceleration. The acceleration of the particle is obtained by taking
the first time derivative of Eq. 12–11 (or the second time derivative of
Eq. 12–10). We have
s
a ! axi # ayj # azk
dv
a =
= axi + ay j + azk
dt
(12–13)
x
Acceleration
(c)
where
#
\$
ax = vx = x
#
\$
ay = vy = y
#
\$
az = vz = z
(12–14)
Here ax , ay , a z represent, respectively, the first time derivatives of
vx = vx1t2, vy = vy1t2, vz = vz1t2, or the second time derivatives of the
functions x = x1t2, y = y1t2, z = z1t2.
The acceleration has a magnitude
a = 4a2x + a2y + a2z
and a direction specified by the unit vector ua = a>a. Since a represents
the time rate of change in both the magnitude and direction of the
velocity, in general a will not be tangent to the path, Fig. 12–17c.
y
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Important Points
• Curvilinear motion can cause changes in both the magnitude and
direction of the position, velocity, and acceleration vectors.
• The velocity vector is always directed tangent to the path.
• In general, the acceleration vector is not tangent to the path, but
rather, it is tangent to the hodograph.
• If the motion is described using rectangular coordinates, then the
components along each of the axes do not change direction, only
their magnitude and sense (algebraic sign) will change.
• By considering the component motions, the change in magnitude
and direction of the particle’s position and velocity are
automatically taken into account.
Procedure for Analysis
Coordinate System.
• A rectangular coordinate system can be used to solve problems
for which the motion can conveniently be expressed in terms of
its x, y, z components.
Kinematic Quantities.
• Since rectilinear motion occurs along each coordinate axis, the
motion along each axis is found using v = ds>dt and a = dv>dt;
or in cases where the motion is not expressed as a function of
time, the equation a ds = v dv can be used.
• In two dimensions, the equation of the path y = f(x) can be used
to relate the x and y components of velocity and acceleration by
applying the chain rule of calculus. A review of this concept is
given in Appendix C.
• Once the x, y, z components of v and a have been determined, the
magnitudes of these vectors are found from the Pythagorean
theorem, Eq. B–3, and their coordinate direction angles from the
components of their unit vectors, Eqs. B–4 and B–5.
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CURVILINEAR MOTION: RECTANGULAR COMPONENTS
37
EXAMPLE 12.9
12
At any instant the horizontal position of the weather balloon in
Fig. 12–18a is defined by x = 18t2 ft, where t is in seconds. If the
equation of the path is y = x2>10, determine the magnitude and
direction of the velocity and the acceleration when t = 2 s.
y
B
y!
SOLUTION
Velocity. The velocity component in the x direction is
d
#
vx = x =
18t2 = 8 ft>s :
dt
x2
10
x
A
16 ft
To find the relationship between the velocity components we will use
the chain rule of calculus. (See Appendix A for a full explanation.)
(a)
d 2
#
#
1x >102 = 2xx>10 = 21162182>10 = 25.6 ft>s c
vy = y =
dt
When t = 2 s, the magnitude of velocity is therefore
v = 4(8 ft>s22 + (25.6 ft>s22 = 26.8 ft>s
Ans.
v ! 26.8 ft/s
The direction is tangent to the path, Fig. 12–18b, where
uv = tan-1
vy
vx
= tan-1
25.6
= 72.6°
8
Ans.
Acceleration. The relationship between the acceleration components
is determined using the chain rule. (See Appendix C.) We have
uv ! 72.6%
B
(b)
d
#
ax = vx =
182 = 0
dt
d
#
#
# #
\$
a y = vy =
12xx>102 = 21x2x>10 + 2x1x2>10
dt
Thus,
= 21822>10 + 21162102>10 = 12.8 ft>s2 c
a = 4(02 2 + (12.822 = 12.8 ft> s2
a ! 12.8 ft/s2
B
The direction of a, as shown in Fig. 12–18c, is
-1 12.8
ua = tan
0
= 90°
ua ! 90%
Ans.
(c)
Ans.
NOTE: It is also possible to obtain vy and a y by first expressing
y = f1t2 = 18t22>10 = 6.4t2 and then taking successive time derivatives.
Fig. 12–18
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EXAMPLE 12.10
For a short time, the path of the plane in Fig. 12–19a is described by
y = (0.001x2) m. If the plane is rising with a constant velocity of 10 m>s ,
determine the magnitudes of the velocity and acceleration of the plane
when it is at y = 100 m.
y
x
SOLUTION
When y = 100 m, then 100 = 0.001x2 or x = 316.2 m. Also, since
vy = 10 m>s, then
y = vyt;
100 m = (10 m>s) t
t = 10 s
Velocity. Using the chain rule (see Appendix C) to find the
relationship between the velocity components, we have
d
#
#
vy = y =
(0.001x2) = (0.002x)x = 0.002xvx
dt
y
(1)
Thus
y ! 0.001x2
100 m
x
10 m>s = 0.002(316.2 m)(vx)
vx = 15.81 m>s
The magnitude of the velocity is therefore
(a)
v = 4vx2 + vy2 = 4(15.81 m>s)2 + (10 m>s)2 = 18.7 m>s
Ans.
Acceleration. Using the chain rule, the time derivative of Eq. (1)
gives the relation between the acceleration components.
#
#
#
ay = vy = 0.002xvx + 0.002xvx = 0.002(v2x + xax)
#
When x = 316.2 m, vx = 15.81 m>s, vy = ay = 0,
y
a
100 m
0 = 0.002((15.81 m>s)2 + 316.2 m(ax))
ax = - 0.791 m>s2
v
vy
vx
x
(b)
Fig. 12–19
The magnitude of the plane’s acceleration is therefore
a = 4a2x + a2y = 4(- 0.791 m>s222 + (0 m>s222
= 0.791 m>s2
These results are shown in Fig. 12–19b.
Ans.
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MOTION OF A PROJECTILE
39
12.6 Motion of a Projectile
12
The free-flight motion of a projectile is often studied in terms of its
rectangular components. To illustrate the kinematic analysis, consider a
projectile launched at point (x0 , y0), with an initial velocity of v0 , having
components 1v02x and 1v02y , Fig. 12–20. When air resistance is neglected,
the only force acting on the projectile is its weight, which causes the
projectile to have a constant downward acceleration of approximately
ac = g = 9.81 m>s2 or g = 32.2 ft>s2.*
y
a!g
vx
v0
(v0)y
vy
(v0)x
v
r
y
y0
x
x0
x
Fig. 12–20
Horizontal Motion. Since ax = 0, application of the constant
acceleration equations, 12–4 to 12–6, yields
+ 2
1:
v = v + a t;
+ 2
1:
+ 2
1:
0
c
x = x0 + v0t +
1
2
2 a ct ;
v2 = v20 + 2ac1x - x02;
vx = 1v02x
x = x0 + 1v02xt
vx = 1v02x
The first and last equations indicate that the horizontal component of
velocity always remains constant during the motion.
Vertical Motion. Since the positive y axis is directed upward, then
ay = -g. Applying Eqs. 12–4 to 12–6, we get
1+ c 2
1+ c 2
1+ c2
v = v0 + act;
y = y0 + v0t +
1
2
2 a ct ;
v2 = v20 + 2ac1y - y02;
vy = 1v02y - gt
y = y0 + 1v02yt - 12 gt2
v2y = 1v022y - 2g1y - y02
Each picture in this sequence is taken
after the same time interval. The red ball
falls from rest, whereas the yellow ball is
given a horizontal velocity when released.
Both balls accelerate downward at the
same rate, and so they remain at the same
elevation at any instant. This acceleration
causes the difference in elevation between
the balls to increase between successive
photos. Also, note the horizontal distance
between successive photos of the yellow
ball is constant since the velocity in the
horizontal direction remains constant.
Recall that the last equation can be formulated on the basis of eliminating
the time t from the first two equations, and therefore only two of the
above three equations are independent of one another.
* This assumes that the earth’s gravitational field does not vary with altitude.
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To summarize, problems involving the motion of a projectile can have
at most three unknowns since only three independent equations can be
written; that is, one equation in the horizontal direction and two in the
vertical direction. Once vx and vy are obtained, the resultant velocity v,
which is always tangent to the path, can be determined by the vector sum
as shown in Fig. 12–20.
12
Procedure for Analysis
Coordinate System.
• Establish the fixed x, y coordinate axes and sketch the trajectory
•
of the particle. Between any two points on the path specify the
given problem data and identify the three unknowns. In all cases
the acceleration of gravity acts downward and equals 9.81 m>s2
or 32.2 ft>s2. The particle’s initial and final velocities should be
represented in terms of their x and y components.
Remember that positive and negative position, velocity, and
acceleration components always act in accordance with their
associated coordinate directions.
Kinematic Equations.
• Depending upon the known data and what is to be determined, a
choice should be made as to which three of the following four
equations should be applied between the two points on the path
to obtain the most direct solution to the problem.
Horizontal Motion.
• The velocity in the horizontal or x direction is constant, i.e.,
vx = 1v02x , and
x = x0 + 1v02x t
Vertical Motion.
• In the vertical or y direction only two of the following three
equations can be used for solution.
Gravel falling off the end of this conveyor
belt follows a path that can be predicted
using the equations of constant acceleration.
In this way the location of the accumulated
pile can be determined. Rectangular
coordinates are used for the analysis since
the acceleration is only in the vertical
direction.
vy = 1v02y + act
y = y0 + 1v02y t + 12 act2
v2y = 1v022y + 2ac1y - y02
For example, if the particle’s final velocity vy is not needed, then
the first and third of these equations will not be useful.
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12.6
MOTION OF A PROJECTILE
EXAMPLE 12.11
41
12
A sack slides off the ramp, shown in Fig. 12–21, with a horizontal
velocity of 12 m>s. If the height of the ramp is 6 m from the floor,
determine the time needed for the sack to strike the floor and the
range R where sacks begin to pile up.
y
A
12 m/s
x
a!g
6m
B
C
R
Fig. 12–21
SOLUTION
Coordinate System. The origin of coordinates is established at the
beginning of the path, point A, Fig. 12–21.The initial velocity of a sack has
components 1vA2x = 12 m>s and 1vA2y = 0. Also, between points A and
B the acceleration is ay = -9.81 m>s2. Since 1vB2x = 1vA2x = 12 m>s,
the three unknowns are 1vB2y , R, and the time of flight tAB . Here we do
not need to determine 1vB2y .
Vertical Motion. The vertical distance from A to B is known, and
therefore we can obtain a direct solution for tAB by using the equation
1+ c 2
yB = yA + 1vA2ytAB + 12 act2AB
-6 m = 0 + 0 + 121-9.81 m>s22t2AB
tAB = 1.11 s
Ans.
Horizontal Motion. Since tAB has been calculated, R is determined
as follows:
+ 2
1:
xB = xA + 1vA2xtAB
R = 0 + 12 m>s 11.11 s2
R = 13.3 m
Ans.
NOTE: The calculation for tAB also indicates that if a sack were
released from rest at A, it would take the same amount of time to strike
the floor at C, Fig. 12–21.
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EXAMPLE 12.12
The chipping machine is designed to eject wood chips at vO = 25 ft>s
as shown in Fig. 12–22. If the tube is oriented at 30° from the
horizontal, determine how high, h, the chips strike the pile if at this
instant they land on the pile 20 ft from the tube.
y
vO ! 25 ft/s
30%
x
A
O
4 ft
h
20 ft
Fig. 12–22
SOLUTION
Coordinate System. When the motion is analyzed between points
O and A, the three unknowns are the height h, time of flight tOA , and
vertical component of velocity 1vA2y . [Note that 1vA2x = 1vO2x .] With
the origin of coordinates at O, Fig. 12–22, the initial velocity of a chip
has components of
1vO2x = 125 cos 30°2 ft>s = 21.65 ft>s :
1vO2y = 125 sin 30°2 ft>s = 12.5 ft>s c
Also, 1vA2x = 1vO2x = 21.65 ft>s and ay = - 32.2 ft>s2. Since we do
not need to determine 1vA2y , we have
Horizontal Motion.
+ 2
1:
xA = xO + 1vO2xtOA
20 ft = 0 + 121.65 ft>s2tOA
tOA = 0.9238 s
Vertical Motion. Relating tOA to the initial and final elevations of a
chip, we have
1+ c 2
yA = yO + 1vO2ytOA + 12 act2OA
1h - 4 ft2 = 0 + 112.5 ft>s210.9238 s2 + 121-32.2 ft>s2210.9238 s22
h = 1.81 ft
Ans.
NOTE: We can determine 1vA2y by using 1vA2y = 1vO2y + a ctOA .
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12.6
MOTION OF A PROJECTILE
43
EXAMPLE 12.13
12
The track for this racing event was designed so that riders jump off the
slope at 30°, from a height of 1 m. During a race it was observed that
the rider shown in Fig. 12–23a remained in mid air for 1.5 s. Determine
the speed at which he was traveling off the ramp, the horizontal
distance he travels before striking the ground, and the maximum
height he attains. Neglect the size of the bike and rider.
(a)
SOLUTION
Coordinate System. As shown in Fig. 12–23b, the origin of the
coordinates is established at A. Between the end points of the path
AB the three unknowns are the initial speed vA , range R, and the
vertical component of velocity 1vB2y .
y
C
30%
A
Vertical Motion. Since the time of flight and the vertical distance
between the ends of the path are known, we can determine vA .
1+ c 2
yB = yA + 1vA2ytAB + 12 act2AB
-1 m = 0 + vA sin 30°11.5 s2 + 121-9.81 m>s2211.5 s22
vA = 13.38 m>s = 13.4 m>s
1m
Ans.
R = 0 + 13.38 cos 30° m>s11.5 s2
Ans.
= 17.4 m
In order to find the maximum height h we will consider the path
AC, Fig. 12–23b. Here the three unknowns are the time of flight tAC ,
the horizontal distance from A to C, and the height h. At the
maximum height 1vC2y = 0, and since vA is known, we can determine
h directly without considering tAC using the following equation.
1vC22y = 1vA22y + 2ac[yC - yA]
02 = 113.38 sin 30° m>s22 + 21-9.81 m>s22[1h - 1 m2 - 0]
Ans.
NOTE: Show that the bike will strike the ground at B with a velocity
having components of
1vB2x = 11.6 m>s : , 1vB2y = 8.02 m>sT
x
B
R
(b)
Horizontal Motion. The range R can now be determined.
+ 2
xB = xA + 1vA2xtAB
1:
h = 3.28 m
h
Fig. 12–23
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FUNDAMENTAL PROBLEMS
F12–15. If the x and y components of a particle’s velocity
are vx = (32t) m>s and vy = 8 m>s, determine the equation
of the path y = f(x). x = 0 and y = 0 when t = 0.
F12–16. A particle is traveling along the straight path. If its
position along the x axis is x = (8t) m, where t is in
seconds, determine its speed when t = 2 s.
F12–18. A particle travels along a straight-line path
y = 0.5x. If the x component of the particle’s velocity is
vx = (2t2) m>s, where t is in seconds, determine the
magnitude of the particle’s velocity and acceleration
when t = 4 s.
y
y
y ! 0.5x
x
y ! 0.75x
F12–18
3m
x
F12–19. A particle is traveling along the parabolic path
y = 0.25x2. If x = (2t2) m, where t is in seconds, determine
the magnitude of the particle’s velocity and acceleration
when t = 2 s.
y
x ! 8t
4m
y ! 0.25x2
F12–16
F12–17. A particle is constrained to travel along the path.
If x = (4t4) m, where t is in seconds, determine the
magnitude of the particle’s velocity and acceleration when
t = 0.5 s.
y
y2 ! 4x
x
F12–19
F12–20. The position of a box sliding down the spiral can
be described by r = [2 sin (2t)i + 2 cos tj - 2t2k] ft, where
t is in seconds and the arguments for the sine and cosine are
in radians. Determine the velocity and acceleration of the
box when t = 2 s.
x
x ! (4t4) m
F12–17
F12–20
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12.6
F12–21. The ball is kicked from point A with the initial
velocity vA = 10 m>s. Determine the maximum height h it
reaches.
MOTION OF A PROJECTILE
45
F12–25. A ball is thrown from A. If it is required to clear
the wall at B, determine the minimum magnitude of its 12
initial velocity vA.
F12–22. The ball is kicked from point A with the initial
velocity vA = 10 m>s. Determine the range R, and the
speed when the ball strikes the ground.
y
y
B
xB
B
vA ! 10 m/s
A
v
h
30%
C
30%
A
x
8 ft
x
3 ft
x
F12–21/22
12 ft
F12–23. Determine the speed at which the basketball at A
must be thrown at the angle of 30° so that it makes it to the
y
B
vA
30%
A
F12–25
x
3m
1.5 m
F12–26. A projectile is fired with an initial velocity of
vA = 150 m>s off the roof of the building. Determine the
range R where it strikes the ground at B.
10 m
F12–23
F12–24. Water is sprayed at an angle of 90° from the slope
at 20 m>s. Determine the range R.
y
vA ! 150 m/s
vB ! 20 m/s
A
5
4
5
3
4
x
150 m
3
R
B
R
F12–24
F12–26
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PROBLEMS
12–71. The position of a particle is r = 5(3t 3 - 2t)i
– (4t1/2 + t)j + (3t2 - 2)k6 m, where t is in seconds.
Determine the magnitude of the particle’s velocity and
acceleration when t = 2 s.
*12–72. The velocity of a particle is v = 53i + (6 - 2t)j6 m>s,
where t is in seconds. If r = 0 when t = 0, determine the
displacement of the particle during the time interval
t = 1 s to t = 3 s.
•12–73. A particle travels along the parabolic path y = bx2 .
If its component of velocity along the y axis is vy = ct2,
determine the x and y components of the particle’s
acceleration. Here b and c are constants.
•12–77. The position of a particle is defined by
r = 55 cos 2t i + 4 sin 2t j6 m, where t is in seconds and the
arguments for the sine and cosine are given in radians.
Determine the magnitudes of the velocity and acceleration
of the particle when t = 1 s. Also, prove that the path of the
particle is elliptical.
12–78. Pegs A and B are restricted to move in the elliptical
slots due to the motion of the slotted link. If the link moves
with a constant speed of 10 m/s, determine the magnitude of
the velocity and acceleration of peg A when x = 1 m.
y
12–74. The velocity of a particle is given by
v = 516t2i + 4t3j + (5t + 2)k6 m>s, where t is in seconds. If
the particle is at the origin when t = 0, determine the
magnitude of the particle’s acceleration when t = 2 s. Also,
what is the x, y, z coordinate position of the particle at this
instant?
12–75. A particle travels along the circular path
x 2 + y2 = r2. If the y component of the particle’s velocity is
vy = 2r cos 2t, determine the x and y components of its
acceleration at any instant.
*12–76. The box slides down the slope described by the
equation y = (0.05x2) m, where x is in meters. If the box has
x components of velocity and acceleration of vx = –3 m>s
and ax = –1.5 m>s2 at x = 5 m, determine the y components
of the velocity and the acceleration of the box at this instant.
A
C
B
D
x
v ! 10 m/s
x2
# y2 ! 1
4
Prob. 12–78
12–79. A particle travels along the path y2 = 4x with a
constant speed of v = 4 m>s. Determine the x and y
components of the particle’s velocity and acceleration when
the particle is at x = 4 m.
*12–80. The van travels over the hill described by
y = ( -1.5(10–3) x2 + 15) ft. If it has a constant speed of
75 ft>s, determine the x and y components of the van’s
velocity and acceleration when x = 50 ft.
y
y ! 0.05 x2
y
x
15 ft
y ! ("1.5 (10"3) x2 # 15) ft
x
100 ft
Prob. 12–76
Prob. 12–80
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12.6
•12–81. A particle travels along the circular path from A to
B in 1 s. If it takes 3 s for it to go from A to C, determine its
average velocity when it goes from B to C.
y
MOTION OF A PROJECTILE
47
*12–84. The path of a particle is defined by y2 = 4kx, and
the component of velocity along the y axis is vy = ct, where 12
both k and c are constants. Determine the x and y
components of acceleration when y = y0.
•12–85. A particle moves along the curve y = x - (x2>400),
where x and y are in ft. If the velocity component in the x
direction is vx = 2 ft>s and remains constant, determine the
magnitudes of the velocity and acceleration when x = 20 ft .
30%
C
45%
12–86. The motorcycle travels with constant speed v0 along
the path that, for a short distance, takes the form of a sine
curve. Determine the x and y components of its velocity at
any instant on the curve.
30 m
B
y
x
A
v0
Prob. 12–81
12–82. A car travels east 2 km for 5 minutes, then north 3 km
for 8 minutes, and then west 4 km for 10 minutes. Determine
the total distance traveled and the magnitude of displacement
of the car. Also, what is the magnitude of the average velocity
and the average speed?
12–83. The roller coaster car travels down the helical path at
constant speed such that the parametric equations that define
its position are x = c sin kt, y = c cos kt, z = h - bt, where
c, h, and b are constants. Determine the magnitudes of its
velocity and acceleration.
π x)
y ! c sin ( ––
L
x
c
L
c
L
Prob. 12–86
12–87. The skateboard rider leaves the ramp at A with an
initial velocity vA at a 30° angle. If he strikes the ground at
B, determine vA and the time of flight.
vA
A
30%
1m
B
z
5m
Prob. 12–87
*12–88. The pitcher throws the baseball horizontally with a
speed of 140 ft>s from a height of 5 ft. If the batter is 60 ft
away, determine the time for the ball to arrive at the batter
and the height h at which it passes the batter.
y
5 ft
h
x
60 ft
Prob. 12–83
Prob. 12–88
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•12–89. The ball is thrown off the top of the building. If it
strikes the ground at B in 3 s, determine the initial velocity
vA and the inclination angle uA at which it was thrown. Also,
find the magnitude of the ball’s velocity when it strikes the
ground.
12–91. The fireman holds the hose at an angle u = 30° with
horizontal, and the water is discharged from the hose at A
with a speed of vA = 40 ft>s. If the water stream strikes the
building at B, determine his two possible distances s from
the building.
vA
uA
A
vA ! 40 ft/s
75 ft
A
B
u
8 ft
4 ft
B
s
60 ft
Prob. 12–89
Prob. 12–91
12–90. A projectile is fired with a speed of v = 60 m>s at an
angle of 60°. A second projectile is then fired with the same
speed 0.5 s later. Determine the angle u of the second
projectile so that the two projectiles collide. At what
position (x, y) will this happen?
*12–92. Water is discharged from the hose with a speed of
40 ft>s. Determine the two possible angles u the fireman can
hold the hose so that the water strikes the building at B.
Take s = 20 ft.
vA ! 40 ft/s
y
A
B
u
8 ft
v ! 60 m/s
4 ft
60%
u
y
v ! 60 m/s
x
x
Prob. 12–90
s
Prob. 12–92
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12.6
•12–93. The pitching machine is adjusted so that the
baseball is launched with a speed of vA = 30 m>s . If the ball
strikes the ground at B, determine the two possible angles uA
at which it was launched.
MOTION OF A PROJECTILE
49
*12–96. The baseball player A hits the baseball with
vA = 40 ft>s and uA = 60°. When the ball is directly above 12
of player B he begins to run under it. Determine the
constant speed vB and the distance d at which B must run in
order to make the catch at the same elevation at which the
ball was hit.
vA ! 30 m/s
A
uA
1.2 m
B
30 m
vA ! 40 ft/s
Prob. 12–93
uA
A
B
15 ft
12–94. It is observed that the time for the ball to strike the
ground at B is 2.5 s. Determine the speed vA and angle uA at
which the ball was thrown.
C
vB
d
Prob. 12–96
vA
A
uA
1.2 m
B
50 m
Prob. 12–94
•12–97. A boy throws a ball at O in the air with a speed v0
at an angle u1. If he then throws another ball with the same
speed v0 at an angle u2 6 u1, determine the time between
the throws so that the balls collide in mid air at B.
12–95. If the motorcycle leaves the ramp traveling at
110 ft>s, determine the height h ramp B must have so that
the motorcycle lands safely.
B
O
110 ft/s
30%
30 ft
A
u1
y
u2
h
350 ft
Prob. 12–95
B
x
Prob. 12–97
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12–98. The golf ball is hit at A with a speed of vA = 40 m>s
12 and directed at an angle of 30° with the horizontal as
shown. Determine the distance d where the ball strikes the
slope at B.
*12–100. The velocity of the water jet discharging from the
orifice can be obtained from v = 22 gh , where h = 2 m is
the depth of the orifice from the free water surface.
Determine the time for a particle of water leaving the orifice
to reach point B and the horizontal distance x where it hits
the surface.
B
vA ! 40 m/s
2m
30%
5
A
1
vA
d
A
1.5 m
B
x
Prob. 12–100
Prob. 12–98
12–99. If the football is kicked at the 45° angle, determine
its minimum initial speed vA so that it passes over the goal
post at C. At what distance s from the goal post will the
football strike the ground at B?
•12–101. A projectile is fired from the platform at B. The
shooter fires his gun from point A at an angle of 30°.
Determine the muzzle speed of the bullet if it hits the
projectile at C.
B
C
vA
A
20 ft
45%
A
vA
C
160 ft
Prob. 12–99
s
B
10 m
30%
1.8 m
20 m
Prob. 12–101
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12–102. A golf ball is struck with a velocity of 80 ft>s as
shown. Determine the distance d to where it will land.
MOTION OF A PROJECTILE
51
•12–105. The boy at A attempts to throw a ball over the
roof of a barn with an initial speed of vA = 15 m>s. 12
Determine the angle uA at which the ball must be thrown so
that it reaches its maximum height at C. Also, find the
distance d where the boy should stand to make the throw.
C
vA ! 80 ft/s
A
45%
8m
vA
B
uA
A
1m
10%
d
d
4m
Prob. 12–102
Prob. 12–105
12–103. The football is to be kicked over the goalpost,
which is 15 ft high. If its initial speed is vA = 80 ft>s,
determine if it makes it over the goalpost, and if so, by how
much, h.
12–106. The boy at A attempts to throw a ball over the roof
of a barn such that it is launched at an angle uA = 40°.
Determine the minimum speed vA at which he must throw
the ball so that it reaches its maximum height at C. Also,
find the distance d where the boy must stand so that he can
make the throw.
*12–104. The football is kicked over the goalpost with an
initial velocity of vA = 80 ft>s as shown. Determine the
point B (x, y) where it strikes the bleachers.
C
h
vA ! 80 ft/s
15 ft
60%
25 ft
45%
30 ft
Probs. 12–103/104
x
8m
vA
B
y
A
uA
1m
d
4m
Prob. 12–106
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12–107. The fireman wishes to direct the flow of water
from his hose to the fire at B. Determine two possible
angles u1 and u2 at which this can be done. Water flows from
the hose at vA = 80 ft>s.
•12–109. Determine the horizontal velocity vA of a tennis
ball at A so that it just clears the net at B. Also, find the
distance s where the ball strikes the ground.
A
u
vA
vA
20 ft
B
B
A
7.5 ft
3 ft
C
s
21 ft
35 ft
Prob. 12–107
Prob. 12–109
*12–108. Small packages traveling on the conveyor belt fall
off into a l-m-long loading car. If the conveyor is running at a
constant speed of vC = 2 m>s, determine the smallest and
largest distance R at which the end A of the car may be
placed from the conveyor so that the packages enter the car.
12–110. It is observed that the skier leaves the ramp A at an
angle uA = 25° with the horizontal. If he strikes the ground
at B, determine his initial speed vA and the time of flight tAB .
vA
uA
A
vc ! 2 m/s
4m
30%
3m
A
3
B
5
4
100 m
R
1m
B
Prob. 12–108
Prob. 12–110
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53
12.7 Curvilinear Motion: Normal and
12
Tangential Components
When the path along which a particle travels is known, then it is often
convenient to describe the motion using n and t coordinate axes which
act normal and tangent to the path, respectively, and at the instant
considered have their origin located at the particle.
O¿
n
O
s
un
Planar Motion. Consider the particle shown in Fig. 12–24a, which
moves in a plane along a fixed curve, such that at a given instant it is at
position s, measured from point O. We will now consider a coordinate
system that has its origin at a fixed point on the curve, and at the instant
considered this origin happens to coincide with the location of the
particle. The t axis is tangent to the curve at the point and is positive in
the direction of increasing s. We will designate this positive direction with
the unit vector ut . A unique choice for the normal axis can be made by
noting that geometrically the curve is constructed from a series of
differential arc segments ds, Fig. 12–24b. Each segment ds is formed from
the arc of an associated circle having a radius of curvature r (rho) and
center of curvature O¿. The normal axis n is perpendicular to the t axis with
its positive sense directed toward the center of curvature O¿, Fig. 12–24a.
This positive direction, which is always on the concave side of the curve,
will be designated by the unit vector un . The plane which contains the n
and t axes is referred to as the embracing or osculating plane, and in this
case it is fixed in the plane of motion.*
ut
(a)
O¿
O¿
r
r
r
ds
ds
r
#
v = s
(12–16)
O¿
ds
(b)
indicated in Sec. 12.4, the particle’s velocity v has a direction that is
always tangent to the path, Fig. 12–24c, and a magnitude that is
determined by taking the time derivative of the path function s = s1t2,
i.e., v = ds>dt (Eq. 12–8). Hence
(12–15)
r
r
Velocity. Since the particle moves, s is a function of time. As
v = vut
t
Position
O¿
r
r
where
v
Velocity
(c)
Fig. 12–24
*The osculating plane may also be defined as the plane which has the greatest contact
with the curve at a point. It is the limiting position of a plane contacting both the point and
the arc segment ds. As noted above, the osculating plane is always coincident with a plane
curve; however, each point on a three-dimensional curve has a unique osculating plane.
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Acceleration. The acceleration of the particle is the time rate of
O¿
du
12
10:30 AM
change of the velocity. Thus,
r
r
#
#
#
a = v = vut + vut
un
ds
u¿t
ut
(d)
(12–17)
#
In order to determine the time derivative ut , note that as the particle
moves along the arc ds in time dt, ut preserves its magnitude of unity;
however, its direction changes, and becomes utœ , Fig. 12–24d. As shown in
Fig. 12–24e, we require utœ = ut + dut . Here dut stretches between the
arrowheads of ut and utœ , which lie on an infinitesimal arc of radius ut = 1.
Hence, dut has a magnitude of dut = 112 du, and its direction is defined by
un . Consequently, dut = duun , and therefore the time derivative becomes
#
#
#
#
ut = uun . Since ds = rdu, Fig. 12–24d, then u = s>r, and therefore
#
#
s
v
#
ut = uun = un = un
r
r
un
Substituting into Eq. 12–17, a can be written as the sum of its two
components,
du u¿
t
dut
ut
a = atut + anun
(e)
(12–18)
where
#
at = v
atds = v dv
or
(12–19)
O¿
and
an
a
an =
P
at
v2
r
(12–20)
Acceleration
(f)
Fig. 12–24 (cont.)
These two mutually perpendicular components are shown in Fig. 12–24f.
Therefore, the magnitude of acceleration is the positive value of
a = 4a2t + a2n
(12–21)
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55
To better understand these results, consider the following two special
cases of motion.
12
1. If the particle moves along a straight line, then r : q and from
#
Eq. 12–20, an = 0. Thus a = at = v, and we can conclude that the
tangential component of acceleration represents the time rate of
change in the magnitude of the velocity.
2. If the particle moves along a curve with a constant speed, then
#
at = v = 0 and a = an = v2>r. Therefore, the normal component
of acceleration represents the time rate of change in the direction of
the velocity. Since a n always acts towards the center of curvature,
this component is sometimes referred to as the centripetal (or center
seeking) acceleration.
As a result of these interpretations, a particle moving along the curved
path in Fig. 12–25 will have accelerations directed as shown.
a ! at
Change in
direction of
velocity
at
Increasing
speed
an
an
a
a
at
Change in
magnitude of
velocity
Fig. 12–25
Three-Dimensional Motion. If the particle moves along a space
curve, Fig. 12–26, then at a given instant the t axis is uniquely specified;
however, an infinite number of straight lines can be constructed normal
to the tangent axis. As in the case of planar motion, we will choose the
positive n axis directed toward the path’s center of curvature O¿. This axis
is referred to as the principal normal to the curve. With the n and t axes so
defined, Eqs. 12–15 through 12–21 can be used to determine v and a. Since
ut and un are always perpendicular to one another and lie in the
osculating plane, for spatial motion a third unit vector, ub , defines the
binormal axis b which is perpendicular to ut and un , Fig. 12–26.
Since the three unit vectors are related to one another by the vector
cross product, e.g., ub = ut * un , Fig. 12–26, it may be possible to use this
relation to establish the direction of one of the axes, if the directions of
the other two are known. For example, if no motion occurs in the ub
direction, and this direction and ut are known, then un can be
determined, where in this case un = ub * ut , Fig. 12–26. Remember,
though, that un is always on the concave side of the curve.
b
osculating plane
O
n
s
O¿
ub
un
ut
Fig. 12–26
t
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Procedure for Analysis
12
Coordinate System.
• Provided the path of the particle is known, we can establish a set
of n and t coordinates having a fixed origin, which is coincident
with the particle at the instant considered.
• The positive tangent axis acts in the direction of motion and the
positive normal axis is directed toward the path’s center of
curvature.
Velocity.
• The particle’s velocity is always tangent to the path.
• The magnitude of velocity is found from the time derivative of
the path function.
#
v = s
Tangential Acceleration.
• The tangential component of acceleration is the result of the time
rate of change in the magnitude of velocity. This component acts
in the positive s direction if the particle’s speed is increasing or in
the opposite direction if the speed is decreasing.
• The relations between at , v, t and s are the same as for rectilinear
motion, namely,
#
at = v at ds = v dv
• If at is constant, at = 1at2c , the above equations, when integrated,
yield
s = s0 + v0t + 121at2ct2
v = v0 + 1at2ct
v2 = v20 + 21at2c1s - s02
Motorists traveling along this cloverleaf interchange experience a normal
acceleration due to the change in
direction of their velocity. A tangential
component of acceleration occurs when
the cars’ speed is increased or decreased.
Normal Acceleration.
• The normal component of acceleration is the result of the time
rate of change in the direction of the velocity. This component is
always directed toward the center of curvature of the path, i.e.,
along the positive n axis.
• The magnitude of this component is determined from
v2
an =
r
• If the path is expressed as y = f1x2, the radius of curvature r at
any point on the path is determined from the equation
r =
[1 + 1dy>dx22]3>2
ƒ d2y>dx2 ƒ
The derivation of this result is given in any standard calculus text.
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57
EXAMPLE 12.14
12
When the skier reaches point A along the parabolic path in Fig. 12–27a,
he has a speed of 6 m>s which is increasing at 2 m>s2. Determine the
direction of his velocity and the direction and magnitude of his
acceleration at this instant. Neglect the size of the skier in the calculation.
SOLUTION
Coordinate System. Although the path has been expressed in terms
of its x and y coordinates, we can still establish the origin of the n, t axes
at the fixed point A on the path and determine the components of v
and a along these axes, Fig. 12–27a.
Velocity. By definition, the velocity is always directed tangent to the
1 2
1
path. Since y = 20
x , dy>dx = 10
x, then at x = 10 m, dy>dx = 1.
Hence, at A, v makes an angle of u = tan-11 = 45° with the x axis,
Fig. 12–27a. Therefore,
vA = 6 m>s
Ans.
45°d
#
2
The acceleration is determined from a = vut + 1v >r2un . However,
it is first necessary to determine the radius of curvature of the path at
1
A (10 m, 5 m). Since d 2y>dx2 = 10
, then
2 3>2
r =
[1 + 1dy>dx2 ]
2
=
2
ƒ d y>dx ƒ
The acceleration becomes
C1 +
A
1
2 3>2
10 x
1
ƒ 10
ƒ
B D
`
x = 10 m
y ! 1 x2
20
y
n
u
= 28.28 m
vA
10 m
(a)
As shown in Fig. 12–27b,
2 2
n
a
2 m/s2
Thus, 45° + 90° + 57.5° - 180° = 12.5° so that,
a = 2.37 m>s
12.5°d
1.273 m/s2
90%
45%
f
2
a = 4(2 m>s 2 + (1.273 m>s 2 = 2.37 m>s
2
f = tan-1
= 57.5°
1.273
2
5m
x
v2
#
a A = vut +
u
r n
16 m>s22
= 2ut +
u
28.28 m n
= 52ut + 1.273un6m>s2
2 2
A
t
t
Ans.
NOTE: By using n, t coordinates, we were able to readily solve this
problem through the use of Eq. 12–18, since it accounts for the separate
changes in the magnitude and direction of v.
(b)
Fig. 12–27
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EXAMPLE 12.15
A race car C travels around the horizontal circular track that has a
radius of 300 ft, Fig. 12–28. If the car increases its speed at a constant
rate of 7 ft>s2, starting from rest, determine the time needed for it to
reach an acceleration of 8 ft>s2. What is its speed at this instant?
C
an
at
n
t
a
r ! 300 ft
Fig. 12–28
SOLUTION
Coordinate System. The origin of the n and t axes is coincident
with the car at the instant considered. The t axis is in the direction of
motion, and the positive n axis is directed toward the center of the
circle. This coordinate system is selected since the path is known.
Acceleration. The magnitude of acceleration can be related to its
components using a = 4a2t + a2n . Here a t = 7 ft>s2. Since an = v2>r,
the velocity as a function of time must be determined first.
v = v0 + 1at2ct
v = 0 + 7t
Thus
an =
17t22
v2
=
= 0.163t2 ft>s2
r
300
The time needed for the acceleration to reach 8 ft>s2 is therefore
a = 4a2t + a2n
8 ft>s2 = 4(7 ft>s2)2 + (0.163t2)2
Solving for the positive value of t yields
0.163t2 = 4(8 ft>s2)2 - (7 ft>s2)2
t = 4.87 s
Velocity. The speed at time t = 4.87 s is
v = 7t = 714.872 = 34.1 ft>s
Ans.
Ans.
Remember the velocity will always be tangent to the path,
whereas the acceleration will be directed within the curvature of the path.
NOTE:
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59
EXAMPLE 12.16
12
The boxes in Fig. 12–29a travel along the industrial conveyor. If a box
as in Fig. 12–29b starts from rest at A and increases its speed such that
at = 10.2t2 m>s2, where t is in seconds, determine the magnitude of its
acceleration when it arrives at point B.
SOLUTION
Coordinate System. The position of the box at any instant is
defined from the fixed point A using the position or path coordinate s,
Fig. 12–29b. The acceleration is to be determined at B, so the origin of
the n, t axes is at this point.
#
Acceleration. To determine the acceleration components at = v
#
and an = v2>r, it is first necessary to formulate v and v so that they
may be evaluated at B. Since vA = 0 when t = 0, then
#
(1)
a t = v = 0.2t
L0
v
(a)
A
s
t
0.2t dt
L0
v = 0.1t2
dv =
3m
(2)
The time needed for the box to reach point B can be determined by
realizing that the position of B is sB = 3 + 2p122>4 = 6.142 m,
Fig. 12–29b, and since sA = 0 when t = 0 we have
v =
L0
6.142 m
ds =
2m
n
ds
= 0.1t2
dt
L0
tB
t
B
2
(b)
0.1t dt
6.142 m = 0.0333t3B
tB = 5.690s
Substituting into Eqs. 1 and 2 yields
#
1aB2t = vB = 0.215.6902 = 1.138 m>s2
n
vB = 0.115.6922 = 3.238 m>s
At B, rB = 2 m, so that
1aB2n =
5.242 m/s2
13.238 m>s22
v2B
=
= 5.242 m>s2
rB
2m
t
B
The magnitude of aB , Fig. 12–29c, is therefore
aB = 4(1.138 m>s2)2 + (5.242 m>s2)2 = 5.36 m>s2
aB
1.138 m/s
(c)
Ans.
2
Fig. 12–29
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FUNDAMENTAL PROBLEMS
F12–27. The boat is traveling along the circular path with a
speed of v = (0.0625t2) m>s, where t is in seconds. Determine
the magnitude of its acceleration when t = 10 s.
t
F12–30. When x = 10 ft, the crate has a speed of 20 ft>s
which is increasing at 6 ft>s2. Determine the direction of the
crate’s velocity and the magnitude of the crate’s acceleration
at this instant.
v ! 0.0625t2
40 m
y
y ! 1 x2
24
n
O
20 ft/s
F12–27
F12–28. The car is traveling along the road with a speed of
v = (300>s) m>s, where s is in meters. Determine the
magnitude of its acceleration when t = 3 s if t = 0 at s = 0.
x
10 ft
F12–30
v ! ( 300
s )m/s
t
F12–31. If the motorcycle has a deceleration of
at = -(0.001s) m>s2 and its speed at position A is 25 m>s,
determine the magnitude of its acceleration when it passes
point B.
s
100 m n
A
300 m n
F12–28
F12–29. If the car decelerates uniformly along the curved
road from 25 m>s at A to 15 m>s at C, determine the
acceleration of the car at B.
A
90%
s
O
B
t
F12–31
F12–32. The car travels up the hill with a speed of
v = (0.2s) m>s, where s is in meters, measured from A.
Determine the magnitude of its acceleration when it is at
point s = 50 m, where r = 500 m.
250 m
y
rB ! 300 m
n
50 m
B
C
A
s ! 50 m
t
x
O
F12–29
F12–32
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61
PROBLEMS
12
12–111. When designing a highway curve it is required that
cars traveling at a constant speed of 25 m>s must not have
an acceleration that exceeds 3 m>s2. Determine the
minimum radius of curvature of the curve.
•12–117. Starting from rest the motorboat travels around
the circular path, r = 50 m, at a speed v = (0.8t) m>s,
where t is in seconds. Determine the magnitudes of the
boat’s velocity and acceleration when it has traveled 20 m.
*12–112. At a given instant, a car travels along a circular
curved road with a speed of 20 m>s while decreasing its speed
at the rate of 3 m>s2. If the magnitude of the car’s acceleration
12–118. Starting from rest, the motorboat travels around
the circular path, r = 50 m, at a speed v = (0.2t2) m>s,
where t is in seconds. Determine the magnitudes of the
boat’s velocity and acceleration at the instant t = 3 s.
•12–113. Determine the maximum constant speed a race
car can have if the acceleration of the car cannot exceed
7.5 m>s2 while rounding a track having a radius of curvature
of 200 m.
12–114. An automobile is traveling on a horizontal circular
curve having a radius of 800 ft. If the acceleration of the
automobile is 5 ft>s2, determine the constant speed at
which the automobile is traveling.
12–115. A car travels along a horizontal circular curved
road that has a radius of 600 m. If the speed is uniformly
increased at a rate of 2000 km>h2, determine the magnitude
of the acceleration at the instant the speed of the car is
60 km>h.
*12–116. The automobile has a speed of 80 ft>s at point A
and an acceleration a having a magnitude of 10 ft>s2, acting
in the direction shown. Determine the radius of curvature
of the path at point A and the tangential component of
acceleration.
t
A
r ! 50 m
v
Probs. 12–117/118
12–119. A car moves along a circular track of radius 250 ft,
and its speed for a short period of time 0 … t … 2 s is
v = 3(t + t2) ft>s, where t is in seconds. Determine the
magnitude of the car’s acceleration when t = 2 s. How far
has it traveled in t = 2 s?
*12–120. The car travels along the circular path such that its
speed is increased by at = (0.5et) m>s2, where t is in
seconds. Determine the magnitudes of its velocity and
acceleration after the car has traveled s = 18 m starting
from rest. Neglect the size of the car.
s ! 18 m
u ! 30%
a
n
Prob. 12–116
ρ ! 30 m
Prob. 12–120
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•12–121. The train passes point B with a speed of 20 m>s
12 which is decreasing at at = – 0.5 m>s2. Determine the
magnitude of acceleration of the train at this point.
12–122. The train passes point A with a speed of 30 m>s
and begins to decrease its speed at a constant rate of
at = – 0.25 m>s2. Determine the magnitude of the
acceleration of the train when it reaches point B, where
sAB = 412 m.
•12–125. When the car reaches point A it has a speed of
25 m>s. If the brakes are applied, its speed is reduced by
1>2
at = ( - 14 t ) m>s2. Determine the magnitude of acceleration
of the car just before it reaches point C.
12–126. When the car reaches point A, it has a speed of
25 m>s. If the brakes are applied, its speed is reduced by
at = (0.001s - 1) m>s2. Determine the magnitude of
acceleration of the car just before it reaches point C.
y
r ! 250 m
x
y ! 200 e 1000
C
B
B
A
A
200 m
30%
Probs. 12–125/126
x
400 m
12–127. Determine the magnitude of acceleration of the
airplane during the turn. It flies along the horizontal
circular path AB in 40 s, while maintaining a constant speed
of 300 ft>s.
Probs. 12–121/122
12–123. The car passes point A with a speed of 25 m>s after
which its speed is defined by v = (25 - 0.15s) m>s.
Determine the magnitude of the car’s acceleration when it
reaches point B, where s = 51.5 m.
*12–128. The airplane flies along the horizontal circular path
AB in 60 s. If its speed at point A is 400 ft>s, which decreases
at a rate of at = (–0.1t) ft>s2, determine the magnitude of the
plane’s acceleration when it reaches point B.
*12–124. If the car passes point A with a speed of 20 m>s
and begins to increase its speed at a constant rate of
at = 0.5 m>s2, determine the magnitude of the car’s
acceleration when s = 100 m.
A
B
y
y ! 16 "
16 m
1 2
x
625
B
60%
s
A x
Probs. 12–123/124
Probs. 12–127/128
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•12–129. When the roller coaster is at B, it has a speed of
25 m>s, which is increasing at at = 3 m>s2. Determine the
magnitude of the acceleration of the roller coaster at this
instant and the direction angle it makes with the x axis.
12–130. If the roller coaster starts from rest at A and its
speed increases at at = (6 – 0.06s) m>s2, determine the
magnitude of its acceleration when it reaches B where
sB = 40 m.
y
63
•12–133. A particle is traveling along a circular curve
having a radius of 20 m. If it has an initial speed of 20 m>s 12
and then begins to decrease its speed at the rate of
at = ( -0.25s) m>s2, determine the magnitude of the
acceleration of the particle two seconds later.
12–134. A racing car travels with a constant speed of
240 km>h around the elliptical race track. Determine the
acceleration experienced by the driver at A.
12–135. The racing car travels with a constant speed of
240 km>h around the elliptical race track. Determine the
acceleration experienced by the driver at B.
y ! 1 x2
100
A
s
y
B
x2 # ––
y2 ! 1
––
16
4
x
30 m
B
Probs. 12–129/130
A
x
2 km
12–131. The car is traveling at a constant speed of 30 m>s.
The driver then applies the brakes at A and thereby reduces
the car’s speed at the rate of at = (– 0.08v) m>s2, where v is
in m>s. Determine the acceleration of the car just before it
reaches point C on the circular curve. It takes 15 s for the
car to travel from A to C.
4 km
Probs. 12–134/135
*12–132. The car is traveling at a speed of 30 m>s. The
driver applies the brakes at A and thereby reduces the
speed at the rate of at = A - 18t B m>s2, where t is in seconds.
Determine the acceleration of the car just before it reaches
point C on the circular curve. It takes 15 s for the car to
travel from A to C.
45%
A
C
B
100 m
s
Probs. 12–131/132
*12–136. The position of a particle is defined by
r = 52 sin (p4 )t i + 2 cos (p4 )t j + 3 t k6 m, where t is in
seconds. Determine the magnitudes of the velocity and
acceleration at any instant.
•12–137. The position of a particle is defined by
r = 5t 3 i + 3t 2 j + 8t k6 m, where t is in seconds. Determine
the magnitude of the velocity and acceleration and the
radius of curvature of the path when t = 2 s.
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12–138. Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when the arm AB rotates u = 30°.
Neglect the size of the car.
12–139. Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when t = 2 s. Neglect the size of
the car.
12–142. Two cyclists, A and B, are traveling counterclockwise
around a circular track at a constant speed of 8 ft>s at
the instant shown. If the speed of A is increased at
(at)A = (sA) ft>s2, where sA is in feet, determine the
distance measured counterclockwise along the track from B
to A between the cyclists when t = 1 s. What is the
magnitude of the acceleration of each cyclist at this instant?
sA
A
B
sB
u ! 120%
B
r ! 50 ft
5m
A
Prob. 12–142
u
Probs. 12–138/139
*12–140. The truck travels at a speed of 4 m>s along a
circular road that has a radius of 50 m. For a short distance
from s = 0, its speed is then increased by at = (0.05s) m>s2 ,
where s is in meters. Determine its speed and the magnitude
of its acceleration when it has moved s = 10 m.
12–143. A toboggan is traveling down along a curve which
can be approximated by the parabola y = 0.01x2.
Determine the magnitude of its acceleration when it
reaches point A, where its speed is vA = 10 m>s, and it is
increasing at the rate of (at)A = 3 m>s2.
•12–141. The truck travels along a circular road that has a
radius of 50 m at a speed of 4 m>s. For a short distance when
t = 0, its speed is then increased by at = (0.4t) m>s2, where
t is in seconds. Determine the speed and the magnitude of
the truck’s acceleration when t = 4 s.
y
y ! 0.01x2
A
36 m
50 m
x
60 m
Probs. 12–140/141
Prob. 12–143
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CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS
*12–144. The jet plane is traveling with a speed of 120 m>s
which is decreasing at 40 m>s2 when it reaches point A.
Determine the magnitude of its acceleration when it is at
this point. Also, specify the direction of flight, measured
from the x axis.
y
65
12–146. The motorcyclist travels along the curve at a
constant speed of 30 ft>s. Determine his acceleration when 12
he is located at point A. Neglect the size of the motorcycle
and rider for the calculation.
y
x)
y ! 15 ln ( ––
80
500
v ! 30 ft/s
y ! –—
x
80 m
A
x
x
A
100 ft
Prob. 12–144
Prob. 12–146
•12–145. The jet plane is traveling with a constant speed of
110 m>s along the curved path. Determine the magnitude of
the acceleration of the plane at the instant it reaches point
A (y = 0).
12–147. The box of negligible size is sliding down along a
curved path defined by the parabola y = 0.4x2. When it is at
A (xA = 2 m, yA = 1.6 m), the speed is vB = 8 m>s and the
increase in speed is dvB>dt = 4 m>s2. Determine the
magnitude of the acceleration of the box at this instant.
y
y
x)
y ! 15 ln ( ––
80
A
80 m
A
x
y ! 0.4x2
x
2m
Prob. 12–145
Prob. 12–147
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*12–148. A spiral transition curve is used on railroads to
12 connect a straight portion of the track with a curved
portion. If the spiral is defined by the equation
y = (10 - 6)x3, where x and y are in feet, determine the
magnitude of the acceleration of a train engine moving with
a constant speed of 40 ft>s when it is at point x = 600 ft.
12–150. Particles A and B are traveling around a circular
track at a speed of 8 m>s at the instant shown. If the speed of
B is increasing by (at)B = 4 m>s2, and at the same instant A
has an increase in speed of (at)A = 0.8t m>s2, determine how
long it takes for a collision to occur. What is the magnitude of
the acceleration of each particle just before the collision
occurs?
y
A
y ! (10"6)x3
sA
u ! 120%
v ! 40 ft/s
sB
B
r!5m
x
600 ft
Prob. 12–148
Prob. 12–150
•12–149. Particles A and B are traveling counter-clockwise
around a circular track at a constant speed of 8 m>s. If at
the instant shown the speed of A begins to increase by
(at)A = (0.4sA) m>s2, where sA is in meters, determine the
distance measured counterclockwise along the track from B
to A when t = 1 s. What is the magnitude of the
acceleration of each particle at this instant?
12–151. The race car travels around the circular track with a
speed of 16 m>s. When it reaches point A it increases its
speed at at = (43 v1>4) m>s2, where v is in m>s . Determine the
magnitudes of the velocity and acceleration of the car when
it reaches point B. Also, how much time is required for it to
travel from A to B?
y
A
A
sA
u ! 120%
sB
B
200 m
B
x
r!5m
Prob. 12–149
Prob. 12–151
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CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
67
*12–152. A particle travels along the path y = a + bx + cx 2,
where a, b, c are constants. If the speed of the particle is
constant, v = v0, determine the x and y components
of velocity and the normal component of acceleration
when x = 0.
12–154. The motion of a particle is defined by the
equations x = (2t + t2) m and y = (t2) m, where t is in 12
seconds. Determine the normal and tangential components
of the particle’s velocity and acceleration when t = 2 s.
•12–153. The ball is kicked with an initial speed
vA = 8 m>s at an angle uA = 40° with the horizontal. Find
the equation of the path, y = f(x), and then determine the
normal and tangential components of its acceleration when
t = 0.25 s.
12–155. The motorcycle travels along the elliptical track at
a constant speed v. Determine the greatest magnitude of
the acceleration if a 7 b.
y
y
b
vA = 8 m/s
uA ! 40%
A
x 2 y2
#
!1
a2 b2
y
x
x
Prob. 12–153
12.8
x
a
Prob. 12–155
Curvilinear Motion: Cylindrical
Components
Sometimes the motion of the particle is constrained on a path that is best
described using cylindrical coordinates. If motion is restricted to the plane,
then polar coordinates are used.
Polar Coordinates. We can specify the location of the particle
shown in Fig. 12–30a using a radial coordinate r, which extends outward
from the fixed origin O to the particle, and a transverse coordinate u,
which is the counterclockwise angle between a fixed reference line and
the r axis. The angle is generally measured in degrees or radians, where
1 rad = 180°>p. The positive directions of the r and u coordinates are
defined by the unit vectors ur and uu , respectively. Here ur is in the
direction of increasing r when u is held fixed, and uu is in a direction of
increasing u when r is held fixed. Note that these directions are
perpendicular to one another.
u
uu
r
ur
r
u
O
Position
(a)
Fig. 12–30
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Position. At any instant the position of the particle, Fig. 12–30a, is
u
12
Page 68
defined by the position vector
uu
r
r = rur
(12–22)
ur
Velocity. The instantaneous velocity v is obtained by taking the time
derivative of r. Using a dot to represent the time derivative, we have
r
u
O
#
#
#
v = r = rur + rur
Position
#
To evaluate ur , notice that ur only changes its direction with respect to
time, since by definition the magnitude of this vector is always one unit.
Hence, during the time ¢t, a change ¢r will not cause a change in the
direction of ur ; however, a change ¢u will cause ur to become urœ , where
urœ = ur + ¢ur , Fig. 12–30b. The time change in ur is then ¢ur . For small
angles ¢u this vector has a magnitude ¢ur L 11¢u2 and acts in the uu
direction. Therefore, ¢ur = ¢uuu , and so
(a)
uu
u¿r
\$ur
ur
\$u
¢ur
¢u
#
ur = lim
= a lim
b uu
¢t : 0 ¢t
¢t : 0 ¢t
(b)
#
#
ur = uuu
(12–23)
Substituting into the above equation, the velocity can be written in
component form as
v = vrur + vuuu
(12–24)
#
vr = r
#
vu = ru
(12–25)
where
v
vu
vr
r
u
O
Velocity
(c)
Fig. 12–30 (cont.)
These components are shown graphically in Fig. 12–30c. The radial
component vr is a measure of the rate of increase or decrease in the
#
length of the radial coordinate, i.e., r; whereas the transverse component
vu can be interpreted as the rate of motion along # the circumference of a
circle having a radius r. In particular, the term u = du>dt is called the
angular velocity, since it indicates the time rate of change of the angle u.
Common units used for this measurement are rad>s.
Since vr and vu are mutually perpendicular, the magnitude of velocity
or speed is simply the positive value of
#
#
v = 4(r22 + (ru22
(12–26)
and the direction of v is, of course, tangent to the path, Fig. 12–30c.
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CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
69
Acceleration. Taking the time derivatives of Eq. 12–24, using
12
Eqs. 12–25, we obtain the particle’s instantaneous acceleration,
\$
# #
#
\$
##
##
a = v = rur + rur + ruuu + ruuu + ruuu
#
To evaluate uu , it is necessary only to find the change in the direction of uu
since its magnitude is always unity. During the time ¢t, a change ¢r will not
change the direction of uu , however, a change ¢u will cause uu to become
uuœ , where uuœ = uu + ¢uu , Fig. 12–30d. The time change in uu is thus ¢uu .
For small angles this vector has a magnitude ¢uu L 11¢u2 and acts in the
-ur , direction; i.e., ¢uu = - ¢uur . Thus,
\$uu
uu
u¿u
ur
\$u
(d)
¢uu
¢u
#
uu = lim
= - a lim
bur
¢t : 0 ¢t
¢t : 0 ¢t
#
#
uu = -uur
(12–27)
Substituting this result and Eq. 12–23 into the above equation for a, we
can write the acceleration in component form as
a = arur + auuu
(12–28)
#
\$
ar = r - ru2
# #
\$
au = r u + 2 ru
(12–29)
where
\$
The term u = d2u>dt2 = d>dt1du>dt2 is called the angular acceleration
since it measures the change made in the angular velocity during an
instant of time. Units for this measurement are rad>s2.
Since a r , and au are always perpendicular, the magnitude of
acceleration is simply the positive value of
\$
#
# #
\$
a = 4(r - ru 2)2 + (r u + 2r u)2
a
au
ar
r
u
(12–30)
O
The direction is determined from the vector addition of its two
components. In general, a will not be tangent to the path, Fig. 12–30e.
Acceleration
(e)
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uz
12
uu
ur
rP
z
O
u
r
Page 70
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PA R T I C L E
Cylindrical Coordinates. If the particle moves along a space
curve as shown in Fig. 12–31, then its location may be specified by the
three cylindrical coordinates, r, u, z. The z coordinate is identical to that
used for rectangular coordinates. Since the unit vector defining its
direction, uz , is constant, the time derivatives of this vector are zero, and
therefore the position, velocity, and acceleration of the particle can be
written in terms of its cylindrical coordinates as follows:
rP = rur + zuz
#
#
#
v = rur + ruuu + zuz
\$
#
##
\$
\$
a = 1r - ru22ur + 1ru + 2ru2uu + zuz
(12–31)
(12–32)
Time Derivatives.
The
\$ above equations require that we obtain the
# \$ #
time derivatives r, r, u, and u in order to evaluate the r and u components
of v and a. Two types of problems generally occur:
Fig. 12–31
1. If the polar coordinates are specified as time parametric equations,
r = r1t2 and u = u1t2, then the time derivatives can be found directly.
2. If the time-parametric equations are not given, then the path
r = f1u2 must be known. Using the chain rule of calculus we can
#
\$
#
\$
then find the relation between r and u, and between r and u.
Application of the chain rule, along with some examples, is
explained in Appendix C.
Procedure for Analysis
z
u r
The spiral motion of this boy can be
followed by using cylindrical components.
Here the radial coordinate r is constant,
the transverse coordinate u will increase
with time as the boy rotates about the
vertical, and his altitude z will decrease
with time.
Coordinate System.
• Polar coordinates are a suitable choice for solving problems when
data regarding the angular motion of the radial coordinate r is
given to describe the particle’s motion. Also, some paths of motion
can conveniently be described in terms of these coordinates.
• To use polar coordinates, the origin is established at a fixed point,
and the radial line r is directed to the particle.
• The transverse coordinate u is measured from a fixed reference
Velocity and Acceleration.
\$
# \$ #
• Once r and the four time derivatives r, r, u, and u have been
evaluated at the instant considered, their values can be
substituted into Eqs. 12–25 and 12–29 to obtain the radial and
transverse components of v and a.
• If it is necessary to take the time derivatives of r = f1u2, then the
chain rule of calculus must be used. See Appendix C.
a simple extension of the
• Motion in three dimensions# requires
\$
above procedure to include z and z.
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CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
EXAMPLE 12.17
12
The amusement park ride shown in Fig. 12–32a consists of a chair that
is rotating in a horizontal circular
path of radius r such \$that the arm
#
OB has an angular velocity u and angular acceleration u. Determine
the radial and transverse components of velocity and acceleration of
the passenger. Neglect his size in the calculation.
n
O
· ··
u, u
r
·
v ! ru
·
"ar ! ru2
u
B
··
au ! ru
r
r
u,t
(b)
(a)
Fig. 12–32
SOLUTION
Coordinate System. Since the angular motion of the arm is
reported, polar coordinates are chosen for the solution, Fig. 12–32a.
Here u is not related to r, since the radius is constant for all u.
Velocity and Acceleration. It is first necessary to specify the first
and second time derivatives of r and u. Since r is constant, we have
#
\$
r = r
r = 0
r = 0
Thus,
71
#
vr = r = 0
#
vu = ru
#
#
\$
ar = r - ru2 = - ru2
\$
\$
##
au = ru + 2ru = ru
Ans.
Ans.
Ans.
Ans.
These results are shown in Fig. 12–32b.
The n, t axes are also shown in Fig. 12–32b, which in this
special case of circular motion happen to# be collinear with the r and u
axes, respectively. Since v = vu = vt = ru, then by comparison,
#
#
1ru22
v2
=
= ru2
-ar = an =
r
r
#
\$
du
dv
d #
dr #
au = at =
=
1ru2 =
u + r
= 0 + ru
dt
dt
dt
dt
NOTE:
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EXAMPLE 12.18
The rod OA in Fig. 12–33a rotates in the horizontal plane such that
u = 1t32 rad. At the same time, the collar B is sliding outward along
OA so that r = 1100t22 mm. If in both cases t is in seconds, determine
the velocity and acceleration of the collar when t = 1 s.
SOLUTION
Coordinate System. Since time-parametric equations of the path
are given, it is not necessary to relate r to u.
O
r
Velocity and Acceleration. Determining the time derivatives and
evaluating them when t = 1 s, we have
B
u
A
r = 100t2 `
(a)
u ! 57.3%
#
r = 200t `
vu ! 300 mm/s
\$
r = 200 `
u
t=1 s
t=1 s
t=1 s
= 100 mm u = t3 `
#
= 200 mm>s u = 3t2 `
= 200 mm>s2
As shown in Fig. 12–33b,
#
#
v = rur + ruuu
v
d
vr ! 200 mm/s
t=1 s
\$
u = 6t `
t=1 s
t=1 s
= 200ur + 100132uu = 5200ur + 300uu6 mm>s
r
The magnitude of v is
(b)
v = 4120022 + 130022 = 361 mm>s
d = tan-1 a
u ! 57.3%
a
au ! 1800 mm/s2
ar ! 700 mm/s2
r
(c)
Fig. 12–33
d + 57.3° = 114°
Ans.
As shown in Fig. 12–33c,
#
\$
\$
##
a = 1r - ru22ur + 1ru + 2ru2uu
u
f
300
b = 56.3°
200
Ans.
= [200 - 1001322]ur + [100162 + 2120023]uu
= 5- 700ur + 1800uu6 mm>s2
The magnitude of a is
f = tan-1 a
a = 2(700)2 + (1800)2 = 1930 mm>s2
1800
b = 68.7°
700
1180° - f2 + 57.3° = 169°
Ans.
Ans.
NOTE: The velocity is tangent to the path; however, the acceleration
is directed within the curvature of the path, as expected.
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73
EXAMPLE 12.19
12
The searchlight in Fig. 12–34a casts a spot of light along the face of a wall
that is located 100 m from the searchlight. Determine the magnitudes of
the velocity and acceleration at which the spot appears to travel across
the# wall at the instant u = 45°. The searchlight rotates at a constant rate
Velocity and Acceleration. Using the chain rule of calculus, noting
that d1sec u2 = sec u tan u du, and d1tan u2 = sec2 u du, we have
#
#
r = 1001sec u tan u2u
#
#
# #
\$
r = 1001sec u tan u2u1tan u2u + 100 sec u1sec2 u2u1u2
\$
+ 100 sec u tan u1u2
#
#
\$
= 100 sec u tan2 u 1u22 + 100 sec3u 1u22 + 1001sec u tan u2u
#
\$
Since u = 4 rad>s = constant, then u = 0, and the above equations,
when u = 45°, become
r = 100 sec 45° = 141.4
#
r = 400 sec 45° tan 45° = 565.7
\$
r = 1600 1sec 45° tan2 45° + sec3 45°2 = 6788.2
As shown in Fig. 12–34b,
#
#
v = rur + ruuu
= 565.7ur + 141.4142uu
= 5565.7ur + 565.7uu6 m>s
2
4vr
v =
+
= 800 m>s
v2u
2
·
100 m
(a)
r
vr
v
u
r
u
100 m
u
100 m
vu
u
(b)
r
ar
a
u
r
2
= 4(565.7) + (565.7)
As shown in Fig. 12–34c,
#
\$
\$
##
a = 1r - ru22ur + 1ru + 2ru2uu
= [6788.2 - 141.41422]ur + [141.4102 + 21565.724]uu
= 54525.5ur + 4525.5uu6 m>s2
a = 4a2r + a2u = 4(4525.5)2 + (4525.5)2
= 6400 m>s2
u
r
SOLUTION
Coordinate System. Polar coordinates will be used to solve this
problem since the angular rate of the searchlight is given. To find the
necessary time derivatives it is first necessary to relate r to u. From
Fig. 12–34a,
r = 100>cos u = 100 sec u
Ans.
au
u
(c)
a
u ! 45%
ar
Ans.
\$
NOTE: It is also possible to find a without having to calculate r (or a r ).
As shown in Fig. 12–34d, since au = 4525.5 m>s2, then by vector
resolution, a = 4525.5>cos 45° = 6400 m>s2.
au ! 4525.5 m/s2
(d)
Fig. 12–34
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EXAMPLE 12.20
r ! 0.5 (1 " cos u) ft
u
r
· ··
u, u
Due to the rotation of the forked rod, the ball in Fig. 12–35a travels
around the slotted path, a portion of which is in the shape of a
cardioid, r = 0.511 - cos u2 ft, where u is in radians. If the ball’s
velocity is v = 4 ft>s and its acceleration is a = 30 ft>s2 at the instant
#
\$
u = 180°, determine the angular velocity u and angular acceleration u
of the fork.
SOLUTION
Coordinate System. This path is most unusual, and mathematically
it is best expressed using polar coordinates,
#
\$ as done here, rather than
rectangular coordinates. Also, since u and u must be determined, then
r, u coordinates are an obvious choice.
(a)
Velocity and Acceleration. The time derivatives of r and u can be
determined using the chain rule.
r = 0.511 - cos u2
#
#
r = 0.51sin u2u
# #
\$
\$
r = 0.51cos u2u1u2 + 0.51sin u2u
Evaluating these results at u = 180°, we have
#
#
\$
r = 1 ft
r = 0
r = - 0.5u2
#
Since v = 4 ft>s, using Eq. 12–26 to determine u yields
#
#
v = 4(r)2 + (ru)2
#
4 = 4(0)2 + (1u)2
#
Ans.
\$
In a similar manner, u can be found using Eq. 12–30.
r
v ! 4 ft/s
a ! 30 ft/s2
#
\$
\$
##
a = 41 r - ru222 + 1 ru + 2ru22
\$
30 = 4[ -0.5(4)2 - 1(4)2]2 + [1u + 2(0)(4)]2
\$
(30)2 = 1- 24)2 + u 2
\$
Ans.
u
(b)
Fig. 12–35
Vectors a and v are shown in Fig. 12–35b.
NOTE: At this location, the u and t (tangential) axes will coincide. The
+n (normal) axis is directed to the right, opposite to + r.
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CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
75
FUNDAMENTAL PROBLEMS
F12–33. The car# has a speed of 55 ft>s. Determine the
angular velocity u of the radial line OA at this instant.
12
F12–36. Peg P is driven by the forked link OA along the
path described by r = eu. When u = p4 rad, the link
has an
#
angular
velocity
and
angular
acceleration
of
u
=
2
\$
components of the peg’s acceleration at this instant.
A
r ! eu
A
r ! 400 ft
u
r
O
F12–33
F12–34. The platform is rotating about the vertical axis
such that at any instant its angular position is
u = (4t3/2) rad, where t is in seconds. A ball rolls outward
along the radial groove so that its position is r = (0.1t3) m,
where t is in seconds. Determine the magnitudes of the
velocity and acceleration of the ball when t = 1.5 s.
u, u
P
u
u, u
O
F12–36
F12–37. The collars are pin-connected at B and are free
to move along rod OA and the curved guide OC having
the shape of a cardioid, r = [0.2(1
# + cos u)] m. At u = 30°,
the angular velocity of OA is u = 3 rad>s. Determine the
magnitudes of the velocity of the collars at this point.
A
r ! 0.2(l + cos u) m
u
B
r
r
F12–34
F12–35. Peg P is driven by the fork link OA along the
curved path described by r = (2u) ft. At the instant
\$ and angular acceleration
# angular velocity
magnitude of the peg’s acceleration at this instant.
O
u
C
F12–37
F12–38. At the instant u = 45°, the athlete is running with
a constant speed of 2 m>s. Determine the angular velocity
at which the camera must turn in order to follow the
motion.
r ! (30 csc u) m
A
v
P
u, u
r
r
30 m
u
O
u
F12–35
A
u
F12–38
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PA R T I C L E
PROBLEMS
*12–156. A particle moves along
# a circular path of radius
300 mm. If its angular velocity is u = (2t2) rad>s , where t is in
seconds, determine the magnitude of the particle’s
acceleration when t = 2 s.
•12–157. A particle moves along
# a circular path of radius
300 mm. If its angular velocity is u = (3t2) rad>s , where t is in
seconds, determine the magnitudes of the particle’s velocity
and acceleration when u = 45°. The particle starts from rest
when u = 0°.
*12–164. A particle travels around a lituus, defined by the
equation r2u = a2, where a is a constant. Determine the
particle’s radial and transverse components of velocity and
acceleration as a function of u and its time derivatives.
•12–165. A car travels along the circular curve of radius
r = # 300 ft. At the instant shown, its angular rate of rotation
is
\$ u = 0.4 rad>s, which is increasing at the rate of
u = 0.2 rad>s2. Determine the magnitudes of the car’s
velocity and acceleration at this instant.
12–158. A particle moves along a circular path of radius
5 ft. If its position is u = (e0.5t) rad, where t is in seconds,
determine the magnitude of the particle’s acceleration
when u = 90°.
A
r ! 300 ft
.
12–159. The position of a particle is described by
r = (t3 + 4t - 4) m and u = (t3>2) rad, where t is in seconds.
Determine the magnitudes of the particle’s velocity and
acceleration at the instant t = 2 s.
*12–160. The position of a particle is described by
r = (300e–0.5t) mm and u = (0.3t2) rad, where t is in seconds.
Determine the magnitudes of the particle’s velocity and
acceleration at the instant t = 1.5 s.
•12–161. An airplane is flying in a straight line with a
velocity of 200 mi>h and an acceleration of 3 mi>h2. If the
propeller has a diameter of 6 ft and is rotating at an angular
rate of 120 rad>s, determine the magnitudes of velocity and
acceleration of a particle located on the tip of the propeller.
..
u
Prob. 12–165
12–166. The slotted arm OA rotates counterclockwise
#
about O with a constant angular velocity of u . The motion of
pin B is constrained such that it moves on the fixed circular
surface and along the slot in OA. Determine the magnitudes
of the velocity and acceleration of pin B as a function of u.
12–167. The slotted arm OA rotates counterclockwise
about O such that when# u = p>4, arm OA is rotating with
\$
an angular velocity of u and an angular acceleration of u.
Determine the magnitudes of the velocity and acceleration
of pin B at this instant. The motion of pin B is constrained
such that it moves on the fixed circular surface and along
the slot in OA.
12–162. A particle moves along a circular path having a
radius of 4 in. such that its position as a function of time is
given by u = (cos 2t) rad, where t is in seconds. Determine
the magnitude of the acceleration of the particle when
u = 30°.
B
r ! 2 a cos u
u
O
12–163. A particle travels around a limaçon, defined by the
equation r = b - a cos u, where a and b are constants.
Determine the particle’s radial and transverse components
of velocity and acceleration as a function of u and its time
derivatives.
a
Probs. 12–166/167
A
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12.8
*12–168. The car travels along the circular curve having a
ft. At the instant shown, its angular rate of
#
rotation
is
u
=
0.025
rad>s, which is decreasing at the rate
\$
components of the car’s velocity and acceleration at this
instant and sketch these components on the curve.
CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
77
12–171. The small washer slides down the cord OA. When
it is at the midpoint, its speed is 200 mm>s and its 12
acceleration is 10 mm>s2. Express the velocity and
acceleration of the washer at this point in terms of its
cylindrical components.
•12–169. The car travels along the circular curve of radius
r = 400 ft with a constant speed
# of v = 30 ft>s. Determine
the angular rate of rotation u of the radial line r and the
magnitude of the car’s acceleration.
z
A
v, a
z
700 mm
O
r
u
r ! 400 ft
x
300 mm
400 mm
.
u
Prob. 12–171
*12–172. If arm OA rotates# counterclockwise with a
constant angular velocity of u = 2 rad>s, determine the
magnitudes of the velocity and acceleration of peg P at
u = 30°. The peg moves in the fixed groove defined by the
lemniscate, and along the slot in the arm.
Probs. 12–168/169
12–170. Starting from rest, the boy runs outward in the
radial direction from the center of the platform with a
2
constant acceleration
# of 0.5 m>s . If the platform is rotating
at a constant rate u = 0.2 rad>s, determine the radial and
transverse components of the velocity and acceleration of
the boy when t = 3 s. Neglect his size.
•12–173. The peg moves in the curved slot defined by the
lemniscate, and through
the slot in the arm. At u = 30°, the
#
u
=
2
angular
velocity
is
\$
is u = 1.5 rad>s2. Determine the magnitudes of the velocity
and acceleration of peg P at this instant.
r2 ! (4 sin 2 u)m2
r
y
P
0.5 m/s2
r
u
Prob. 12–170
O
u
Probs. 12–172/173
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12–174. The airplane on the amusement park ride moves
along a path defined by the equations r = 4 m,
u = (0.2t) rad, and z = (0.5 cos u) m, where t is in seconds.
Determine the cylindrical components of the velocity and
acceleration of the airplane when t = 6 s.
z
u
•12–177. The driver of the car maintains a constant speed
of 40 m>s. Determine the angular velocity of the camera
tracking the car when u = 15°.
12–178. When u = 15°, the car has a speed of 50 m>s which
is increasing at 6 m>s2. Determine the angular velocity of
the camera tracking the car at this instant.
r ! (100 cos 2u) m
r!4m
u
Prob. 12–174
12–175. The motion of peg P is constrained by the
lemniscate curved slot in OB and by the slotted arm OA. If
OA rotates # counterclockwise with a constant angular
velocity of u = 3 rad>s, determine the magnitudes of the
velocity and acceleration of peg P at u = 30°.
*12–176. The motion of peg P is constrained by the
lemniscate curved slot in OB and by the slotted arm OA.
If
# OA rotates counterclockwise with an angular velocity of
u = (3t3/2) rad>s, where t is in seconds, determine the
magnitudes of the velocity and acceleration of peg P at
u = 30°. When t = 0, u = 0°.
A
P
r
O
u
r2 ! (4 cos 2 u)m2
Probs. 12–175/176
B
Probs. 12–177/178
12–179. If the cam# rotates clockwise with a constant
angular velocity of u = 5 rad>s, determine the magnitudes
of the velocity and acceleration of the follower rod AB at
the instant u = 30°. The surface of the cam has a shape of
limaçon defined by r = (200 + 100 cos u) mm.
*12–180. At the instant u = #30°, the cam rotates with a
clockwise angular
\$ velocity of u = 5 rad>s and and angular
acceleration of u = 6 rad>s2. Determine the magnitudes of
the velocity and acceleration of the follower rod AB at this
instant. The surface of the cam has a shape of a limaçon
defined by r = (200 + 100 cos u) mm.
r ! (200 # 100 cos u) mm
u
A
Probs. 12–179/180
B
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12.8
•12–181. The automobile travels from a parking deck
down along a cylindrical spiral ramp at a constant speed of
v = 1.5 m>s. If the ramp descends a distance of 12 m for
every full revolution, u = 2p rad, determine the magnitude
of the car’s acceleration as it moves along the ramp,
r = 10 m. Hint: For part of the solution, note that the
tangent to the ramp at any point is at an angle of
f = tan - 1 (12>32p(10)4) = 10.81° from the horizontal.
Use this to determine the velocity components
vu and vz,
#
#
which in turn are used to determine u and z.
CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
79
*12–184. Rod OA rotates
counterclockwise with a constant
#
angular velocity of u = 6 rad>s. Through mechanical means 12
#
collar B moves along the rod with a speed of r = (4t2) m>s ,
where t is in seconds. If r = 0 when t = 0, determine the
magnitudes of velocity and acceleration of the collar when
t = 0.75 s.
•12–185. Rod
OA is rotating counterclockwise with an angular
#
velocity of u = (2t2) rad>s.Through mechanical means collar B
#
moves along the rod with a speed of r = (4t2) m>s. If u = 0 and
r = 0 when t = 0, determine the magnitudes of velocity and
acceleration of the collar at u = 60°.
A
10 m
B
r
12 m
u
O
Probs. 12–184/185
Prob. 12–181
12–182. The box slides down the helical ramp with a
constant speed of v = 2 m>s. Determine the magnitude of
its acceleration. The ramp descends a vertical distance of
1 m for every full revolution. The mean radius of the ramp is
r = 0.5 m.
12–183. The box slides down the helical ramp which is
defined by r = 0.5 m, u = (0.5t3) rad, and z = (2 – 0.2t2) m,
where t is in seconds. Determine the magnitudes of the
velocity and acceleration of the box at the instant
0.5 m
12–186. The slotted arm AB drives pin C through the spiral
groove described by the
equation r = a u. If the angular
#
velocity is constant at u, determine the radial and transverse
components of velocity and acceleration of the pin.
12–187. The slotted arm AB drives pin C through the spiral
groove described by the equation r = (1.5 u) ft , where u is in
radians. If the arm starts from # rest when u = 60° and is
driven at an angular velocity of u = (4t) rad>s , where t is in
seconds, determine the radial and transverse components of
velocity and acceleration of the pin C when t = 1 s.
B
C
r
1m
u
A
Probs. 12–182/183
Probs. 12–186/187
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*12–188. The partial surface of the cam is that of a
12 logarithmic spiral r = (40e0.05u) mm, where u is in
the cam rotates at a constant angular velocity of u = 4 rad>s ,
determine the magnitudes of the velocity and acceleration of
the point on the cam that contacts the follower rod at the
instant u = 30°.
*12–192. The boat moves along a path defined by
r2 = 310(103) cos 2u4 ft2, where u is in radians. If
u = (0.4t2) rad, where t is in seconds, determine the radial
and transverse components of the boat’s velocity and
acceleration at the instant t = 1 s.
•12–189. Solve \$Prob. 12–188, if the cam has an angular
acceleration
of u = 2 rad>s2 when its angular velocity is
#
u = 4 rad>s at u = 30°.
r
u
r ! 40e0.05u
u
Prob. 12–192
·
Probs. 12–188/189
12–190. A particle moves along an Archimedean
spiral
#
r = (8u) ft, where u is given in radians. If u = 4 rad>s
(constant), determine the radial and transverse components
of the particle’s velocity and acceleration at the instant
u = p>2 rad. Sketch the curve and show the components on
the curve.
•12–193. A car travels along a road, which for a short
distance is defined by r = (200>u) ft, where u is in radians. If
it maintains a constant speed of v = 35 ft>s, determine the
radial and transverse components of its velocity when
12–194. For a short time the jet plane moves along a path
in the shape of a lemniscate, r2 = (2500 cos 2u) km2. At the
instant u = 30°, the radar tracking device is rotating at
\$
#
u = 5(10 - 3) rad>s with u = 2(10 - 3) rad>s2. Determine the
radial and transverse components of velocity and
acceleration of the plane at this instant.
12–191. Solve
particle has an angular
\$ Prob. 12–190 if the
#
r2 ! 2500 cos 2 u
y
r
r ! (8 u) ft
u
r
u
x
Probs. 12–190/191
Prob. 12–194
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12.9
12.9
ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES
81
Absolute Dependent Motion
Analysis of Two Particles
12
In some types of problems the motion of one particle will depend on the
corresponding motion of another particle. This dependency commonly
occurs if the particles, here represented by blocks, are interconnected by
inextensible cords which are wrapped around pulleys. For example, the
movement of block A downward along the inclined plane in Fig. 12–36
will cause a corresponding movement of block B up the other incline. We
can show this mathematically by first specifying the location of the blocks
using position coordinates sA and sB . Note that each of the coordinate axes
is (1) measured from a fixed point (O) or fixed datum line, (2) measured
along each inclined plane in the direction of motion of each block, and
(3) has a positive sense from C to A and D to B. If the total cord length is
lT , the two position coordinates are related by the equation
Datum
sA
C D
Datum
sB
O
A
B
Fig. 12–36
sA + lCD + sB = lT
Here lCD is the length of the cord passing over arc CD. Taking the time
derivative of this expression, realizing that lCD and lT remain constant,
while sA and sB measure the segments of the cord that change in length.
We have
dsA
dsB
+
= 0
or
vB = - vA
dt
dt
The negative sign indicates that when block A has a velocity downward,
i.e., in the direction of positive sA , it causes a corresponding upward
velocity of block B; i.e., B moves in the negative sB direction.
In a similar manner, time differentiation of the velocities yields the
relation between the accelerations, i.e.,
a B = -aA
A more complicated example is shown in Fig. 12–37a. In this case, the
position of block A is specified by sA , and the position of the end of the
cord from which block B is suspended is defined by sB . As above, we
have chosen position coordinates which (1) have their origin at fixed
points or datums, (2) are measured in the direction of motion of each
block, and (3) are positive to the right for sA and positive downward for
sB. During the motion, the length of the red colored segments of the cord
in Fig. 12–37a remains constant. If l represents the total length of cord
minus these segments, then the position coordinates can be related by
the equation
Datum
sB
B
h
A
2sB + h + sA = l
Since l and h are constant during the motion, the two time derivatives yield
2vB = -vA
2aB = - aA
Hence, when B moves downward 1+sB2, A moves to the left 1-sA2 with
twice the motion.
Datum
sA
(a)
Fig. 12–37
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This example can also be worked by defining the position of block B
from the center of the bottom pulley (a fixed point), Fig. 12–37b. In
this case
12
Datum
21h - sB2 + h + sA = l
Time differentiation yields
2vB = vA
sB
B
Here the signs are the same. Why?
h
Datum
Procedure for Analysis
A
Datum
2aB = aA
sA
(b)
Fig. 12–37 (cont.)
The above method of relating the dependent motion of one particle
to that of another can be performed using algebraic scalars or
position coordinates provided each particle moves along a
rectilinear path. When this is the case, only the magnitudes of the
velocity and acceleration of the particles will change, not their line
of direction.
Position-Coordinate Equation.
• Establish each position coordinate with an origin located at a
fixed point or datum.
• It is not necessary that the origin be the same for each of the
coordinates; however, it is important that each coordinate axis
selected be directed along the path of motion of the particle.
• Using geometry or trigonometry, relate the position coordinates
to the total length of the cord, lT , or to that portion of cord, l,
which excludes the segments that do not change length as the
particles move—such as arc segments wrapped over pulleys.
• If a problem involves a system of two or more cords wrapped
around pulleys, then the position of a point on one cord must be
related to the position of a point on another cord using the above
procedure. Separate equations are written for a fixed length of
each cord of the system and the positions of the two particles are
then related by these equations (see Examples 12.22 and 12.23).
Time Derivatives.
The motion of the traveling block on this
oil rig depends upon the motion of the
cable connected to the winch which
operates it. It is important to be able to
relate these motions in order to determine
the power requirements of the winch and
the force in the cable caused by any
accelerated motion.
• Two successive time derivatives of the position-coordinate
•
equations yield the required velocity and acceleration equations
which relate the motions of the particles.
The signs of the terms in these equations will be consistent with
those that specify the positive and negative sense of the position
coordinates.
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12.9
ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES
EXAMPLE 12.21
83
12
Determine the speed of block A in Fig. 12–38 if block B has an
upward speed of 6 ft>s.
D Datum
C
sB
sA
E
B
6 ft/s
A
Fig. 12–38
SOLUTION
Position-Coordinate Equation. There is one cord in this system
having segments which change length. Position coordinates sA and sB
will be used since each is measured from a fixed point (C or D) and
extends along each block’s path of motion. In particular, sB is directed
to point E since motion of B and E is the same.
The red colored segments of the cord in Fig. 12–38 remain at a
constant length and do not have to be considered as the blocks move.
The remaining length of cord, l, is also constant and is related to the
changing position coordinates sA and sB by the equation
sA + 3sB = l
Time Derivative. Taking the time derivative yields
vA + 3vB = 0
so that when vB = -6 ft>s (upward),
vA = 18 ft>sT
Ans.
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EXAMPLE 12.22
Determine the speed of A in Fig. 12–39 if B has an upward speed
of 6 ft>s.
Datum
sA
sC
A
sB
C
D
6 ft/s
B
Fig. 12–39
SOLUTION
Position-Coordinate Equation. As shown, the positions of blocks
A and B are defined using coordinates sA and sB . Since the system has
two cords with segments that change length, it will be necessary to use
a third coordinate, sC , in order to relate sA to sB . In other words, the
length of one of the cords can be expressed in terms of sA and sC , and
the length of the other cord can be expressed in terms of sB and sC .
The red colored segments of the cords in Fig. 12–39 do not have to
be considered in the analysis. Why? For the remaining cord lengths,
say l1 and l2 , we have
sA + 2sC = l1
sB + 1sB - sC2 = l2
Time Derivative. Taking the time derivative of these equations yields
vA + 2vC = 0
2vB - vC = 0
Eliminating vC produces the relationship between the motions of each
cylinder.
vA + 4vB = 0
so that when vB = - 6 ft>s (upward),
vA = + 24 ft>s = 24 ft>s T
Ans.
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12.9
ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES
EXAMPLE 12.23
85
12
Determine the speed of block B in Fig. 12–40 if the end of the cord at
A is pulled down with a speed of 2 m>s.
D
Datum
sC
C
sB
sA
A
E
B
2 m/s
Fig. 12–40
SOLUTION
Position-Coordinate Equation. The position of point A is defined
by sA , and the position of block B is specified by sB since point E on
the pulley will have the same motion as the block. Both coordinates
are measured from a horizontal datum passing through the fixed pin
at pulley D. Since the system consists of two cords, the coordinates sA
and sB cannot be related directly. Instead, by establishing a third
position coordinate, sC , we can now express the length of one of the
cords in terms of sB and sC , and the length of the other cord in terms
of sA , sB , and sC .
Excluding the red colored segments of the cords in Fig. 12–40, the
remaining constant cord lengths l1 and l2 (along with the hook and
link dimensions) can be expressed as
sC + sB = l1
1sA - sC2 + 1sB - sC2 + sB = l2
Time Derivative. The time derivative of each equation gives
vC + vB = 0
vA - 2vC + 2vB = 0
Eliminating vC, we obtain
vA + 4vB = 0
so that when vA = 2 m>s (downward),
vB = -0.5 m>s = 0.5 m>s c
Ans.
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EXAMPLE 12.24
A man at A is hoisting a safe S as shown in Fig. 12–41 by walking to
the right with a constant velocity vA = 0.5 m>s. Determine the
velocity and acceleration of the safe when it reaches the elevation of
10 m. The rope is 30 m long and passes over a small pulley at D.
SOLUTION
Position-Coordinate Equation. This problem is unlike the previous
examples since rope segment DA changes both direction and
magnitude. However, the ends of the rope, which define the positions
of S and A, are specified by means of the x and y coordinates since
they must be measured from a fixed point and directed along the paths
of motion of the ends of the rope.
The x and y coordinates may be related since the rope has a fixed
length l = 30 m, which at all times is equal to the length of segment DA
plus CD. Using the Pythagorean theorem to determine lDA , we have
lDA = 4(15)2 + x2 ; also, lCD = 15 - y. Hence,
D
E
15 m
C
10 m
S
l = lDA + lCD
y
30 = 4(15)2 + x2 + (15 - y)
A vA ! 0.5 m/s
y = 4225 + x2 - 15
x
(1)
Time Derivatives. Taking the time derivative, using the chain rule
(see Appendix C), where vS = dy>dt and vA = dx>dt, yields
Fig. 12–41
dy
1
2x
dx
= B
R
2 dt
dt
2 2225 + x
x
(2)
=
vA
2
225
+
x
4
At y = 10 m, x is determined from Eq. 1, i.e., x = 20 m. Hence, from
Eq. 2 with vA = 0.5 m>s,
20
vS =
(0.5) = 0.4m>s = 400 mm> s c Ans.
2
225
+
(20)
4
The acceleration is determined by taking the time derivative of Eq. 2.
Since vA is constant, then a A = dvA>dt = 0, and we have
vS =
aS =
d2y
dt2
= c
-x1dx>dt2
dxvA + c
1225 + x223>2
1
da
dx
bv + c
dt A
1
dx
dvA
225v2A
=
dt
1225 + x223>2
2
2
4225 + x
4225 + x
At x = 20 m, with vA = 0.5 m>s, the acceleration becomes
22510.5 m>s22
aS =
= 0.00360 m>s2 = 3.60 mm>s2 c Ans.
[225 + 120 m22]3>2
NOTE: The constant velocity at A causes the other end C of the rope
to have an acceleration since vA causes segment DA to change its
direction as well as its length.
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12.10
RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES
87
12.10 Relative-Motion of Two Particles
12
Using Translating Axes
Throughout this chapter the absolute motion of a particle has been
determined using a single fixed reference frame. There are many cases,
however, where the path of motion for a particle is complicated, so that it
may be easier to analyze the motion in parts by using two or more frames
of reference. For example, the motion of a particle located at the tip of an
airplane propeller, while the plane is in flight, is more easily described if
one observes first the motion of the airplane from a fixed reference and
then superimposes (vectorially) the circular motion of the particle
measured from a reference attached to the airplane.
In this section translating frames of reference will be considered for the
analysis. Relative-motion analysis of particles using rotating frames of
reference will be treated in Secs. 16.8 and 20.4, since such an analysis
depends on prior knowledge of the kinematics of line segments.
Position. Consider particles A and B, which move along the
arbitrary paths shown in Fig. 12–42. The absolute position of each
particle, rA and rB , is measured from the common origin O of the fixed x,
y, z reference frame. The origin of a second frame of reference x¿, y¿, z¿ is
attached to and moves with particle A. The axes of this frame are only
permitted to translate relative to the fixed frame. The position of B
measured relative to A is denoted by the relative-position vector rB>A .
Using vector addition, the three vectors shown in Fig. 12–42 can be
related by the equation
z¿
z
A
rA
Fixed
observer
(12–33)
x¿
x
Velocity. An equation that relates the velocities of the particles is
determined by taking the time derivative of the above equation; i.e.,
Fig. 12–42
vB = vA + vB>A
(12–34)
Here vB = drB>dt and vA = drA>dt refer to absolute velocities, since
they are observed from the fixed frame; whereas the relative velocity
vB>A = drB>A>dt is observed from the translating frame. It is important
to note that since the x¿, y¿, z¿ axes translate, the components of rB>A
will not change direction and therefore the time derivative of these
components will only have to account for the change in their
magnitudes. Equation 12–34 therefore states that the velocity of B is
equal to the velocity of A plus (vectorially) the velocity of “B with
respect to A,” as measured by the translating observer fixed in the x¿, y¿,
z¿ reference frame.
y¿
B
y
O
rB = rA + rB>A
Translating
observer
rB/A
rB
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Acceleration. The time derivative of Eq. 12–34 yields a similar
vector relation between the absolute and relative accelerations of
particles A and B.
a B = aA + aB>A
(12–35)
Here aB>A is the acceleration of B as seen by the observer located at A
and translating with the x¿, y¿, z¿ reference frame.*
Procedure For Analysis
• When applying the relative velocity and acceleration equations,
it is first necessary to specify the particle A that is the origin for
the translating x¿, y¿, z¿ axes. Usually this point has a known
velocity or acceleration.
• Since vector addition forms a triangle, there can be at most two
unknowns, represented by the magnitudes and/or directions of
the vector quantities.
• These unknowns can be solved for either graphically, using
trigonometry (law of sines, law of cosines), or by resolving each
of the three vectors into rectangular or Cartesian components,
thereby generating a set of scalar equations.
The pilots of these jet planes flying close
to one another must be aware of their
relative positions and velocities at all
times in order to avoid a collision.
* An easy way to remember the setup of these equations, is to note the “cancellation”
of the subscript A between the two terms, e.g., aB = aA + aB>A .
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89
EXAMPLE 12.25
12
A train travels at a constant speed of 60 mi>h, crosses over a road as
shown in Fig. 12–43a. If the automobile A is traveling at 45 mi>h along
the road, determine the magnitude and direction of the velocity of the
train relative to the automobile.
SOLUTION I
Vector Analysis. The relative velocity vT>A is measured from the
translating x¿, y¿ axes attached to the automobile, Fig. 12–43a. It is
determined from vT = vA + vT>A . Since vT and vA are known in both
magnitude and direction, the unknowns become the x and y
components of vT>A . Using the x, y axes in Fig. 12–43a, we have
45%
T
vT ! 60 mi/h
y¿
y
x
A
x¿
vA ! 45 mi/h
(a)
vT = vA + vT>A
60i = 145 cos 45°i + 45 sin 45°j2 + vT>A
vT>A = 528.2i - 31.8j6 mi>h
The magnitude of vT>A is thus
vT>A = 2128.222 + 1-31.822 = 42.5 mi>h
Ans.
Ans.
28.2 mi/h
From the direction of each component, Fig. 12–43b, the direction of
vT>A is
1vT>A2y
31.8
tan u =
=
1vT>A2x
28.2
Ans.
u = 48.5° c
Note that the vector addition shown in Fig. 12–43b indicates the
correct sense for vT>A . This figure anticipates the answer and can be
used to check it.
u
vT/A
31.8 mi/h
(b)
SOLUTION II
Scalar Analysis. The unknown components of vT>A can also be
determined by applying a scalar analysis. We will assume these
components act in the positive x and y directions. Thus,
vT = vA + vT>A
c
60 mi>h
45 mi>h
1vT>A2x
1v 2
d = c
d + c T>A y d
45° d + c
c
:
a
:
Resolving each vector into its x and y components yields
+ 2
1:
60 = 45 cos 45° + 1v 2 + 0
vT/A
vA ! 45 mi/h
T>A x
1+ c 2
0 = 45 sin 45° + 0 + 1vT>A2y
Solving, we obtain the previous results,
1vT>A2x = 28.2 mi>h = 28.2 mi>h :
1vT>A2y = -31.8 mi>h = 31.8 mi>h T
45%
u
vT ! 60 mi/h
(c)
Fig. 12–43
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EXAMPLE 12.26
y¿
y
700 km/h
600 km/h
A
B
x¿
50 km/h2
x
100 km/h2
400 km
4 km
(a)
Plane A in Fig. 12–44a is flying along a straight-line path, whereas
plane B is flying along a circular path having a radius of curvature of
rB = 400 km. Determine the velocity and acceleration of B as
measured by the pilot of A.
SOLUTION
Velocity. The origin of the x and y axes are located at an arbitrary
fixed point. Since the motion relative to plane A is to be determined,
the translating frame of reference x¿, y¿ is attached to it, Fig. 12–44a.
Applying the relative-velocity equation in scalar form since the velocity
vectors of both planes are parallel at the instant shown, we have
1+ c 2
vB/A
vB = vA + vB>A
600 km>h = 700 km>h + vB>A
vB>A = - 100 km>h = 100 km>h T
vA ! 700 km/h v ! 600 km/h
B
Ans.
The vector addition is shown in Fig. 12–44b.
Acceleration. Plane B has both tangential and normal components
of acceleration since it is flying along a curved path. From Eq. 12–20,
the magnitude of the normal component is
(b)
1aB2n =
1600 km>h22
v2B
=
= 900 km>h2
r
400 km
Applying the relative-acceleration equation gives
aB = aA + aB>A
900i - 100j = 50j + aB>A
Thus,
a B>A = 5900i - 150j6 km>h2
From Fig. 12–44c, the magnitude and direction of a B>A are therefore
aB>A = 912 km>h2
900 km/h2
u
aB/A
150 km/h2
(c)
Fig. 12–44
u = tan-1
150
= 9.46° c
900
Ans.
NOTE: The solution to this problem was possible using a translating
frame of reference, since the pilot in plane A is “translating.”
Observation of the motion of plane A with respect to the pilot of
plane B, however, must be obtained using a rotating set of axes
attached to plane B. (This assumes, of course, that the pilot of B is
fixed in the rotating frame, so he does not turn his eyes to follow the
motion of A.) The analysis for this case is given in Example 16.21.
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91
EXAMPLE 12.27
12
At the instant shown in Fig. 12–45a, cars A and B are traveling with
speeds of 18 m>s and 12 m>s, respectively. Also at this instant, A has a
decrease in speed of 2 m>s2, and B has an increase in speed of 3 m>s2.
Determine the velocity and acceleration of B with respect to A.
SOLUTION
Velocity. The fixed x, y axes are established at an arbitrary point on
the ground and the translating x¿, y¿ axes are attached to car A, Fig.
12–45a. Why? The relative velocity is determined from
vB = vA + vB>A . What are the two unknowns? Using a Cartesian
vector analysis, we have
y¿
A
3 m/s2
18 m/s
r ! 100 m
12 m/s
B
y
vB = vA + vB>A
60%
-12j = 1-18 cos 60°i - 18 sin 60°j2 + vB>A
Thus,
2 m/s2
60%
x¿
vB>A = 59i + 3.588j6 m>s
vB>A = 41922 + 13.58822 = 9.69 m>s
x
(a)
Ans.
Noting that vB>A has +i and +j components, Fig. 12–45b, its direction is
tan u =
1vB>A2y
=
1vB>A2x
u = 21.7° a
3.588
9
3.588 m/s
vB/A
Ans.
u
Acceleration. Car B has both tangential and normal components of
acceleration. Why? The magnitude of the normal component is
1aB2n =
9 m/s
(b)
112 m>s22
v2B
=
= 1.440 m>s2
r
100 m
Applying the equation for relative acceleration yields
a B = aA + aB>A
1-1.440i - 3j2 = 12 cos 60°i + 2 sin 60°j2 + aB>A
2.440 m/s2
a B>A = 5- 2.440i - 4.732j6 m>s2
f
Here aB>A has -i and -j components. Thus, from Fig. 12–45c,
aB>A = 212.44022 + 14.73222 = 5.32 m>s2
tan f =
1aB>A2y
1aB>A2x
=
f = 62.7° d
Ans.
4.732
2.440
aB/A
Ans.
NOTE: Is it possible to obtain the relative acceleration of a A>B using
this method? Refer to the comment made at the end of Example 12.26.
4.732 m/s2
(c)
Fig. 12–45
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FUNDAMENTAL PROBLEMS
F12–39. Determine the speed of block D if end A of the
rope is pulled down with a speed of vA = 3 m>s.
B
F12–42. Determine the speed of block A if end F of the
rope is pulled down with a speed of vF = 3 m>s.
C
C
B
E
D
A
A
vA ! 3 m/s
F
D
F12–39
F12–40. Determine the speed of block A if end B of the
rope is pulled down with a speed of 6 m>s.
vF ! 3 m/s
F12–42
F12–43. Determine the speed of car A if point P on the
cable has a speed of 4 m>s when the motor M winds the
cable in.
M
6 m/s
B
P
A
A
F12–40
F12–41. Determine the speed of block A if end B of the
rope is pulled down with a speed of 1.5 m>s.
F12–43
F12–44. Determine the speed of cylinder B if cylinder A
moves downward with a speed of vA = 4 ft>s.
F
E
C
1.5 m/s
B
A
D
A vA ! 4 ft/s
F12–41
B
F12–44
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F12–45. Car A is traveling with a constant speed of
80 km>h due north, while car B is traveling with a constant
speed of 100 km>h due east. Determine the velocity of car B
relative to car A.
93
F12–47. The boats A and B travel with constant speeds of
vA = 15 m>s and vB = 10 m>s when they leave the pier at 12
O at the same time. Determine the distance between them
when t = 4 s.
y
vB ! 10 m/s
B
B
100 km/h
2 km
A
30%
45%
O
A
vA ! 15 m/s
30%
x
80 km/h
F12–45
F12–47
F12–46. Two planes A and B are traveling with the
constant velocities shown. Determine the magnitude and
direction of the velocity of plane B relative to plane A.
F12–48. At the instant shown, cars A and B are traveling at
the speeds shown. If B is accelerating at 1200 km>h2 while
A maintains a constant speed, determine the velocity and
acceleration of A with respect to B.
45%
vB ! 800 km/h
B
vA ! 650 km/h
A
60%
B
A
20 km/h
65 km/h
F12–46
F12–48
100 m
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PROBLEMS
12–195. The mine car C is being pulled up the incline using
the motor M and the rope-and-pulley arrangement shown.
Determine the speed vP at which a point P on the cable
must be traveling toward the motor to move the car up the
plane with a constant speed of v = 2 m>s.
12–198. If end A of the rope moves downward with a speed
of 5 m>s, determine the speed of cylinder B.
M
5 m/s
P
v
A
C
B
Prob. 12–198
12–199. Determine the speed of the elevator if each motor
draws in the cable with a constant speed of 5 m>s.
Prob. 12–195
*12–196. Determine the displacement of the log if the
truck at C pulls the cable 4 ft to the right.
B
C
C
B
Prob. 12–196
•12–197. If the hydraulic cylinder H draws in rod BC at
2 ft>s, determine the speed of slider A.
A
B
C
H
Prob. 12–197
Prob. 12–199
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*12–200. Determine the speed of cylinder A, if the rope is
drawn towards the motor M at a constant rate of 10 m>s.
95
12–203. Determine the speed of B if A is moving
downwards with a speed of vA = 4 m>s at the instant shown. 12
•12–201. If the rope is drawn towards the motor M at a
speed of vM = (5t3>2) m>s, where t is in seconds, determine
the speed of cylinder A when t = 1 s.
B
A
A
M
h
Prob. 12–203
Probs. 12–200/201
12–202. If the end of the cable at A is pulled down with a
speed of 2 m>s, determine the speed at which block B rises.
vA ! 4 m/s
*12–204. The crane is used to hoist the load. If the motors
at A and B are drawing in the cable at a speed of 2 ft>s and
4 ft>s, respectively, determine the speed of the load.
D
2 ft/s
C
2 m/s
A
B
4 ft/s
A
B
Prob. 12–202
Prob. 12–204
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•12–205. The cable at B is pulled downwards at 4 ft>s, and
12 the speed is decreasing at 2 ft>s2. Determine the velocity
and acceleration of block A at this instant.
*12–208. If the end of the cable at A is pulled down with a
speed of 2 m>s, determine the speed at which block E rises.
B
D
2 m/s
A
C
A
C
D
B
4 ft/s
E
Prob. 12–205
Prob. 12–208
12–206. If block A is moving downward with a speed of
4 ft>s while C is moving up at 2 ft>s, determine the speed of
block B.
•12–209. If motors at A and B draw in their attached
cables with an acceleration of a = (0.2t) m>s2, where t is in
seconds, determine the speed of the block when it reaches a
height of h = 4 m, starting from rest at h = 0. Also, how
much time does it take to reach this height?
12–207. If block A is moving downward at 6 ft>s while
block C is moving down at 18 ft>s, determine the speed of
block B.
D
C
B
C
A
A
Probs. 12–206/207
h
Prob. 12–209
B
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12–210. The motor at C pulls in the cable with an
acceleration aC = (3t2) m>s2, where t is in seconds. The
motor at D draws in its cable at aD = 5 m>s2. If both motors
start at the same instant from rest when d = 3 m, determine
(a) the time needed for d = 0, and (b) the velocities of
blocks A and B when this occurs.
97
*12–212. The man pulls the boy up to the tree limb C by
walking backward at a constant speed of 1.5 m>s. 12
Determine the speed at which the boy is being lifted at the
instant xA = 4 m. Neglect the size of the limb. When
xA = 0, yB = 8 m, so that A and B are coincident, i.e., the
rope is 16 m long.
•12–213. The man pulls the boy up to the tree limb C by
walking backward. If he starts from rest when xA = 0 and
moves backward with a constant acceleration aA = 0.2 m>s2 ,
determine the speed of the boy at the instant yB = 4 m.
Neglect the size of the limb.When xA = 0, yB = 8 m, so that A
and B are coincident, i.e., the rope is 16 m long.
D
B
A
C
C
d!3m
Prob. 12–210
yB
8m
B
A
12–211. The motion of the collar at A is controlled by a motor
at B such that when the collar is at sA = 3 ft it is moving
upwards at 2 ft>s and decreasing at 1 ft>s2. Determine the
velocity and acceleration of a point on the cable as it is drawn
into the motor B at this instant.
xA
Probs. 12–212/213
12–214. If the truck travels at a constant speed of
vT = 6 ft>s, determine the speed of the crate for any angle u
of the rope. The rope has a length of 100 ft and passes over
a pulley of negligible size at A. Hint: Relate the coordinates
xT and xC to the length of the rope and take the time
derivative. Then substitute the trigonometric relation
between xC and u.
4 ft
sA
xC
xT
A
A
20 ft
B
Prob. 12–211
C
u
Prob. 12–214
T
vT
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12–215. At the instant shown, car A travels along the
12 straight portion of the road with a speed of 25 m>s. At this
same instant car B travels along the circular portion of the
road with a speed of 15 m>s. Determine the velocity of car B
relative to car A.
12–218. The ship travels at a constant speed of vs = 20 m>s
and the wind is blowing at a speed of vw = 10 m>s , as shown.
Determine the magnitude and direction of the horizontal
component of velocity of the smoke coming from the smoke
stack as it appears to a passenger on the ship.
vs ! 20 m/s
30%
15%
15%
45%
vw ! 10 m/s
y
r ! 200 m
x
A
C
30%
B
Prob. 12–215
Prob. 12–218
*12–216. Car A travels along a straight road at a speed of
25 m>s while accelerating at 1.5 m>s2. At this same instant
car C is traveling along the straight road with a speed of
30 m>s while decelerating at 3 m>s2. Determine the velocity
and acceleration of car A relative to car C.
•12–217. Car B is traveling along the curved road with a
speed of 15 m>s while decreasing its speed at 2 m>s2. At this
same instant car C is traveling along the straight road with a
speed of 30 m>s while decelerating at 3 m>s2. Determine the
velocity and acceleration of car B relative to car C.
A
25 m/s
45%
12–219. The car is traveling at a constant speed of
100 km>h. If the rain is falling at 6 m>s in the direction
shown, determine the velocity of the rain as seen by the
driver.
30%
vr
vC ! 100 km/h
Prob. 12–219
*12–220. The man can row the boat in still water with a
speed of 5 m>s. If the river is flowing at 2 m>s, determine
the speed of the boat and the angle u he must direct the
boat so that it travels from A to B.
2
1.5 m/s
r ! 100 m
2 m/s2
3 m/s2
30%
B
B
C
30 m/s
vw ! 2 m/s
5 m/s
15 m/s
u
50 m
A
25 m
Probs. 12–216/217
Prob. 12–220
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•12–221. At the instant shown, cars A and B travel at
speeds of 30 mi>h and 20 mi>h, respectively. If B is
increasing its speed by 1200 mi>h2, while A maintains a
constant speed, determine the velocity and acceleration of
B with respect to A.
12–222. At the instant shown, cars A and B travel at speeds
of 30 m>h and 20 mi>h, respectively. If A is increasing its
speed at 400 mi>h2 whereas the speed of B is decreasing at
800 mi>h2, determine the velocity and acceleration of B
with respect to A.
B
99
*12–224. At the instant shown, cars A and B travel at
speeds of 70 mi>h and 50 mi>h, respectively. If B is 12
increasing its speed by 1100 mi>h2, while A maintains a
constant speed, determine the velocity and acceleration of
B with respect to A. Car B moves along a curve having a
radius of curvature of 0.7 mi.
•12–225. At the instant shown, cars A and B travel at
speeds of 70 mi>h and 50 mi>h, respectively. If B is
decreasing its speed at 1400 mi>h2 while A is increasing its
speed at 800 mi>h2, determine the acceleration of B with
respect to A. Car B moves along a curve having a radius of
curvature of 0.7 mi.
30%
0.3 mi
vB ! 20 mi/h
vA ! 70 mi/h
A
vA ! 30 mi/h
A
vB ! 50 mi/h
B
30%
Probs. 12–221/222
Probs. 12–224/225
12–223. Two boats leave the shore at the same time and
travel in the directions shown. If vA = 20 ft>s and
vB = 15 ft>s, determine the velocity of boat A with respect
to boat B. How long after leaving the shore will the boats be
800 ft apart?
12–226. An aircraft carrier is traveling forward with a
velocity of 50 km>h. At the instant shown, the plane at A
has just taken off and has attained a forward horizontal air
speed of 200 km>h, measured from still water. If the plane
at B is traveling along the runway of the carrier at 175 km>h
in the direction shown, determine the velocity of A with
respect to B.
15%
vA
B
A
A
B
50 km/h
vB
Prob. 12–226
30%
O
45%
Prob. 12–223
12–227. A car is traveling north along a straight road at
50 km>h. An instrument in the car indicates that the wind is
directed towards the east. If the car’s speed is 80 km>h, the
instrument indicates that the wind is directed towards the
north-east. Determine the speed and direction of the wind.
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*12–228. At the instant shown car A is traveling with a
12 velocity of 30 m>s and has an acceleration of 2 m>s2 along
the highway. At the same instant B is traveling on the
trumpet interchange curve with a speed of 15 m>s, which is
decreasing at 0.8 m>s2. Determine the relative velocity and
relative acceleration of B with respect to A at this instant.
12–230. A man walks at 5 km>h in the direction of a
20-km>h wind. If raindrops fall vertically at 7 km>h in still air,
determine the direction in which the drops appear to fall with
respect to the man. Assume the horizontal speed of the
raindrops is equal to that of the wind.
vw ! 20 km/h
vm ! 5 km/h
A
B
r ! 250 m
60%
Prob. 12–228
•12–229. Two cyclists A and B travel at the same constant
speed v. Determine the velocity of A with respect to B if A
travels along the circular track, while B travels along the
diameter of the circle.
Prob. 12–230
12–231. A man can row a boat at 5 m>s in still water. He
wishes to cross a 50-m-wide river to point B, 50 m
downstream. If the river flows with a velocity of 2 m>s,
determine the speed of the boat and the time needed to
make the crossing.
50 m
2 m/s
v
A
f
A
r
B
u
u
v
50 m
B
Prob. 12–229
Prob. 12–231
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101
CONCEPTUAL PROBLEMS
P12–1. If you measured the time it takes for the
construction elevator to go from A to B, then B to C, and
then C to D, and you also know the distance between each
of the points, how could you determine the average
velocity and average acceleration of the elevator as it
ascends from A to D? Use numerical values to explain how
this can be done.
12
P12–3. The basketball was thrown at an angle measured
from the horizontal to the man’s outstretched arms. If the
basket is 10 ft from the ground, make appropriate
measurements in the photo and determine if the ball
located as shown will pass through the basket.
D
C
B
P12–3
A
P12–1
P12–4. The pilot tells you the wingspan of her plane and her
constant airspeed. How would you determine the acceleration
of the plane at the moment shown? Use numerical values and
take any necessary measurements from the photo.
P12–2. If the sprinkler at A is 1 m from the ground, then
scale the necessary measurements from the photo to
determine the approximate velocity of the water jet as it
flows from the nozzle of the sprinkler.
P12–2
P12–4
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CHAPTER REVIEW
Rectilinear Kinematics
Rectilinear kinematics refers to motion
along a straight line.A position coordinate
s specifies the location of the particle on
the line, and the displacement ¢s is the
change in this position.
s
O
s
The average velocity is a vector quantity,
defined as the displacement divided by
the time interval.
vavg =
- ¢s
¢t
\$s
\$s
s
O
sT
The average speed is a scalar, and is the
total distance traveled divided by the time
of travel.
The time, position, velocity, and
acceleration are related by three
differential equations.
If the acceleration is known to be
constant, then the differential equations
relating time, position, velocity, and
acceleration can be integrated.
1vsp2avg =
a =
dv
,
dt
v =
sT
¢t
ds
,
dt
a ds = v dv
v = v0 + act
s = s0 + v0t + 12 act2
v2 = v20 + 2ac1s - s02
Graphical Solutions
If the motion is erratic, then it can be
described by a graph. If one of these
graphs is given, then the others can be
established using the differential relations
between a, v, s, and t.
a =
dv
,
dt
ds
,
dt
a ds = v dv
v =
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12
Curvilinear Motion, x , y, z
#
vx = x
#
vy = y
#
vz = z
Curvilinear motion along the path can
be resolved into rectilinear motion
along the x, y, z axes. The equation of the
path is used to relate the motion along
each axis.
#
ax = vx
#
a y = vy
#
az = vz
z
s
k
i
a
r ! xi # yj # zk
j
Projectile Motion
Free-flight motion of a projectile follows
a parabolic path. It has a constant
velocity in the horizontal direction, and a
constant downward acceleration of
g = 9.81 m>s2 or 32.2 ft>s2 in the vertical
direction. Any two of the three equations
for constant acceleration apply in the
vertical direction, and in the horizontal
direction only one equation applies.
1+ c 2
vy = 1v02y + act
1+ c2
v2y = 1v022y + 2ac1y - y02
1+ c 2
+ 2
1:
y = y0 + 1v02yt + 12 act2
x = x0 + 1v02xt
y
a!g
vx
v0
(v0)y
vy
(v0)x
v
r
y
y0
x
x0
x
y
x
y
x
v
z
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PA R T I C L E
Curvilinear Motion n, t
If normal and tangential axes are used
for the analysis, then v is always in the
positive t direction.
The acceleration has two components.
The tangential component, at, accounts
for the change in the magnitude of the
velocity; a slowing down is in the
negative t direction, and a speeding up is
in the positive t direction. The normal
component an accounts for the change in
the direction of the velocity. This
component is always in the positive n
direction.
O¿
n
O
an
s
#
at = v
a
a tds = v dv
or
at
2
v
t
v
an =
r
Curvilinear Motion r, U
If the path of motion is expressed in
polar coordinates, then the velocity and
acceleration components can be related
to the time derivatives of r and u.
To apply the time-derivative equations,
it
# \$ # \$
is necessary to determine r, r, r, u, u at
the instant considered. If the path
r = f1u2 is given, then the chain rule of
calculus must be used to obtain time
derivatives. (See Appendix C.)
#
vr = r
#
vu = ru
#
\$
a r = r - ru2
\$
##
au = ru + 2ru
v
vu
vr
P
r
u
O
Velocity
Once the data are substituted into the
equations, then the algebraic sign of
the results will indicate the direction of
the components of v or a along each axis.
a
au
ar
r
u
O
Acceleration
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12
Absolute Dependent Motion of Two
Particles
The dependent motion of blocks that are
suspended from pulleys and cables can
be related by the geometry of the
system. This is done by first establishing
position coordinates, measured from a
fixed origin to each block. Each
coordinate must be directed along the
line of motion of a block.
Using geometry and/or trigonometry,
the coordinates are then related to the
cable length in order to formulate a
position coordinate equation.
The first time derivative of this equation
gives a relationship between the
velocities of the blocks, and a second time
derivative gives the relation between
their accelerations.
Datum
sB
B
h
2sB + h + sA = l
A
2vB = - vA
2aB = - aA
sA
Datum
Relative-Motion Analysis Using
Translating Axes
If two particles A and B undergo
independent motions, then these
motions can be related to their relative
motion using a translating set of axes
attached to one of the particles (A).
For planar motion, each vector equation
produces two scalar equations, one in the
x, and the other in the y direction. For
solution, the vectors can be expressed in
Cartesian form, or the x and y scalar
components can be written directly.
z¿
a
z
rB = rA + rB>A
vB = vA + vB>A
a B = aA + aB>A
A
a
Translating
observer
rB/A
rA
Fixed
observer
x¿
x
b
B
y
O
y¿
rB
b
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The design of conveyors for a bottling plant requires knowledge of the forces that act
on them and the ability to predict the motion of the bottles they transport.
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Kinetics of a Particle:
Force and Acceleration
CHAPTER OBJECTIVES
• To state Newton’s Second Law of Motion and to define mass
and weight.
• To analyze the accelerated motion of a particle using the equation
of motion with different coordinate systems.
• To investigate central-force motion and apply it to problems in
space mechanics.
13.1 Newton’s Second Law of Motion
Kinetics is a branch of dynamics that deals with the relationship between
the change in motion of a body and the forces that cause this change. The
basis for kinetics is Newton’s second law, which states that when an
unbalanced force acts on a particle, the particle will accelerate in the
direction of the force with a magnitude that is proportional to the force.
This law can be verified experimentally by applying a known
unbalanced force F to a particle, and then measuring the acceleration
a. Since the force and acceleration are directly proportional, the
constant of proportionality, m, may be determined from the ratio
m = F>a. This positive scalar m is called the mass of the particle.
Being constant during any acceleration, m provides a quantitative
measure of the resistance of the particle to a change in its velocity, that
is its inertia.
13
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If the mass of the particle is m, Newton’s second law of motion may be
written in mathematical form as
F = ma
13
The above equation, which is referred to as the equation of motion, is
one of the most important formulations in mechanics.* As previously
stated, its validity is based solely on experimental evidence. In 1905,
however, Albert Einstein developed the theory of relativity and
placed limitations on the use of Newton’s second law for describing
general particle motion. Through experiments it was proven that time
is not an absolute quantity as assumed by Newton; and as a result, the
equation of motion fails to predict the exact behavior of a particle,
especially when the particle’s speed approaches the speed of light
10.3 Gm>s2. Developments of the theory of quantum mechanics by
Erwin Schrödinger and others indicate further that conclusions drawn
from using this equation are also invalid when particles are the size of
an atom and move close to one another. For the most part, however,
these requirements regarding particle speed and size are not
encountered in engineering problems, so their effects will not be
considered in this book.
Newton’s Law of Gravitational Attraction. Shortly after
formulating his three laws of motion, Newton postulated a law governing
the mutual attraction between any two particles. In mathematical form
this law can be expressed as
F = G
m1m2
r2
(13–1)
where
F = force of attraction between the two particles
G = universal constant of gravitation; according to
experimental evidence G = 66.73110-122 m3>1kg # s22
m1 , m2 = mass of each of the two particles
r = distance between the centers of the two particles
*Since m is constant, we can also write F = d1mv2>dt, where mv is the particle’s linear
momentum. Here the unbalanced force acting on the particle is proportional to the time
rate of change of the particle’s linear momentum.
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13.1
109
NEWTON’S SECOND LAW OF MOTION
In the case of a particle located at or near the surface of the earth, the
only gravitational force having any sizable magnitude is that between the
earth and the particle. This force is termed the “weight” and, for our
purpose, it will be the only gravitational force considered.
From Eq. 13–1, we can develop a general expression for finding the
weight W of a particle having a mass m1 = m. Let m2 = Me be the mass
of the earth and r the distance between the earth’s center and the
particle. Then, if g = GMe>r2, we have
13
W = mg
By comparison with F = ma, we term g the acceleration due to gravity.
For most engineering calculations g is a point on the surface of the earth
at sea level, and at a latitude of 45°, which is considered the “standard
location.” Here the values g = 9.81 m>s2 = 32.2 ft>s2 will be used for
calculations.
In the SI system the mass of the body is specified in kilograms, and the
weight must be calculated using the above equation, Fig. 13–1a. Thus,
m (kg)
W = mg 1N2
1g = 9.81 m>s22
(13–2)
As a result, a body of mass 1 kg has a weight of 9.81 N; a 2-kg body
weighs 19.62 N; and so on.
a ! g (m/s2)
W ! mg (N)
SI system
(a)
In the FPS system the weight of the body is specified in pounds. The
mass is measured in slugs, a term derived from “sluggish” which refers to
the body’s inertia. It must be calculated, Fig. 13–1b, using
m! W
g (slug)
m =
W
1slug2 1g = 32.2 ft>s22
g
a ! g (ft/s2)
(13–3)
Therefore, a body weighing 32.2 lb has a mass of 1 slug; a 64.4-lb body
has a mass of 2 slugs; and so on.
W (lb)
FPS system
(b)
Fig. 13–1
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13.2 The Equation of Motion
F2
a
When more than one force acts on a particle, the resultant force is
determined by a vector summation of all the forces; i.e., FR = ©F. For
this more general case, the equation of motion may be written as
F1
13
OF A
Page 110
(a)
To illustrate application of this equation, consider the particle shown
in Fig. 13–2a, which has a mass m and is subjected to the action of two
forces, F1 and F2 . We can graphically account for the magnitude and
direction of each force acting on the particle by drawing the particle’s
free-body diagram, Fig. 13–2b. Since the resultant of these forces
produces the vector ma, its magnitude and direction can be represented
graphically on the kinetic diagram, shown in Fig. 13–2c.* The equal sign
written between the diagrams symbolizes the graphical equivalency
between the free-body diagram and the kinetic diagram; i.e., ©F = ma.†
In particular, note that if FR = ©F = 0, then the acceleration is also
zero, so that the particle will either remain at rest or move along a
straight-line path with constant velocity. Such are the conditions of static
equilibrium, Newton’s first law of motion.
F2
FR ! "F
ma
!
F1
Kinetic
diagram
Free-body
diagram
Fig. 13–2
Inertial Reference Frame. When applying the equation of
y
a
Path of particle
vO
O
(13–4)
Inertial frame of reference
Fig. 13–3
x
motion, it is important that the acceleration of the particle be measured
with respect to a reference frame that is either fixed or translates with a
constant velocity. In this way, the observer will not accelerate and
measurements of the particle’s acceleration will be the same from any
reference of this type. Such a frame of reference is commonly known as a
Newtonian or inertial reference frame, Fig. 13–3.
When studying the motions of rockets and satellites, it is justifiable to
consider the inertial reference frame as fixed to the stars, whereas
dynamics problems concerned with motions on or near the surface of the
earth may be solved by using an inertial frame which is assumed fixed to
the earth. Even though the earth both rotates about its own axis and
revolves about the sun, the accelerations created by these rotations are
relatively small and so they can be neglected for most applications.
*Recall the free-body diagram considers the particle to be free of its surrounding supports
and shows all the forces acting on the particle. The kinetic diagram pertains to the particle’s
motion as caused by the forces.
†The equation of motion can also be rewritten in the form ©F - ma = 0. The vector
- ma is referred to as the inertia force vector. If it is treated in the same way as a “force
vector,” then the state of “equilibrium” created is referred to as dynamic equilibrium. This
method of application is often referred to as the D’Alembert principle, named after the
French mathematician Jean le Rond d’Alembert.
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THE EQUATION OF MOTION
111
We are all familiar with the sensation one feels when sitting in a car that is subjected to a forward acceleration. Often people
think this is caused by a “force” which acts on them and tends to push them back in their seats; however, this is not the case.
Instead, this sensation occurs due to their inertia or the resistance of their mass to a change in velocity.
Consider the passenger who is strapped to the seat of a rocket sled. Provided the sled is at rest or is moving with constant
velocity, then no force is exerted on his back as shown on his free-body diagram.
13
W
N2
N1
At rest or constant velocity
When the thrust of the rocket engine causes the sled to accelerate, then the seat upon which he is sitting exerts a force F on
him which pushes him forward with the sled. In the photo, notice that the inertia of his head resists this change in motion
(acceleration), and so his head moves back against the seat and his face, which is nonrigid, tends to distort backward.
F
W
N2
N1
Acceleration
Upon deceleration the force of the seatbelt F¿ tends to pull his body to a stop, but his head leaves contact with the back of the
seat and his face distorts forward, again due to his inertia or tendency to continue to move forward. No force is pulling him
forward, although this is the sensation he receives.
F¿
W
N2
N1
Deceleration
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13.3 Equation of Motion for a System of
Particles
13
The equation of motion will now be extended to include a system of
particles isolated within an enclosed region in space, as shown in Fig. 13–4a.
In particular, there is no restriction in the way the particles are connected,
so the following analysis applies equally well to the motion of a solid,
liquid, or gas system.
At the instant considered, the arbitrary i-th particle, having a mass mi ,
is subjected to a system of internal forces and a resultant external force.
The internal force, represented symbolically as fi , is the resultant of all
the forces the other particles exert on the ith particle. The resultant
external force Fi represents, for example, the effect of gravitational,
electrical, magnetic, or contact forces between the ith particle and
adjacent bodies or particles not included within the system.
The free-body and kinetic diagrams for the ith particle are shown in
Fig. 13–4b. Applying the equation of motion,
Fi + fi = miai
When the equation of motion is applied to each of the other particles of
the system, similar equations will result. And, if all these equations are
z
Fi
fi
i
ri
G
rG
Fi
y
x
fi
!
Free-body
diagram
Inertial coordinate
system
Kinetic
diagram
(b)
(a)
Fig. 13–4
mi ai
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EQUATION OF MOTION FOR A SYSTEM OF PARTICLES
113
The summation of the internal forces, if carried out, will equal zero, since
internal forces between any two particles occur in equal but opposite
collinear pairs. Consequently, only the sum of the external forces will
remain, and therefore the equation of motion, written for the system of
particles, becomes
(13–5)
If rG is a position vector which locates the center of mass G of the
particles, Fig. 13–4a, then by definition of the center of mass,
mrG = ©miri , where m = ©mi is the total mass of all the particles.
Differentiating this equation twice with respect to time, assuming that no
mass is entering or leaving the system, yields
Substituting this result into Eq. 13–5, we obtain
(13–6)
Hence, the sum of the external forces acting on the system of particles is
equal to the total mass of the particles times the acceleration of its center
of mass G. Since in reality all particles must have a finite size to possess
mass, Eq. 13–6 justifies application of the equation of motion to a body
that is represented as a single particle.
Important Points
• The equation of motion is based on experimental evidence and is
•
•
•
•
valid only when applied within an inertial frame of reference.
The equation of motion states that the unbalanced force on a
particle causes it to accelerate.
An inertial frame of reference does not rotate, rather its axes
either translate with constant velocity or are at rest.
Mass is a property of matter that provides a quantitative measure
of its resistance to a change in velocity. It is an absolute quantity
and so it does not change from one location to another.
Weight is a force that is caused by the earth’s gravitation. It is not
absolute; rather it depends on the altitude of the mass from the
earth’s surface.
13
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13.4 Equations of Motion: Rectangular
Coordinates
When a particle moves relative to an inertial x, y, z frame of reference,
the forces acting on the particle, as well as its acceleration, can be expressed
in terms of their i, j, k components, Fig. 13–5. Applying the equation of
motion, we have
13
z
Fz
Fy
Fx
z
For this equation to be satisfied, the respective i, j, k components on the
left side must equal the corresponding components on the right side.
Consequently, we may write the following three scalar equations:
y
x
y
x
Fig. 13–5
©Fx i + ©Fy j + ©Fz k = m1ax i + ay j + az k2
(13–7)
In particular, if the particle is constrained to move only in the x–y plane,
then the first two of these equations are used to specify the motion.
Procedure for Analysis
The equations of motion are used to solve problems which require a
relationship between the forces acting on a particle and the
accelerated motion they cause.
Free-Body Diagram.
● Select the inertial coordinate system. Most often, rectangular or
x, y, z coordinates are chosen to analyze problems for which the
particle has rectilinear motion.
● Once the coordinates are established, draw the particle’s freebody diagram. Drawing this diagram is very important since it
provides a graphical representation that accounts for all the
forces 1©F2 which act on the particle, and thereby makes it
possible to resolve these forces into their x, y, z components.
● The direction and sense of the particle’s acceleration a should also
be established. If the sense is unknown, for mathematical
convenience assume that the sense of each acceleration component
acts in the same direction as its positive inertial coordinate axis.
● The acceleration may be represented as the ma vector on the
kinetic diagram.*
●
Identify the unknowns in the problem.
*It is a convention in this text always to use the kinetic diagram as a graphical aid
when developing the proofs and theory. The particle’s acceleration or its components
will be shown as blue colored vectors near the free-body diagram in the examples.
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EQUATIONS OF MOTION: RECTANGULAR COORDINATES
115
Equations of Motion.
● If the forces can be resolved directly from the free-body diagram,
apply the equations of motion in their scalar component form.
●
●
●
If the geometry of the problem appears complicated, which often
occurs in three dimensions, Cartesian vector analysis can be used
for the solution.
Friction. If a moving particle contacts a rough surface, it may be
necessary to use the frictional equation, which relates the
frictional and normal forces Ff and N acting at the surface of
contact by using the coefficient of kinetic friction, i.e., Ff = mkN.
Remember that Ff always acts on the free-body diagram such
that it opposes the motion of the particle relative to the surface it
contacts. If the particle is on the verge of relative motion, then the
coefficient of static friction should be used.
Spring. If the particle is connected to an elastic spring having
negligible mass, the spring force Fs can be related to the
deformation of the spring by the equation Fs = ks. Here k is the
spring’s stiffness measured as a force per unit length, and s is
the stretch or compression defined as the difference between
the deformed length l and the undeformed length l0 , i.e.,
s = l - l0 .
Kinematics.
● If the velocity or position of the particle is to be found, it will be
necessary to apply the necessary kinematic equations once the
particle’s acceleration is determined from ©F = ma.
●
●
●
●
●
If acceleration is a function of time, use a = dv>dt and v = ds>dt
which, when integrated, yield the particle’s velocity and position,
respectively.
If acceleration is a function of displacement, integrate a ds = v dv
to obtain the velocity as a function of position.
If acceleration is constant, use v = v0 + ac t, s = s0 + v0t + 12 ac t2,
v2 = v20 + 2ac1s - s02 to determine the velocity or position of the
particle.
If the problem involves the dependent motion of several particles,
use the method outlined in Sec. 12.9 to relate their accelerations.
In all cases, make sure the positive inertial coordinate directions
used for writing the kinematic equations are the same as those
used for writing the equations of motion; otherwise, simultaneous
solution of the equations will result in errors.
If the solution for an unknown vector component yields a
negative scalar, it indicates that the component acts in the
direction opposite to that which was assumed.
13
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EXAMPLE 13.1
The 50-kg crate shown in Fig. 13–6a rests on a horizontal surface for
which the coefficient of kinetic friction is mk = 0.3. If the crate is
subjected to a 400-N towing force as shown, determine the velocity of
the crate in 3 s starting from rest.
P ! 400 N
30#
13
SOLUTION
Using the equations of motion, we can relate the crate’s acceleration
to the force causing the motion. The crate’s velocity can then be
determined using kinematics.
(a)
Free-Body Diagram. The weight of the crate is W = mg =
50 kg 19.81 m>s22 = 490.5 N. As shown in Fig. 13–6b, the frictional
force has a magnitude F = mkNC and acts to the left, since it opposes the
motion of the crate. The acceleration a is assumed to act horizontally, in
the positive x direction. There are two unknowns, namely NC and a.
Equations of Motion. Using the data shown on the free-body
diagram, we have
(1)
:
400 cos 30° - 0.3NC = 50a
x
x
c
(2)
NC - 490.5 + 400 sin 30° = 0
y
a
x
490.5 N
400 N
Solving Eq. 2 for NC , substituting the result into Eq. 1, and solving
for a yields
NC = 290.5 N
a = 5.185 m>s2
30#
F ! 0.3 NC
NC
(b)
Fig. 13–6
Kinematics. Notice that the acceleration is constant, since the
applied force P is constant. Since the initial velocity is zero, the
velocity of the crate in 3 s is
+2
v = v + a t = 0 + 5.185132
1:
0
c
= 15.6 m>s :
Ans.
490.5 N
400 N
30#
!
50a
F ! 0.3NC
NC
(c)
NOTE: We can also use the alternative procedure of drawing the
crate’s free-body and kinetic diagrams, Fig. 13–6c, prior to applying
the equations of motion.
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EQUATIONS OF MOTION: RECTANGULAR COORDINATES
EXAMPLE 13.2
A 10-kg projectile is fired vertically upward from the ground, with an
initial velocity of 50 m>s, Fig. 13–7a. Determine the maximum height
to which it will travel if (a) atmospheric resistance is neglected; and
(b) atmospheric resistance is measured as FD = 10.01v22 N, where v is
the speed of the projectile at any instant, measured in m>s.
z
SOLUTION
In both cases the known force on the projectile can be related to its
acceleration using the equation of motion. Kinematics can then be
used to relate the projectile’s acceleration to its position.
Part (a) Free-Body Diagram. As shown in Fig. 13–7b, the projectile’s
weight is W = mg = 1019.812 = 98.1 N. We will assume the
unknown acceleration a acts upward in the positive z direction.
Equation of Motion.
+ c ©Fz = maz ;
(a)
a = - 9.81 m>s2
-98.1 = 10a,
z
The result indicates that the projectile, like every object having freeflight motion near the earth’s surface, is subjected to a constant
downward acceleration of 9.81 m>s2.
Kinematics. Initially, z0 = 0 and v0 = 50 m>s, and at the maximum
height z = h, v = 0. Since the acceleration is constant, then
v2 = v20 + 2ac1z - z02
0 = 15022 + 21- 9.8121h - 02
h = 127 m
1+ c2
a
98.1 N
(b)
Ans.
z
Part (b) Free-Body Diagram. Since the force FD = 10.01v 2 N
tends to retard the upward motion of the projectile, it acts downward
as shown on the free-body diagram, Fig. 13–7c.
FD
Equation of Motion.
+ c ©Fz = maz ;
-0.01v2 - 98.1 = 10a,
98.1 N
2
2
a = -(0.001v + 9.81)
Kinematics. Here the acceleration is not constant since FD depends
on the velocity. Since a = f1v2, we can relate a to position using
1 + c 2 a dz = v dv;
-10.001v2 + 9.812 dz = v dv
Separating the variables and integrating, realizing that initially z0 = 0,
v0 = 50 m>s (positive upward), and at z = h, v = 0, we have
L0
0
h
dz = -
0
v dv
= -500 ln1v2 + 98102 `
2
L50 0.001v + 9.81
50 m>s
h = 114 m
Ans.
The answer indicates a lower elevation than that obtained in
part (a) due to atmospheric resistance or drag.
NOTE:
a
(c)
Fig. 13–7
13
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EXAMPLE 13.3
The baggage truck A shown in the photo has a weight of 900 lb and
tows a 550-lb cart B and a 325-lb cart C. For a short time the driving
frictional force developed at the wheels of the truck is FA = 140t2 lb,
where t is in seconds. If the truck starts from rest, determine its speed
in 2 seconds. Also, what is the horizontal force acting on the coupling
between the truck and cart B at this instant? Neglect the size of the
truck and carts.
13
A
B
C
900 lb
550 lb
325 lb
FA
NC
NB
NA
(a)
SOLUTION
Free-Body Diagram. As shown in Fig. 13–8a, it is the frictional
driving force that gives both the truck and carts an acceleration. Here
we have considered all three vehicles as a single system.
Equation of Motion. Only motion in the horizontal direction has
to be considered.
900 + 550 + 325
;
40t = a
ba
x
x
32.2
a = 0.7256t
Kinematics. Since the acceleration is a function of time, the velocity
of the truck is obtained using a = dv>dt with the initial condition that
v0 = 0 at t = 0. We have
L0
900 lb
T
FA
NA
(b)
Fig. 13–8
v
dv =
L0
2s
0.7256t dt;
v = 0.3628t2 `
2s
0
= 1.45 ft>s
Ans.
Free-Body Diagram. In order to determine the force between the
truck and cart B, we will consider a free-body diagram of the truck so
that we can “expose” the coupling force T as external to the free-body
diagram, Fig. 13–8b.
Equation of Motion. When t = 2 s, then
900
;
b[0.7256122]
40122 - T = a
x
x
32.2
T = 39.4 lb
Ans.
Try and obtain this same result by considering a free-body
diagram of carts B and C as a single system.
NOTE:
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EQUATIONS OF MOTION: RECTANGULAR COORDINATES
EXAMPLE 13.4
A smooth 2-kg collar C, shown in Fig. 13–9a, is attached to a spring
having a stiffness k = 3 N>m and an unstretched length of 0.75 m. If
the collar is released from rest at A, determine its acceleration and the
normal force of the rod on the collar at the instant y = 1 m.
SOLUTION
0.75 m
A
y
13
k ! 3 N/m
Free-Body Diagram. The free-body diagram of the collar when it
is located at the arbitrary position y is shown in Fig. 13–9b.
Furthermore, the collar is assumed to be accelerating so that “a” acts
downward in the positive y direction. There are four unknowns,
namely, NC , Fs , a, and u.
C
Equations of Motion.
(a)
:
x
x
- NC + Fs cos u = 0
(1)
+ T ©Fy = may ;
19.62 - Fs sin u = 2a
(2)
From Eq. 2 it is seen that the acceleration depends on the magnitude
and direction of the spring force. Solution for NC and a is possible
once Fs and u are known.
x
a
s = CB - AB = 4y2 + 10.7522 - 0.75.
Fs = ks = 3 A 4y2 + 10.7522 - 0.75 B
(3)
From Fig. 13–9a, the angle u is related to y by trigonometry.
tan u =
y
0.75
(4)
Substituting y = 1 m into Eqs. 3 and 4 yields Fs = 1.50 N and
u = 53.1°. Substituting these results into Eqs. 1 and 2, we obtain
NC = 0.900 N
Ans.
2
a = 9.21 m>s T
19.62 N
y
The magnitude of the spring force is a function of the stretch s of the
spring; i.e., Fs = ks. Here the unstretched length is AB = 0.75 m,
Fig. 13–9a; therefore,
Since k = 3 N>m, then
B
u
Ans.
This is not a case of constant acceleration, since the spring
force changes both its magnitude and direction as the collar moves
downward.
NOTE:
(b)
Fig. 13–9
u
NC
Fs
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EXAMPLE 13.5
Datum
13
sA
The 100-kg block A shown in Fig. 13–10a is released from rest. If the
masses of the pulleys and the cord are neglected, determine the speed
of the 20-kg block B in 2 s.
SOLUTION
C
sB
A
B
(a)
T T
Equations of Motion. Block A,
+ T ©Fy = may ;
Block B,
981 - 2T = 100aA
(1)
+ T ©Fy = may ;
196.2 - T = 20aB
(2)
Kinematics. The necessary third equation is obtained by relating aA
to aB using a dependent motion analysis, discussed in Sect. 12.9. The
coordinates sA and sB in Fig. 13–10a measure the positions of A and B
from the fixed datum. It is seen that
2T
(b)
2T
aA
sA 981 N
(c)
T
aB
sB
Free-Body Diagrams. Since the mass of the pulleys is neglected,
then for pulley C, ma = 0 and we can apply ©Fy = 0 as shown in
Fig. 13–10b. The free-body diagrams for blocks A and B are shown
in Fig. 13–10c and d, respectively. Notice that for A to remain
stationary T = 490.5 N, whereas for B to remain static T = 196.2 N.
Hence A will move down while B moves up. Although this is the
case, we will assume both blocks accelerate downward, in the
direction of + sA and + sB . The three unknowns are T, aA , and aB .
196.2 N
(d)
Fig. 13–10
2sA + sB = l
where l is constant and represents the total vertical length of cord.
Differentiating this expression twice with respect to time yields
(3)
2aA = - aB
Notice that when writing Eqs. 1 to 3, the positive direction was always
assumed downward. It is very important to be consistent in this
assumption since we are seeking a simultaneous solution of equations.
The results are
T = 327.0 N
aA = 3.27 m>s2
aB = - 6.54 m>s2
Hence when block A accelerates downward, block B accelerates
upward as expected. Since a B is constant, the velocity of block B in 2 s
is thus
v = v0 + aBt
1+ T2
= 0 + 1-6.542122
= - 13.1 m>s
Ans.
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EQUATIONS OF MOTION: RECTANGULAR COORDINATES
FUNDAMENTAL PROBLEMS
F13–1. The motor winds in the cable with a constant
acceleration, such that the 20-kg crate moves a distance
s = 6 m in 3 s, starting from rest. Determine the tension
developed in the cable. The coefficient of kinetic friction
between the crate and the plane is mk = 0.3.
F13–4. The 2-Mg car is being towed by a winch. If the winch
exerts a force of T = (100s) N on the cable, where s is the
displacement of the car in meters, determine the speed of the
13
car when s = 10 m, starting from rest. Neglect rolling
resistance of the car.
M
s
F13–4
A
F13–5. The spring has a stiffness k = 200 N/m and is
unstretched when the 25-kg block is at A. Determine the
acceleration of the block when s = 0.4 m. The contact
surface between the block and the plane is smooth.
30!
F13–1
s
F13–2. If motor M exerts a force of F = (10t2 + 100) N
on the cable, where t is in seconds, determine the velocity of
the 25-kg crate when t = 4 s. The coefficients of static and
kinetic friction between the crate and the plane are
ms = 0.3 and mk = 0.25, respectively. The crate is initially
at rest.
A
F ! 100 N
F ! 100 N
k ! 200 N/m
0.3 m
M
F13–5
F13–2
F13–3. A spring of stiffness k = 500 N>m is mounted
against the 10-kg block. If the block is subjected to the force
of F = 500 N, determine its velocity at s = 0.5 m. When
s = 0, the block is at rest and the spring is uncompressed.
The contact surface is smooth.
F " 500 N
5
3
A
P
s
4
F13–6. Block B rests upon a smooth surface. If the
coefficients of static and kinetic friction between A and B
are ms = 0.4 and mk = 0.3, respectively, determine the
acceleration of each block if P = 6 lb.
20 lb
k " 500 N/m
50 lb
B
F13–3
F13–6
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PROBLEMS
13
•13–1. The casting has a mass of 3 Mg. Suspended in a
vertical position and initially at rest, it is given an upward
speed of 200 mm> s in 0.3 s using a crane hook H. Determine
the tension in cables AC and AB during this time interval if
the acceleration is constant.
*13–4. The 2-Mg truck is traveling at 15 m> s when the brakes
on all its wheels are applied, causing it to skid for a distance of
10 m before coming to rest. Determine the constant horizontal
force developed in the coupling C, and the frictional force
developed between the tires of the truck and the road during
this time.The total mass of the boat and trailer is 1 Mg.
H
C
A
30"
Prob. 13–4
•13–5. If blocks A and B of mass 10 kg and 6 kg,
respectively, are placed on the inclined plane and released,
determine the force developed in the link. The coefficients
of kinetic friction between the blocks and the inclined plane
are mA = 0.1 and mB = 0.3. Neglect the mass of the link.
30"
B
C
A
B
30!
Prob. 13–1
13–2. The 160-Mg train travels with a speed of 80 km>h
when it starts to climb the slope. If the engine exerts a
traction force F of 1>20 of the weight of the train and the
rolling resistance FD is equal to 1>500 of the weight of the
train, determine the deceleration of the train.
Prob. 13–5
13–6. Motors A and B draw in the cable with the
accelerations shown. Determine the acceleration of the
300-lb crate C and the tension developed in the cable. Neglect
the mass of all the pulleys.
13–3. The 160-Mg train starts from rest and begins to
climb the slope as shown. If the engine exerts a traction
force F of 1>8 of the weight of the train, determine the
speed of the train when it has traveled up the slope a
distance of 1 km. Neglect rolling resistance.
C
F
10
1
aP¿ " 2 ft/s2
aP " 3 ft/s2
B
P
P¿
Probs. 13–2/3
Prob. 13–6
A
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13.4
13–7. The van is traveling at 20 km> h when the coupling
of the trailer at A fails. If the trailer has a mass of 250 kg and
coasts 45 m before coming to rest, determine the constant
horizontal force F created by rolling friction which causes
the trailer to stop.
123
EQUATIONS OF MOTION: RECTANGULAR COORDINATES
13–10. The crate has a mass of 80 kg and is being towed by
a chain which is always directed at 20° from the horizontal
as shown. If the magnitude of P is increased until the crate
begins to slide, determine the crate’s initial acceleration if
the coefficient of static friction is ms = 0.5 and the
coefficient of kinetic friction is mk = 0.3.
13–11. The crate has a mass of 80 kg and is being towed by
a chain which is always directed at 20° from the horizontal
as shown. Determine the crate’s acceleration in t = 2 s if
the coefficient of static friction is ms = 0.4, the coefficient of
kinetic friction is mk = 0.3, and the towing force is
P = (90t2) N, where t is in seconds.
20 km/h
A
F
p
Prob. 13–7
20"
*13–8. If the 10-lb block A slides down the plane with a
constant velocity when u = 30°, determine the acceleration
of the block when u = 45°.
Probs. 13–10/11
A
B
u
C
Prob. 13–8
•13–9. Each of the three barges has a mass of 30 Mg,
whereas the tugboat has a mass of 12 Mg. As the barges are
being pulled forward with a constant velocity of 4 m> s, the
tugboat must overcome the frictional resistance of the water,
which is 2 kN for each barge and 1.5 kN for the tugboat. If the
cable between A and B breaks, determine the acceleration of
the tugboat.
*13–12. Determine the acceleration of the system and the
tension in each cable. The inclined plane is smooth, and the
coefficient of kinetic friction between the horizontal surface
and block C is (mk)C = 0.2.
E
B
5 kg
A
4 m/s
25 kg
A B
D
30!
2 kN
2 kN
2 kN
1.5 kN
(mk)C " 0.2
Prob. 13–9
Prob. 13–12
10 kg
C
13
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•13–13. The two boxcars A and B have a weight of 20 000 lb
and 30 000 lb, respectively. If they coast freely down the
incline when the brakes are applied to all the wheels of car A
causing it to skid, determine the force in the coupling C
between the two cars. The coefficient of kinetic friction
between the wheels of A and the tracks is mk = 0.5. The
13 wheels of car B are free to roll. Neglect their mass in the
calculation. Suggestion: Solve the problem by representing
single resultant normal forces acting on A and B, respectively.
AND
A C C E L E R AT I O N
*13–16. The man pushes on the 60-lb crate with a force F.
The force is always directed down at 30° from the
horizontal as shown, and its magnitude is increased until the
crate begins to slide. Determine the crate’s initial
acceleration if the coefficient of static friction is ms = 0.6
and the coefficient of kinetic friction is mk = 0.3.
F
B
A
30"
5"
C
Prob. 13–13
13–14. The 3.5-Mg engine is suspended from a spreader
beam AB having a negligible mass and is hoisted by a crane
which gives it an acceleration of 4 m>s2 when it has a
velocity of 2 m> s. Determine the force in chains CA and CB
during the lift.
13–15. The 3.5-Mg engine is suspended from a spreader
beam having a negligible mass and is hoisted by a crane
which exerts a force of 40 kN on the hoisting cable.
Determine the distance the engine is hoisted in 4 s, starting
from rest.
Prob. 13–16
•13–17. A force of F = 15 lb is applied to the cord.
Determine how high the 30-lb block A rises in 2 s starting
from rest. Neglect the weight of the pulleys and cord.
13–18. Determine the constant force F which must be
applied to the cord in order to cause the 30-lb block A to
have a speed of 12 ft/s when it has been displaced 3 ft
upward starting from rest. Neglect the weight of the pulleys
and cord.
C
60"
A
D
60"
B
C
B
E
F
Probs. 13–14/15
A
Probs. 13–17/18
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13.4
13–19. The 800-kg car at B is connected to the 350-kg car
at A by a spring coupling. Determine the stretch in the
spring if (a) the wheels of both cars are free to roll and
(b) the brakes are applied to all four wheels of
car B, causing the wheels to skid. Take (mk)B = 0.4. Neglect
the mass of the wheels.
125
EQUATIONS OF MOTION: RECTANGULAR COORDINATES
•13–21. Block B has a mass m and is released from rest
when it is on top of cart A, which has a mass of 3m.
Determine the tension in cord CD needed to hold the cart
from moving while B slides down A. Neglect friction.
13–22. Block B has a mass m and is released from rest
when it is on top of cart A, which has a mass of 3m. 13
Determine the tension in cord CD needed to hold the cart
from moving while B slides down A. The coefficient of
kinetic friction between A and B is mk.
B
k ! 600 N/m
A
B
D
A
C
u
5
4
3
Probs. 13–21/22
Prob. 13–19
*13–20. The 10-lb block A travels to the right at
vA = 2 ft>s at the instant shown. If the coefficient of kinetic
friction is mk = 0.2 between the surface and A, determine
the velocity of A when it has moved 4 ft. Block B has a
weight of 20 lb.
13–23. The 2-kg shaft CA passes through a smooth journal
bearing at B. Initially, the springs, which are coiled loosely
around the shaft, are unstretched when no force is applied
to the shaft. In this position s = s¿ = 250 mm and the shaft
is at rest. If a horizontal force of F = 5 kN is applied,
determine the speed of the shaft at the instant s = 50 mm,
s¿ = 450 mm. The ends of the springs are attached to the
bearing at B and the caps at C and A.
A
s¿
C
B
Prob. 13–20
s
A
B
kAB ! 2 kN/m
kCB ! 3 kN/m
Prob. 13–23
F ! 5 kN
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*13–24. If the force of the motor M on the cable is shown
in the graph, determine the velocity of the cart when t = 3 s.
The load and cart have a mass of 200 kg and the car starts
from rest.
13
AND
A C C E L E R AT I O N
13–26. A freight elevator, including its load, has a mass of
500 kg. It is prevented from rotating by the track and wheels
mounted along its sides. When t = 2 s, the motor M draws in
the cable with a speed of 6 m> s, measured relative to the
elevator. If it starts from rest, determine the constant
acceleration of the elevator and the tension in the cable.
Neglect the mass of the pulleys, motor, and cables.
F (N)
450
3
t (s)
M
M
F
C
30!
Prob. 13–24
Prob. 13–26
•13–25. If the motor draws in the cable with an
acceleration of 3 m>s2, determine the reactions at the
supports A and B. The beam has a uniform mass of 30 kg> m,
and the crate has a mass of 200 kg. Neglect the mass of the
motor and pulleys.
13–27. Determine the required mass of block A so that
when it is released from rest it moves the 5-kg block B a
distance of 0.75 m up along the smooth inclined plane in
t = 2 s. Neglect the mass of the pulleys and cords.
0.5 m
2.5 m
3m
A
B
E
3 m/s2
C
D
B
C
A
60"
Prob. 13–25
Prob. 13–27
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13.4
*13–28. Blocks A and B have a mass of mA and mB, where
mA > mB. If pulley C is given an acceleration of a 0,
determine the acceleration of the blocks. Neglect the mass
of the pulley.
EQUATIONS OF MOTION: RECTANGULAR COORDINATES
127
13–31. The 75-kg man climbs up the rope with an
acceleration of 0.25 m>s2, measured relative to the rope.
Determine the tension in the rope and the acceleration of
the 80-kg block.
13
a0
C
A
B
B
A
Prob. 13–28
Prob. 13–31
•13–29. The tractor is used to lift the 150-kg load B with
the 24-m-long rope, boom, and pulley system. If the tractor
travels to the right at a constant speed of 4 m> s, determine
the tension in the rope when sA = 5 m. When sA = 0,
sB = 0.
*13–32. Motor M draws in the cable with an acceleration
of 4 ft>s2, measured relative to the 200-lb mine car.
Determine the acceleration of the car and the tension in the
cable. Neglect the mass of the pulleys.
13–30. The tractor is used to lift the 150-kg load B with the
24-m-long rope, boom, and pulley system. If the tractor
travels to the right with an acceleration of 3 m>s2 and has a
velocity of 4 m> s at the instant sA = 5 m, determine the
tension in the rope at this instant. When sA = 0, sB = 0.
aP/c " 4 ft/s2
12 m
P
M
sB
B
A
30!
sA
Probs. 13–29/30
Prob. 13–32
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•13–33. The 2-lb collar C fits loosely on the smooth shaft.
If the spring is unstretched when s = 0 and the collar is
given a velocity of 15 ft> s, determine the velocity of the
collar when s = 1 ft.
13
15 ft/s
AND
A C C E L E R AT I O N
13–35. The 2-kg collar C is free to slide along the smooth
shaft AB. Determine the acceleration of collar C if (a) the
shaft is fixed from moving, (b) collar A, which is fixed to
shaft AB, moves to the left at constant velocity along the
horizontal guide, and (c) collar A is subjected to an
acceleration of 2 m>s2 to the left. In all cases, the motion
occurs in the vertical plane.
s
C
A
45"
1 ft
k ! 4 lb/ft
C
B
Prob. 13–33
Prob. 13–35
13–34. In the cathode-ray tube, electrons having a mass m
are emitted from a source point S and begin to travel
horizontally with an initial velocity v0. While passing
between the grid plates a distance l, they are subjected to a
vertical force having a magnitude eV> w, where e is the
charge of an electron, V the applied voltage acting across
the plates, and w the distance between the plates. After
passing clear of the plates, the electrons then travel in
straight lines and strike the screen at A. Determine the
deflection d of the electrons in terms of the dimensions of
the voltage plate and tube. Neglect gravity which causes a
slight vertical deflection when the electron travels from S to
the screen, and the slight deflection between the plates.
*13–36. Blocks A and B each have a mass m. Determine
the largest horizontal force P which can be applied to B so
that A will not move relative to B. All surfaces are smooth.
•13–37. Blocks A and B each have a mass m. Determine
the largest horizontal force P which can be applied to B so
that A will not slip on B. The coefficient of static friction
between A and B is ms. Neglect any friction between B and C.
A
d
S
e
v0
+
+
–
–
w
l
A
P
u B
C
L
Prob. 13–34
Probs. 13–36/37
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13.4
13–38. If a force F = 200 N is applied to the 30-kg cart,
show that the 20-kg block A will slide on the cart. Also
determine the time for block A to move on the cart 1.5 m.
The coefficients of static and kinetic friction between the
block and the cart are ms = 0.3 and mk = 0.25. Both the cart
and the block start from rest.
EQUATIONS OF MOTION: RECTANGULAR COORDINATES
129
*13–40. The 30-lb crate is being hoisted upward with a
constant acceleration of 6 ft>s2. If the uniform beam AB has
a weight of 200 lb, determine the components of reaction at
the fixed support A. Neglect the size and mass of the pulley
at B. Hint: First find the tension in the cable, then analyze
the forces in the beam using statics.
1.5 m
A
F " 200 N
y
5 ft
Prob. 13–38
B
A
13–39. Suppose it is possible to dig a smooth tunnel
through the earth from a city at A to a city at B as shown. By
the theory of gravitation, any vehicle C of mass m placed
within the tunnel would be subjected to a gravitational
force which is always directed toward the center of the
earth D. This force F has a magnitude that is directly
proportional to its distance r from the earth’s center. Hence,
if the vehicle has a weight of W = mg when it is located on
the earth’s surface, then at an arbitrary location r the
magnitude of force F is F = (mg>R)r, where R = 6328 km,
the radius of the earth. If the vehicle is released from rest
when it is at B, x = s = 2 Mm, determine the time needed
for it to reach A, and the maximum velocity it attains.
Neglect the effect of the earth’s rotation in the calculation
and assume the earth has a constant density. Hint: Write the
equation of motion in the x direction, noting that r cos
u = x. Integrate, using the kinematic relation v dv = a dx,
then integrate the result using v = dx>dt.
x
6 ft/s2
Prob. 13–40
•13–41. If a horizontal force of P = 10 lb is applied to
block A, determine the acceleration of block B. Neglect
friction. Hint: Show that a B = aA tan 15°.
s
x
B
s
C
r
F
D
u
A
15 lb
R
B
P
8 lb
A
Prob. 13–39
15"
Prob. 13–41
13
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13–42. Block A has a mass mA and is attached to a spring
having a stiffness k and unstretched length l0. If another
block B, having a mass mB, is pressed against A so that the
spring deforms a distance d, determine the distance both
blocks slide on the smooth surface before they begin to
separate. What is their velocity at this instant?
13–43. Block A has a mass mA and is attached to a spring
having a stiffness k and unstretched length l0. If another
block B, having a mass mB, is pressed against A so that the
spring deforms a distance d, show that for separation to
occur it is necessary that d 7 2mkg(mA + mB)>k, where mk
is the coefficient of kinetic friction between the blocks and
the ground. Also, what is the distance the blocks slide on the
surface before they separate?
k
A
B
Probs. 13–42/43
AND
A C C E L E R AT I O N
•13–45. The buoyancy force on the 500-kg balloon is
F = 6 kN, and the air resistance is FD = (100v) N, where v is
in m>s. Determine the terminal or maximum velocity of the
balloon if it starts from rest.
FD " (100v)N
F " 6 kN
Prob. 13–45
13–46. The parachutist of mass m is falling with a velocity
of v0 at the instant he opens the parachute. If air resistance
is FD = Cv2, determine her maximum velocity (terminal
velocity) during the descent.
FD " Cv2
*13–44. The 600-kg dragster is traveling with a velocity of
125 m>s when the engine is shut off and the braking
parachute is deployed. If air resistance imposed on the
dragster due to the parachute is FD = (6000 + 0.9v2) N,
where v is in m>s, determine the time required for the
dragster to come to rest.
Prob. 13–44
Prob. 13–46
13–47. The weight of a particle varies with altitude such
that W = m(gr20)>r2, where r0 is the radius of the earth and
r is the distance from the particle to the earth’s center. If the
particle is fired vertically with a velocity v0 from the earth’s
surface, determine its velocity as a function of position r.
What is the smallest velocity v0 required to escape the
earth’s gravitational field, what is rmax, and what is the time
required to reach this altitude?
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Equations of Motion: Normal and
Tangential Coordinates
When a particle moves along a curved path which is known, the equation
of motion for the particle may be written in the tangential, normal, and
binormal directions, Fig. 13–11. Note that there is no motion of the particle
in the binormal direction, since the particle is constrained to move along
the path. We have
13
b
This equation is satisfied provided
#Fbub
O
t
n
#Fnun
(13–8)
Recall that at 1= dv>dt2 represents the time rate of change in the
magnitude of velocity. So if ©Ft acts in the direction of motion, the
particle’s speed will increase, whereas if it acts in the opposite
direction, the particle will slow down. Likewise, an 1= v2>r2 represents
the time rate of change in the velocity’s direction. It is caused by ©Fn ,
which always acts in the positive n direction, i.e., toward the path’s
center of curvature. From this reason it is often referred to as the
centripetal force.
The centrifuge is used to subject a passenger to a very large
normal acceleration caused by rapid rotation. Realize that
this acceleration is caused by the unbalanced normal force
exerted on the passenger by the seat of the centrifuge.
#Ftut
P
Inertial coordinate
system
Fig. 13–11
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Procedure for Analysis
13
When a problem involves the motion of a particle along a known
curved path, normal and tangential coordinates should be
considered for the analysis since the acceleration components can
be readily formulated. The method for applying the equations of
motion, which relate the forces to the acceleration, has been
outlined in the procedure given in Sec. 13.4. Specifically, for t, n, b
coordinates it may be stated as follows:
Free-Body Diagram.
●
●
●
Establish the inertial t, n, b coordinate system at the particle and
draw the particle’s free-body diagram.
The particle’s normal acceleration a n always acts in the positive n
direction.
If the tangential acceleration at is unknown, assume it acts in the
positive t direction.
●
There is no acceleration in the b direction.
●
Identify the unknowns in the problem.
Equations of Motion.
●
Apply the equations of motion, Eqs. 13–8.
Kinematics.
●
●
Formulate the tangential and normal components
acceleration; i.e., at = dv>dt or at = v dv>ds and an = v2>r.
of
If the path is defined as y = f1x2, the radius of curvature at the
point where the particle is located can be obtained from
r = [1 + 1dy>dx22]3>2> ƒ d2y>dx2 ƒ .
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EXAMPLE 13.6
Determine the banking angle u for the race track so that the wheels of
the racing cars shown in Fig. 13–12a will not have to depend upon
friction to prevent any car from sliding up or down the track. Assume
the cars have negligible size, a mass m, and travel around the curve of
radius r with a constant speed v.
13
u
(a)
SOLUTION
Before looking at the following solution, give some thought as to why
it should be solved using t, n, b coordinates.
Free-Body Diagram. As shown in Fig. 13–12b, and as stated in the
problem, no frictional force acts on the car. Here NC represents the
resultant of the ground on all four wheels. Since an can be calculated,
the unknowns are NC and u.
Equations of Motion. Using the n, b axes shown,
:
n
n
v2
NC sin u = m
r
b
an
n
NC
u
W ! mg
(1)
(2)
NC cos u - mg = 0
Eliminating NC and m from these equations by dividing Eq. 1 by Eq. 2,
we obtain
v2
tan u =
gr
u = tan-1 a
v2
b
gr
Ans.
The result is independent of the mass of the car. Also, a force
summation in the tangential direction is of no consequence to the
solution. If it were considered, then at = dv>dt = 0, since the car
moves with constant speed. A further analysis of this problem is
discussed in Prob. 21–47.
NOTE:
(b)
Fig. 13–12
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EXAMPLE 13.7
The 3-kg disk D is attached to the end of a cord as shown in
Fig. 13–13a. The other end of the cord is attached to a ball-and-socket
joint located at the center of a platform. If the platform rotates
rapidly, and the disk is placed on it and released from rest as shown,
determine the time it takes for the disk to reach a speed great enough
to break the cord. The maximum tension the cord can sustain is 100 N,
and the coefficient of kinetic friction between the disk and the
platform is mk = 0.1.
13
Motion of
platform
D
1m
(a)
SOLUTION
Free-Body Diagram. The frictional force has a magnitude
F = mkND = 0.1ND and a sense of direction that opposes the relative
motion of the disk with respect to the platform. It is this force that
gives the disk a tangential component of acceleration causing v to
increase, thereby causing T to increase until it reaches 100 N. The
weight of the disk is W = 319.812 = 29.43 N. Since a n can be related
to v, the unknowns are ND , a t , and v.
b
29.43 N
T
F ! 0.1 ND
t
at
ND
(b)
Fig. 13–13
an
Equations of Motion.
n
T = 3a
v2
b
1
0.1ND = 3at
ND - 29.43 = 0
(1)
(2)
(3)
Setting T = 100 N, Eq. 1 can be solved for the critical speed vcr of the
disk needed to break the cord. Solving all the equations, we obtain
ND = 29.43 N
at = 0.981 m>s2
vcr = 5.77 m>s
Kinematics.
Since at is constant, the time needed to break the cord is
vcr = v0 + att
5.77 = 0 + 10.9812t
t = 5.89 s
Ans.
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EXAMPLE 13.8
Design of the ski jump shown in the photo requires knowing the type
of forces that will be exerted on the skier and her approximate
trajectory. If in this case the jump can be approximated by the parabola
shown in Fig. 13–14a, determine the normal force on the 150-lb skier
the instant she arrives at the end of the jump, point A, where her
velocity is 65 ft>s. Also, what is her acceleration at this point?
13
y
SOLUTION
Why consider using n, t coordinates to solve this problem?
y!
Free-Body Diagram. Since dy>dx = x>100 |x = 0 = 0, the slope at A
is horizontal. The free-body diagram of the skier when she is at A is
shown in Fig. 13–14b. Since the path is curved, there are two
components of acceleration, a n and a t . Since an can be calculated, the
unknowns are at and NA .
Equations of Motion.
+ c ©Fn = man ;
;
t
t
0 =
[1 + 1dy>dx22]3>2
2
2
ƒ d y>dx ƒ
`
200 ft
A
(a)
n
150
a
32.2 t
(1)
x=0
=
[1 + 1022]3>2
1
ƒ 100
ƒ
an
150 lb
(2)
The radius of curvature r for the path must be determined at point
1
1
1
x2 - 200, dy>dx = 100
x, d2y>dx2 = 100
,
A(0, -200 ft). Here y = 200
so that at x = 0,
r =
x
2
150 1652
NA - 150 =
a
b
r
32.2
t
= 100 ft
Fig. 13–14
Ans.
Kinematics. From Eq. 2,
at = 0
Thus,
an =
16522
v2
=
= 42.2 ft>s2
r
100
aA = an = 42.2 ft>s2 c
at
(b)
Substituting this into Eq. 1 and solving for NA , we obtain
NA = 347 lb
1 x2 \$ 200
200
Ans.
Apply the equation of motion in the y direction and show
that when the skier is in midair her acceleration is 32.2 ft>s2.
NOTE:
NA
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EXAMPLE 13.9
The 60-kg skateboarder in Fig.13–15a coasts down the circular track.
If he starts from rest when u = 0°, determine the magnitude of the
normal reaction the track exerts on him when u = 60°. Neglect his
size for the calculation.
O
u
4m
13
SOLUTION
(a)
Free-Body Diagram. The free-body diagram of the skateboarder
when he is at an arbitrary position u is shown in Fig. 13–15b. At
u = 60° there are three unknowns, Ns , at , and an (or v).
Equations of Motion.
60 (9.81) N
n
u
T ©Fn = man; NS - [60(9.81)N] sin u = (60 kg)a
an
[60(9.81)N] cos u = (60 kg) at
at
v2
b
4m
(1)
at = 9.81 cos u
Ns
t
(b)
O
4m
u
du
Kinematics. Since at is expressed in terms of u, the equation
v dv = at ds must be used to determine the speed of the
skateboarder when u = 60°. Using the geometric relation s = ur,
where ds = r du = (4 m) du, Fig. 13–15c, and the initial condition
v = 0 at u = 0°, we have,
v dv = at ds
ds ! 4du
(c)
Fig. 13–15
L0
v
v dv =
L0
60°
9.81 cos u(4 du)
60°
v2 v
` = 39.24 sin u `
2 0
0
v2
- 0 = 39.24(sin 60° - 0)
2
v2 = 67.97 m2/s2
Substituting this result and u = 60° into Eq. (1), yields
Ns = 1529.23 N = 1.53 kN
Ans.
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FUNDAMENTAL PROBLEMS
F13–7. The block rests at a distance of 2 m from the center
of the platform. If the coefficient of static friction between
the block and the platform is ms = 0.3, determine the
maximum speed which the block can attain before it begins
to slip. Assume the angular motion of the disk is slowly
z
increasing.
F13–10. The sports car is traveling along a 30° banked road
having a radius of curvature of r = 500 ft. If the coefficient
of static friction between the tires and the road is ms = 0.2, 13
determine the maximum safe speed so no slipping occurs.
Neglect the size of the car.
r ! 500 ft
u ! 30"
2m
F13–7
F13–8. Determine the maximum speed that the jeep can
travel over the crest of the hill and not lose contact with
r ! 250 ft
F13–8
F13–9. A pilot weighs 150 lb and is traveling at a constant
speed of 120 ft>s. Determine the normal force he exerts on
the seat of the plane when he is upside down at A. The loop
has a radius of curvature of 400 ft.
A
F13–10
F13–11. If the 10-kg ball has a velocity of 3 m>s when it is
at the position A, along the vertical path, determine the
tension in the cord and the increase in the speed of the ball
at this position.
2m
O
u " 45!
A
F13–11
F13–12. The motorcycle has a mass of 0.5 Mg and a
negligible size. It passes point A traveling with a speed of
15 m>s, which is increasing at a constant rate of 1.5 m>s2 .
Determine the resultant frictional forcce exerted by the
road on the tires at this instant.
400 ft
rA " 200 m
A
F13–9
3 m/s
F13–12
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PROBLEMS
*13–48. The 2-kg block B and 15-kg cylinder A are
connected to a light cord that passes through a hole in the
center of the smooth table. If the block is given a speed of
13
v = 10 m>s, determine the radius r of the circular path
along which it travels.
•13–49. The 2-kg block B and 15-kg cylinder A are
connected to a light cord that passes through a hole in the
center of the smooth table. If the block travels along a
circular path of radius r = 1.5 m, determine the speed of
the block.
r
*13–52. Determine the mass of the sun, knowing that the
distance from the earth to the sun is 149.6(106) km. Hint: Use
Eq. 13–1 to represent the force of gravity acting on the earth.
•13–53. The sports car, having a mass of 1700 kg, travels
horizontally along a 20° banked track which is circular and
has a radius of curvature of r = 100 m. If the coefficient of
static friction between the tires and the road is ms = 0.2 ,
determine the maximum constant speed at which the car can
travel without sliding up the slope. Neglect the size of the car.
13–54. Using the data in Prob. 13–53, determine the
minimum speed at which the car can travel around the track
without sliding down the slope.
u ! 20"
B
v
A
Probs. 13–48/49
13–50. At the instant shown, the 50-kg projectile travels in
the vertical plane with a speed of v = 40 m>s. Determine
the tangential component of its acceleration and the radius
of curvature r of its trajectory at this instant.
Probs. 13–53/54
13–55. The device shown is used to produce the
experience of weightlessness in a passenger when he
reaches point A, u = 90°, along the path. If the passenger
has a mass of 75 kg, determine the minimum speed he
should have when he reaches A so that he does not exert a
normal reaction on the seat. The chair is pin-connected to
the frame BC so that he is always seated in an upright
position. During the motion his speed remains constant.
*13–56. A man having the mass of 75 kg sits in the chair
which is pin-connected to the frame BC. If the man is
always seated in an upright position, determine the
horizontal and vertical reactions of the chair on the man at
the instant u = 45°. At this instant he has a speed of 6 m> s,
which is increasing at 0.5 m>s2.
A
13–51. At the instant shown, the radius of curvature of the
vertical trajectory of the 50-kg projectile is r = 200 m.
Determine the speed of the projectile at this instant.
B
30"
10 m
r
u
C
Probs. 13–50/51
Probs. 13–55/56
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EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
•13–57. Determine the tension in wire CD just after wire
AB is cut. The small bob has a mass m.
A
139
13–59. An acrobat has a weight of 150 lb and is sitting on a
chair which is perched on top of a pole as shown. If by a
mechanical drive the pole rotates downward at a constant
rate from u = 0°, such that the acrobat’s center of mass G
maintains a constant speed of va = 10 ft>s, determine the
angle u at which he begins to “fly” out of the chair. Neglect
friction and assume that the distance from the pivot O to G 13
is r = 15 ft.
D
u
G
u
va
B
C
u
O
Prob. 13–57
Prob. 13–59
13–58. Determine the time for the satellite to complete its
orbit around the earth. The orbit has a radius r measured
from the center of the earth. The masses of the satellite and
the earth are ms and Me, respectively.
*13–60. A spring, having an unstretched length of 2 ft, has
one end attached to the 10-lb ball. Determine the angle u of
the spring if the ball has a speed of 6 ft> s tangent to the
horizontal circular path.
6 in.
A
r
Prob. 13–58
u
Prob. 13–60
k " 20 lb/ft
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•13–61. If the ball has a mass of 30 kg and a speed
v = 4 m>s at the instant it is at its lowest point, u = 0°,
determine the tension in the cord at this instant. Also,
determine the angle u to which the ball swings and
momentarily stops. Neglect the size of the ball.
13 13–62. The ball has a mass of 30 kg and a speed
v = 4 m>s at the instant it is at its lowest point, u = 0°.
Determine the tension in the cord and the rate at which the
ball’s speed is decreasing at the instant u = 20°. Neglect
the size of the ball.
AND
A C C E L E R AT I O N
*13–64. The ball has a mass m and is attached to the cord
of length l. The cord is tied at the top to a swivel and the ball
is given a velocity v0. Show that the angle u which the cord
makes with the vertical as the ball travels around the
circular path must satisfy the equation tan u sin u = v20>gl.
Neglect air resistance and the size of the ball.
O
u
l
u
4m
v0
Probs. 13–61/62
Prob. 13–64
13–63. The vehicle is designed to combine the feel of a
motorcycle with the comfort and safety of an automobile. If
the vehicle is traveling at a constant speed of 80 km> h along
angle u of the vehicle so that only a normal force from the
seat acts on the driver. Neglect the size of the driver.
•13–65. The smooth block B, having a mass of 0.2 kg, is
attached to the vertex A of the right circular cone using a
light cord. If the block has a speed of 0.5 m> s around the cone,
determine the tension in the cord and the reaction which
the cone exerts on the block. Neglect the size of the block.
z
u
A
200 mm
B
400 mm
300 mm
Prob. 13–63
Prob. 13–65
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13–66. Determine the minimum coefficient of static
friction between the tires and the road surface so that the
1.5-Mg car does not slide as it travels at 80 km> h on the
curved road. Neglect the size of the car.
13–67. If the coefficient of static friction between the tires
and the road surface is ms = 0.25, determine the maximum
speed of the 1.5-Mg car without causing it to slide when it
travels on the curve. Neglect the size of the car.
r " 200 m
13–70. A 5-Mg airplane is flying at a constant speed of
350 km> h along a horizontal circular path of radius
r = 3000 m. Determine the uplift force L acting on the
airplane and the banking angle u. Neglect the size of the
airplane.
13–71. A 5-Mg airplane is flying at a constant speed of 13
350 km> h along a horizontal circular path. If the banking
angle u = 15°, determine the uplift force L acting on the
airplane and the radius r of the circular path. Neglect the
size of the airplane.
L
u
Probs. 13–66/67
*13–68. At the instant shown, the 3000-lb car is traveling
with a speed of 75 ft> s, which is increasing at a rate of 6 ft>s2 .
Determine the magnitude of the resultant frictional force
the road exerts on the tires of the car. Neglect the size of the
car.
r " 600 ft
r
Probs. 13–70/71
*13–72. The 0.8-Mg car travels over the hill having the
shape of a parabola. If the driver maintains a constant speed
of 9 m> s, determine both the resultant normal force and the
resultant frictional force that all the wheels of the car exert
on the road at the instant it reaches point A. Neglect the
size of the car.
Prob. 13–68
•13–69. Determine the maximum speed at which the car
with mass m can pass over the top point A of the vertical
car maintains this speed, what is the normal reaction the
road exerts on the car when it passes the lowest point B on
•13–73. The 0.8-Mg car travels over the hill having the
shape of a parabola. When the car is at point A, it is traveling
at 9 m> s and increasing its speed at 3 m>s2 . Determine both
the resultant normal force and the resultant frictional force
that all the wheels of the car exert on the road at this instant.
Neglect the size of the car.
y
r
A
r
y ! 20 (1 \$
B
r
r
x2 )
6400
A
80 m
Prob. 13–69
Probs. 13–72/73
x
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13–74. The 6-kg block is confined to move along the
smooth parabolic path. The attached spring restricts the
motion and, due to the roller guide, always remains
horizontal as the block descends. If the spring has a stiffness
of k = 10 N>m, and unstretched length of 0.5 m, determine
the normal force of the path on the block at the instant
13 x = 1 m when the block has a speed of 4 m> s. Also, what is
the rate of increase in speed of the block at this point?
Neglect the mass of the roller and the spring.
AND
A C C E L E R AT I O N
*13–76. A toboggan and rider of total mass 90 kg travel
down along the (smooth) slope defined by the equation
y = 0.08x2. At the instant x = 10 m, the toboggan’s speed
is 5 m> s. At this point, determine the rate of increase in
speed and the normal force which the slope exerts on the
toboggan. Neglect the size of the toboggan and rider for the
calculation.
y
y
v ! 5 m/s
y ! 2 \$ 0.5 x2
y ! 0.08x2
B
k ! 10 N/m
A
x
10 m
x
Prob. 13–74
Prob. 13–76
13–75. Prove that if the block is released from rest at point
B of a smooth path of arbitrary shape, the speed it attains
when it reaches point A is equal to the speed it attains when
it falls freely through a distance h; i.e., v = 22gh.
•13–77. The skier starts from rest at A(10 m, 0) and
descends the smooth slope, which may be approximated by
a parabola. If she has a mass of 52 kg, determine the normal
force the ground exerts on the skier at the instant she
arrives at point B. Neglect the size of the skier. Hint: Use
the result of Prob. 13–75.
y
B
1
10 m
y ! ––
x2 \$ 5
20
h
A
5m
B
Prob. 13–75
Prob. 13–77
A
x
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13–78. The 5-lb box is projected with a speed of 20 ft> s at
A up the vertical circular smooth track. Determine the
angle u when the box leaves the track.
13–79. Determine the minimum speed that must be given
to the 5-lb box at A in order for it to remain in contact with
the circular path. Also, determine the speed of the box when
it reaches point B.
13–82. Determine the maximum speed the 1.5-Mg car can
have and still remain in contact with the road when it passes
point A. If the car maintains this speed, what is the normal
reaction of the road on it when it passes point B? Neglect
the size of the car.
13
B 30!
v
143
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
y
A
u 4 ft
B
y " 25 # 1 x2
200
x
A
25 m
Probs. 13–78/79
Prob. 13–82
*13–80. The 800-kg motorbike travels with a constant
speed of 80 km> h up the hill. Determine the normal force
the surface exerts on its wheels when it reaches point A.
Neglect its size.
y
A
y2 " 2x
x
13–83. The 5-lb collar slides on the smooth rod, so that
when it is at A it has a speed of 10 ft> s. If the spring to which
it is attached has an unstretched length of 3 ft and a stiffness
of k = 10 lb>ft, determine the normal force on the collar
and the acceleration of the collar at this instant.
100 m
Prob. 13–80
•13–81. The 1.8-Mg car travels up the incline at a constant
speed of 80 km> h. Determine the normal reaction of the
road on the car when it reaches point A. Neglect its size.
y
y
10 ft/s
y"
x
20e 100
A
1 x2
y ! 8 \$ ––
2
A
O
x
x
2 ft
50 m
Prob. 13–81
Prob. 13–83
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Coordinates
#Fuuu
P
#Frur
When all the forces acting on a particle are resolved into cylindrical
components, i.e., along the unit-vector directions ur , uu , uz , Fig. 13–16, the
equation of motion can be expressed as
z
O
A C C E L E R AT I O N
13.6 Equations of Motion: Cylindrical
#Fzuz
13
AND
u
r
To satisfy this equation, we require
Inertial coordinate system
Fig. 13–16
(13–9)
If the particle is constrained to move only in the r–u plane, then only the
first two of Eqs. 13–9 are used to specify the motion.
Tangential and Normal Forces. The most straightforward type
of problem involving cylindrical coordinates requires the determination
to move with a known acceleration. If, however, the particle’s accelerated
motion is not completely specified at the given instant, then some
information regarding the directions or magnitudes of the forces acting
on the particle must be known or computed in order to solve Eqs. 13–9.
For example, the force P causes the particle in Fig. 13–17a to move along
a path r = f1u2. The normal force N which the path exerts on the particle
is always perpendicular to the tangent of the path, whereas the frictional
force F always acts along the tangent in the opposite direction of motion.
The directions of N and F can be specified relative to the radial
coordinate by using the angle c (psi), Fig. 13–17b, which is defined
between the extended radial line and the tangent to the curve.
r ! f (u)
r ! f (u)
Tangent
Tangent
As the car of weight W descends
the spiral track, the resultant
normal force which the track
exerts on the car can be
represented by its three cylindrical
components. - Nr creates a radial
acceleration, - ar , Nu creates a
transverse acceleration au , and the
difference W - Nz creates an
azimuthal acceleration -az .
c
N
r
O
r
F
u P
O
(a)
Fig. 13–17
u
(b)
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145
EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
This angle can be obtained by noting that when the particle is
displaced a distance ds along the path, Fig. 13–17c, the component of
displacement in the radial direction is dr and the component of
displacement in the transverse direction is r du. Since these two
components are mutually perpendicular, the angle c can be determined
from tan c = r du>dr, or
r
tan c =
(13–10)
dr>du
If c is calculated as a positive quantity, it is measured from the extended
radial line to the tangent in a counterclockwise sense or in the positive
direction of u. If it is negative, it is measured in the opposite direction
to positive u. For example, consider the cardioid r = a11 + cos u2,
shown in Fig. 13–18. Because dr>du = - a sin u, then when u = 30°,
tan c = a11 + cos 30°2>1 -a sin 30°2 = - 3.732, or c = - 75°, measured
clockwise, opposite to +u as shown in the figure.
r ! f (u)
Tangent
dr
r du
du
13
ds
rc
u
O
(c)
Fig. 13–17
Procedure for Analysis
Cylindrical or polar coordinates are a suitable choice for the
analysis of a problem for which data regarding the angular motion
of the radial line r are given, or in cases where the path can be
conveniently expressed in terms of these coordinates. Once these
coordinates have been established, the equations of motion can
then be applied in order to relate the forces acting on the particle to
its acceleration components. The method for doing this has been
outlined in the procedure for analysis given in Sec. 13.4. The
following is a summary of this procedure.
Free-Body Diagram.
● Establish the r, u, z inertial coordinate system and draw the
particle’s free-body diagram.
● Assume that a , a , a act in the positive directions of r, u, z if they
r
u
z
are unknown.
● Identify all the unknowns in the problem.
Equations of Motion.
● Apply the equations of motion, Eqs. 13–9.
Kinematics.
● Use the methods of Sec. 12.8 to determine r and the time
# \$ # \$ \$
derivatives r, r, u, u, #z, and then
evaluate
the acceleration
\$
\$
##
\$
components ar = r - ru2, au = ru + 2ru, az = z.
● If any of the acceleration components is computed as a negative
quantity, it indicates that it acts in its negative coordinate direction.
● When taking the time derivatives of r = f1u2, it is very important
to use the chain rule of calculus, which is discussed at the end of
Appendix C.
c
Tangent
u
r
c ! 75"
r
O
u ! 30"
2a
Fig. 13–18
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EXAMPLE 13.10
The smooth 0.5-kg double-collar in Fig. 13–19a can freely slide on arm
AB and the circular
# guide rod. If the arm rotates with a constant
angular velocity of u = 3 rad>s, determine the force the arm exerts on
the collar at the instant u = 45°. Motion is in the horizontal plane.
B
C
13
r " (0.8 cos u)m
SOLUTION
u
A
Free-Body Diagram. The normal reaction NC of the circular guide
rod and the force F of arm AB act on the collar in the plane of motion,
Fig. 13–19b. Note that F acts perpendicular to the axis of arm AB, that
is, in the direction of the u axis, while NC acts perpendicular to the
tangent of the circular path at u = 45°. The four unknowns are
NC, F, ar, au.
0.4 m
(a)
Equations of Motion.
u
NC
au
45!
C
r
ar
- NC cos 45° = (0.5 kg) ar
(1)
F - NC sin 45° = (0.5 kg) au
(2)
tangent
F
Kinematics. Using the chain rule (see Appendix
C), the
#
\$ first and
second time derivatives of r when u = 45°, u = 3 rad>s, u = 0, are
r = 0.8 cos u = 0.8 cos45° = 0.5657 m
#
#
r = - 0.8 sin u u = - 0.8 sin 45°(3) = -1.6971 m>s
\$
#
\$
r = - 0.8 C sin u u + cos u u2 D
(b)
Fig. 13–19
= - 0.8[sin 45°(0)+cos 45°(32)] = -5.091 m>s2
We have
#
\$
ar = r - ru2 = - 5.091 m>s2 - (0.5657 m)(3 rad>s)2 = - 10.18 m>s2
\$
##
au = ru + 2ru = (0.5657 m)(0) + 2(- 1.6971 m>s)(3 rad>s)
= - 10.18 m>s2
Substituting these results into Eqs. (1) and (2) and solving, we get
NC = 7.20 N
F = 0
Ans.
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EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
EXAMPLE 13.11
The smooth 2-kg cylinder C in Fig. 13–20a has a pin P through its
center which passes through the slot in arm OA.
# If the arm is forced to
rotate in the vertical plane at a constant rate u = 0.5 rad>s, determine
the force that the arm exerts on the peg at the instant u = 60°.
13
SOLUTION
Why is it a good idea to use polar coordinates to solve this problem?
Free-Body Diagram. The free-body diagram for the cylinder is
shown in Fig. 13–20b. The force on the peg, FP , acts perpendicular to
the slot in the arm. As usual, a r and a u are assumed to act in the
directions of positive r and u, respectively. Identify the four unknowns.
Equations of Motion. Using the data in Fig. 13–20b, we have
19.62 sin u - NC sin u = 2ar
(1)
19.62 cos u + FP - NC cos u = 2au
·
0.4 m
r
C
A
(a)
0.4
= 0.4 csc u
sin u
Since d1csc u2 = - 1csc u cot u2 du and d1cot u2 = - 1csc2 u2 du, then
r and the necessary time derivatives become
#
u = 0.5
r = 0.4 csc u
\$
#
#
u = 0
r = -0.41csc u cot u2u
= -0.2 csc u cot u
#
#
\$
r = -0.21-csc u cot u21u2 cot u - 0.2 csc u1-csc2 u2u
2
P
(2)
Kinematics. From Fig. 13–20a, r can be related to u by the equation
r =
O
u
2
= 0.1 csc u1cot u + csc u2
19.62 N
FP
u
u
ar
NC
r
Evaluating these formulas at u = 60°, we get
#
u = 0.5
r = 0.462
\$
#
u = 0
r = -0.133
\$
r = 0.192
#
\$
ar = r - ru2 = 0.192 - 0.46210.522 = 0.0770
\$
##
au = ru + 2ru = 0 + 21-0.133210.52 = - 0.133
Substituting these results into Eqs. 1 and 2 with u = 60° and
solving yields
NC = 19.5 N
FP = - 0.356 N
Ans.
shown in Fig. 13–20b.
(b)
Fig. 13–20
au
u
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EXAMPLE 13.12
r ! 0.1 u
13
A
u
r
C
O
·
Top View
(a)
FC
ar
r
f
NC
f
au
Tangent
A can C, having a mass of 0.5 kg, moves along a grooved horizontal
slot shown in Fig. 13–21a. The slot is in the form of a spiral, which is
defined by the equation r = 10.1u2 m, where u is in radians. If the arm
#
OA rotates with a constant rate u = 4 rad>s in the horizontal plane,
determine the force it exerts on the can at the instant u = p rad.
Neglect friction and the size of the can.
SOLUTION
Free-Body Diagram. The driving force FC acts perpendicular to the
arm OA, whereas the normal force of the wall of the slot on the can,
NC , acts perpendicular to the tangent to the curve at u = p rad,
Fig. 13–21b. As usual, ar and au are assumed to act in the positive
directions of r and u, respectively. Since the path is specified, the angle
c which the extended radial line r makes with the tangent, Fig. 13–21c,
can be determined from Eq. 13–10. We have r = 0.1u, so that
dr>du = 0.1, and therefore
r
0.1u
tan c =
=
= u
dr>du
0.1
When u = p, c = tan-1p = 72.3°, so that f = 90° - c = 17.7°, as
shown in Fig. 13–21c. Identify the four unknowns in Fig. 13–21b.
Equations of Motion. Using f = 17.7° and the data shown in
Fig. 13–21b, we have
(1)
;
NC cos 17.7° = 0.5ar
r
r
(2)
FC - NC sin 17.7° = 0.5au
+ T ©Fu = mau ;
u
(b)
Kinematics.
#
\$
u = 0
r ! 0.1 u
r
r = 0.1u
#
#
r = 0.1u = 0.1142 = 0.4 m>s
\$
\$
r = 0.1u = 0
At the instant u = p rad,
#
\$
ar = r - ru2 = 0 - 0.11p21422 = - 5.03 m>s2
u!p
\$
##
au = ru + 2ru = 0 + 210.42142 = 3.20 m>s2
Substituting these results into Eqs. 1 and 2 and solving yields
c
NC = - 2.64 N
FC = 0.800 N
f
Tangent
The time derivatives of r and u are
u
(c)
What does the negative sign for NC indicate?
Fig. 13–21
Ans.
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EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
FUNDAMENTAL PROBLEMS
#
F13–13. Determine the constant angular velocity u of the
vertical shaft of the amusement ride if f = 45°. Neglect the
mass of the cables and the size of the passengers.
F13–15. The 2-Mg car is traveling along the curved road
described by r = (50e2u) m, where u is in radians. If a camera
is# located at A and it rotates with an angular
velocity of 13
\$
u = 0.05 rad>s and an angular acceleration of u = 0.01 rad>s2
at the instant u = p6 rad, determine the resultant friction force
developed between the tires and the road at this instant.
1.5 m
r " (50e2u)m
u
8m
f
r
u
A u, u
F13–13
F13–15
F13–14. The 0.2-kg ball is blown through the smooth
vertical circular tube whose shape is defined by
r = (0.6 sin u) m, where u is in radians. If u = (p t2) rad,
where t is in seconds, determine the magnitude of force F
exerted by the blower on the ball when t = 0.5 s.
F13–16. The 0.2-kg pin P is constrained to move in the
smooth curved slot, which is defined by the lemniscate
r = (0.6 cos 2u) m. Its motion is controlled by the rotation
of the slotted arm # OA, which has a constant clockwise
angular velocity of u = -3 rad>s. Determine the force arm
OA exerts on the pin P when u = 0°. Motion is in the
vertical plane.
F
r " (0.6 cos 2u) m
P
0.3 m
r
u
F13–14
u
O
u
F13–16
A
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PROBLEMS
*13–84. The path of motion of a 5-lb particle in the
horizontal plane is described in terms of polar coordinates
as r = (2t + 1) ft and u = (0.5t2 - t) rad, where t is in
13
seconds. Determine the magnitude of the resultant force
acting on the particle when t = 2 s.
•13–85. Determine the magnitude of the resultant force
acting on a 5-kg particle at the instant t = 2 s, if the particle
is moving along a horizontal path defined by the equations
r = (2t + 10) m and u = (1.5t2 - 6t) rad, where t is in
seconds.
13–86. A 2-kg particle travels along a horizontal smooth
path defined by
•13–89. The 0.5-kg collar C can slide freely along the smooth
rod AB. At a# given instant, rod AB is rotating with an angular
velocity
of u = 2 rad>s and has an angular acceleration of
\$
u = 2 rad>s2. Determine the normal force of rod AB and the
radial reaction of the end plate B on the collar at this instant.
Neglect the mass of the rod and the size of the collar.
A
0.6 m
u, u
B
C
1
t2
r = ¢ t3 + 2 ≤ m, u = ¢ ≤ rad,
4
4
where t is in seconds. Determine the radial and transverse
components of force exerted on the particle when t = 2 s.
13–87.
A 2-kg particle travels along a path defined by
1
r = (3 + 2t2) m, u = ¢ t3 + 2 ≤ rad
3
and z = (5 - 2t2) m, where t is in seconds. Determine the r,
u, z components of force that the path exerts on the particle
at the instant t = 1 s.
*13–88. If the coefficient of static friction between the
block of mass m and the turntable is m s, determine the
maximum constant angular velocity of the platform without
causing the block to slip.
Prob. 13–89
13–90. The 2-kg rod AB moves up and down as its end slides
on the smooth contoured surface of the cam, where r = 0.1 m
and z = (0.02 sin u) m. If the cam is rotating with a constant
angular velocity of 5 rad> s, determine the force on the roller A
when u = 90°. Neglect friction at the bearing C and the mass
of the roller.
13–91. The 2-kg rod AB moves up and down as its end
slides on the smooth contoured surface of the cam, where
r = 0.1 m and z = (0.02 sin u) m. If the cam is rotating at a
constant angular velocity of 5 rad> s, determine the maximum
and minimum force the cam exerts on the roller at A. Neglect
friction at the bearing C and the mass of the roller.
B
C
z
r
u
z ! 0.02 sin u
A
0.1 m
Prob. 13–88
Probs. 13–90/91
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13.6
*13–92. If the coefficient of static friction between the
conical surface and the block of mass m is m# s = 0.2,
determine the minimum constant angular velocity u so that
the block does not slide downwards.
•13–93. If the coefficient of static friction between the conical
surface and the block is m# s = 0.2, determine the maximum
constant angular velocity u without causing the block to slide
upwards.
13–95. The mechanism is rotating about
the vertical axis
#
with a constant angular velocity of u = 6 rad>s. If rod AB
is smooth, determine the constant position r of the 3-kg
collar C. The spring has an unstretched length of 400 mm.
Neglect the mass of the rod and the size of the collar.
13
u
300 mm
r
B
A
A
45!
151
EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
C
300 mm
45!
k " 200 N/m
u
Probs. 13–92/93
Prob. 13–95
13–94. If the position of the 3-kg collar C on the smooth rod
AB is held
# at r = 720 mm, determine the constant angular
velocity u at which the mechanism is rotating about the
vertical axis. The spring has an unstretched length of 400 mm.
Neglect the mass of the rod and the size of the collar.
*13–96. Due to the constraint, the 0.5-kg cylinder C travels
along the path described by r = (0.6 cos u) m. If arm OA
rotates
counterclockwise with an angular
velocity of
\$
#
u = 2 rad>s and an angular acceleration of u = 0.8 rad>s2 at
the instant u = 30°, determine the force exerted by the arm
on the cylinder at this instant. The cylinder is in contact with
only one edge of the smooth slot, and the motion occurs in
the horizontal plane.
u
r
B
A
300 mm
C
k ! 200 N/m
u
O
Prob. 13–94
C
r ! 0.6 cos u
u, u
0.3 m
Prob. 13–96
A
0.3 m
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•13–97. The 0.75-lb smooth can is guided along the
circular path using
the arm guide. If the arm has an
#
angular
velocity
u
=
2 rad>s and an angular acceleration
\$
u = 0.4 rad>s2 at the instant u = 30°, determine the force of
the guide on the can. Motion occurs in the horizontal plane.
13
13–98.
plane.
AND
A C C E L E R AT I O N
13–102. The amusement
park ride rotates with a constant
#
angular velocity of u = 0.8 rad>s. If the path of the ride is
defined by r = (3 sin u + 5) m and z = (3 cos u) m,
determine the r, u, and z components of force exerted by
the seat on the 20-kg boy when u = 120°.
Solve Prob. 13–97 if motion occurs in the vertical
r
u
0.5 ft
z
u
0.5 ft
r
Probs. 13–97/98
13–99. The forked rod is used to move the smooth
2-lb particle around the horizontal path in the
# shape of a
limaçon, r = (2 + cos u) ft. If at all times u = 0.5 rad>s,
determine the force which the rod exerts on the particle at
the instant u = 90°. The fork and path contact the particle
on only one side.
*13–100.
Solve Prob. 13–99 at the instant u = 60°.
•13–101. The forked rod is used to move the smooth
2-lb particle around the horizontal path in the shape of a
limaçon, r = (2 + cos u) ft. If u = (0.5t2) rad, where t is in
seconds, determine the force which the rod exerts on the
particle at the instant t = 1 s. The fork and path contact the
particle on only one side.
2 ft
Prob. 13–102
13–103. The airplane executes the vertical loop defined by
r2 = [810(103)cos 2u] m2. If the pilot maintains a constant
speed v = 120 m>s along the path, determine the normal
force the seat exerts on him at the instant u = 0°. The pilot
has a mass of 75 kg.
r2 ! [810(103) cos 2 u]m2
r
r
u
·
u
u
3 ft
Probs. 13–99/100/101
Prob. 13–103
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13.6
*13–104. A boy standing firmly spins the girl sitting on a
circular “dish” or sled in a circular
path of radius r0 = 3 m
#
such that her angular velocity is u0 = 0.1 rad>s. If the attached
cable OC is drawn inward such that the radial coordinate r
#
changes with a constant speed of r = - 0.5 m>s, determine
the tension it exerts on the sled at the instant r = 2 m .The sled
and girl have a total mass of 50 kg. Neglect the size of the girl
and sled and the effects of friction between the sled and ice.
Hint: First show that\$ the equation
of motion
in the u
#
##
2
direction yields
a
=
ru
+
2r
u
=
(1>r)
d>dt(r
u
)
=
0. When
u
#
integrated, r2u = C, where the constant C is determined from
the problem data.
O
r!2m
1m
153
EQUATIONS OF MOTION: CYLINDRICAL COORDINATES
13–107. The 1.5-kg cylinder C travels along the path
described by r = (0.6 sin u) m. If arm OA rotates
counterclockwise
with a constant angular velocity of
#
u = 3 rad>s, determine the force exerted by the smooth slot
in arm OA on the cylinder at the instant u = 60° . The spring
has a stiffness of 100 N> m and is unstretched when u = 30° .
The cylinder is in contact with only one edge of the slotted 13
arm. Neglect the size of the cylinder. Motion occurs in the
horizontal plane.
*13–108. The 1.5-kg cylinder C travels along the path
described by r = (0.6 sin u) m. If arm OA # is rotating
counterclockwise with an angular velocity of u = 3 rad>s ,
determine the force exerted by the smooth slot in arm OA on
the cylinder at the instant u = 60°. The spring has a stiffness
of 100 N> m and is unstretched when u = 30° . The cylinder is
in contact with only one edge of the slotted arm. Neglect the
size of the cylinder. Motion occurs in the vertical plane.
u
A
C
r!3m
C
Prob. 13–104
13–105. The smooth particle has a mass of 80 g. It is
attached to an elastic cord extending from O to P and due to
the slotted arm guide moves along the horizontal circular
path r = 10.8 sin u2 m. If the cord has a stiffness
k = 30 N>m and an unstretched length of 0.25 m, determine
the force of the guide on the particle
# when u = 60°. The
guide has a constant angular velocity u = 5 rad>s.
\$
#
13–106. Solve Prob. 13–105 if u = 2 rad>s2 when u = 5 rad>s
and u = 60°.
P
r
r " 0.6 sin u
u
u, u
O
Probs. 13–107/108
•13–109. Using air pressure, the 0.5-kg ball is forced to
move through the tube lying in the horizontal plane and
having the shape of a logarithmic spiral. If the tangential
force exerted on the ball due to air pressure is 6 N,
determine the rate of increase in the ball’s speed at the
instant u = p>2. Also, what is the angle c from the extended
radial coordinate r to the line of action of the 6-N force?
·
0.4 m
ru
u
r ! 0.2e0.1u
O
Probs. 13–105/106
Prob. 13–109
F!6N
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13–110. The tube
# rotates in the horizontal plane at a
constant rate of u = 4 rad>s. If a 0.2-kg ball B starts at the
#
origin O with an initial radial velocity of r = 1.5 m>s and
moves outward through the tube, determine the radial and
transverse components of the ball’s velocity at the instant it
leaves the outer end at C, r = 0.5 m. Hint: Show that
13 the equation of motion in the r direction is r\$ - 16r = 0.
The solution is of the form r = Ae -4t + Be4t. Evaluate the
integration constants A and B, and determine the time t
when r = 0.5 m. Proceed to obtain vr and vu .
AND
A C C E L E R AT I O N
*13–112. The 0.5-lb ball is guided along the vertical circular
path r = 2rc cos u# using the arm OA. If the arm has an
angular
velocity u = 0.4 rad>s and an angular acceleration
\$
u = 0.8 rad>s2 at the instant u = 30°, determine the force of
the arm on the ball. Neglect friction and the size of the ball.
Set rc = 0.4 ft.
•13–113. The ball of mass m is guided along the vertical
circular path r = 2rc cos u using
the arm OA. If the arm has
#
a constant angular velocity u0, determine the angle u … 45°
at which the ball starts to leave the surface of the
semicylinder. Neglect friction and the size of the ball.
P
z
·
r
O
u
r
x
0.5 m
y
rc
u
O
B
Probs. 13–112/113
C
Prob. 13–110
13–111. The pilot of an airplane executes a vertical
loop which in part follows the path of a cardioid,
r = 600(1 + cos u) ft . If his speed at A ( u = 0° ) is a
constant vP = 80 ft>s , determine the vertical force the
seat belt must exert on him to hold him to his seat when
the plane is upside down at A. He weighs 150 lb.
13–114. The ball has a mass of 1 kg and is confined to
move along the smooth vertical slot due to the rotation of
the smooth arm OA. Determine the force of the rod on the
ball and the normal force of the slot on the ball when
u# = 30°. The rod is rotating with a constant angular velocity
u = 3 rad>s. Assume the ball contacts only one side of the
slot at any instant.
13–115. Solve \$ Prob. 13–114 if the
# arm has an angular
acceleration of u = 2 rad>s2 when u = 3 rad>s at u = 30°.
0.5 m
A
A
r
r ! 600 (1 + cos u ) ft
u
u
O
Prob. 13–111
A
Probs. 13–114/115
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*13.7 Central-Force Motion and Space
Mechanics
If a particle is moving only under the influence of a force having a line of
action which is always directed toward a fixed point, the motion is called
central-force motion. This type of motion is commonly caused by
electrostatic and gravitational forces.
In order to analyze the motion, we will consider the particle P shown in
Fig. 13–22a, which has a mass m and is acted upon only by the central
force F. The free-body diagram for the particle is shown in Fig. 13–22b.
Using polar coordinates (r, u), the equations of motion, Eqs. 13–9, become
-F = m c
0 = m ar
d2r
du 2
- ra b d
2
dt
dt
P
F
du
r
u
1
dA ! — r 2du
2
O
(a)
(13–11)
d2u
dr du
+ 2
b
dt dt
dt2
The second of these equations may be written in the form
so that integrating yields
1 d 2 du
c ar
bd = 0
r dt
dt
r
2 du
dt
u
r
= h
(13–12)
Here h is the constant of integration.
From Fig. 13–22a notice that the shaded area described by the radius r,
as r moves through an angle du, is dA = 12 r2 du. If the areal velocity is
defined as
dA
1 du
h
= r2
=
dt
2 dt
2
u
F
(b)
Fig. 13–22
(13–13)
then it is seen that the areal velocity for a particle subjected to centralforce motion is constant. In other words, the particle will sweep out equal
segments of area per unit of time as it travels along the path. To obtain
the path of motion, r = f1u2, the independent variable t must be
eliminated from Eqs. 13–11. Using the chain rule of calculus and
Eq. 13–12, the time derivatives of Eqs. 13–11 may be replaced by
dr
dr du
h dr
=
= 2
dt
du dt
r du
d h dr
d h dr du
d h dr
h
d2r
=
a
b =
a
b
= c a 2
bd
dt r2 du
du r2 du dt
du r du r2
dt2
13
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Substituting a new dependent variable (xi) j = 1>r into the second
equation, we have
2
d2r
2 2d j
=
h
j
dt2
du2
13
Also, the square of Eq. 13–12 becomes
a
du 2
b = h2j4
dt
Substituting these two equations into the first of Eqs. 13–11 yields
This satellite is subjected to a central
force and its orbital motion can be
closely predicted using the equations
developed in this section.
d2j
-h2j2
du2
2
du
v0
Satellite
F
F
r ! r0
Launching
Fig. 13–23
Power-flight
trajectory
F
m
or
d2j
Free-flight
trajectory
- h2j3 = -
+ j =
F
mh2j2
(13–14)
This differential equation defines the path over which the particle travels
when it is subjected to the central force F.*
For application, the force of gravitational attraction will be considered.
Some common examples of central-force systems which depend on
gravitation include the motion of the moon and artificial satellites about
the earth, and the motion of the planets about the sun. As a typical
problem in space mechanics, consider the trajectory of a space satellite or
space vehicle launched into free-flight orbit with an initial velocity v0 ,
Fig. 13–23. It will be assumed that this velocity is initially parallel to the
tangent at the surface of the earth, as shown in the figure.† Just after
the satellite is released into free flight, the only force acting on it is the
gravitational force of the earth. (Gravitational attractions involving other
bodies such as the moon or sun will be neglected, since for orbits close to
the earth their effect is small in comparison with the earth’s gravitation.)
According to Newton’s law of gravitation, force F will always act between
the mass centers of the earth and the satellite, Fig. 13–23. From Eq. 13–1,
this force of attraction has a magnitude of
F = G
Mem
r2
where Me and m represent the mass of the earth and the satellite,
respectively, G is the gravitational constant, and r is the distance between
*In the derivation, F is considered positive when it is directed toward point O. If F is
oppositely directed, the right side of Eq. 13–14 should be negative.
†The case where v0 acts at some initial angle u to the tangent is best described using the
conservation of angular momentum (see Prob. 15–100).
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the mass centers. To obtain the orbital path, we set j = 1>r in the
foregoing equation and substitute the result into Eq. 13–14. We obtain
d2j
2
du
+ j =
GMe
h2
(13–15)
13
This second-order differential equation has constant coefficients and is
nonhomogeneous. The solution is the sum of the complementary and
particular solutions given by
j =
GMe
1
= C cos (u - f) +
r
h2
(13–16)
This equation represents the free-flight trajectory of the satellite. It is
the equation of a conic section expressed in terms of polar coordinates.
A geometric interpretation of Eq. 13–16 requires knowledge of the
equation for a conic section. As shown in Fig. 13–24, a conic section is
defined as the locus of a point P that moves in such a way that the ratio
of its distance to a focus, or fixed point F, to its perpendicular distance
to a fixed line DD called the directrix, is constant. This constant ratio
will be denoted as e and is called the eccentricity. By definition
FP
PA
e =
D
directrix
u\$f
r
x
u
p
D
FP = r = e1PA2 = e[p - r cos1u - f2]
Fig. 13–24
or
1
1
1
= cos1u - f2 +
r
p
ep
Comparing this equation with Eq. 13–16, it is seen that the fixed distance
from the focus to the directrix is
p =
1
C
(13–17)
And the eccentricity of the conic section for the trajectory is
e =
Ch2
GMe
(13–18)
focus
F
f
x¿
From Fig. 13–24,
P
A
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directrix
13
P
AND
A C C E L E R AT I O N
u\$f
r
x
u
p
D
Fig. 13–24
GMe
1
= C cos u +
r
h2
focus
F
f
x¿
PA R T I C L E : F O R C E
Provided the polar angle u is measured from the x axis (an axis of
symmetry since it is perpendicular to the directrix), the angle f is zero,
Fig. 13–24, and therefore Eq. 13–16 reduces to
D
A
OF A
Page 158
(13–19)
The constants h and C are determined from the data obtained for the
position and velocity of the satellite at the end of the power-flight
trajectory. For example, if the initial height or distance to the space
vehicle is r0, measured from the center of the earth, and its initial
speed is v0 at the beginning of its free flight, Fig. 13–25, then the
constant h may be obtained from Eq. 13–12. When u = f = 0°, the
velocity v0 has no radial component; therefore, from Eq. 12–25,
v0 = r01du>dt2, so that
h = r20
du
dt
or
h = r0v0
(13–20)
To determine C, use Eq. 13–19 with u = 0°, r = r0 , and substitute
Eq. 13–20 for h:
C =
GMe
1
a1 b
r0
r0v20
(13–21)
The equation for the free-flight trajectory therefore becomes
GMe
GMe
1
1
=
a1 b cos u + 2 2
r
r0
r0v20
r0v0
(13–22)
The type of path traveled by the satellite is determined from the value
of the eccentricity of the conic section as given by Eq. 13–18. If
e = 0 free-flight trajectory is a circle
e = 1 free-flight trajectory is a parabola
e 6 1 free-flight trajectory is an ellipse
e 7 1 free-flight trajectory is a hyperbola
(13–23)
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Parabolic Path. Each of these trajectories is shown in Fig. 13–25. From
the curves it is seen that when the satellite follows a parabolic path, it is
“on the border” of never returning to its initial starting point. The initial
launch velocity, v0 , required for the satellite to follow a parabolic path is
called the escape velocity. The speed, ve , can be determined by using the
second of Eqs. 13–23, e = 1, with Eqs. 13–18, 13–20, and 13–21. It is left
as an exercise to show that
ve =
B
2GMe
r0
(13–24)
Circular Orbit. The speed vc required to launch a satellite into a
circular orbit can be found using the first of Eqs. 13–23, e = 0. Since e
is related to h and C, Eq. 13–18, C must be zero to satisfy this equation
(from Eq. 13–20, h cannot be zero); and therefore, using Eq. 13–21,
we have
vc =
GMe
B r0
(13–25)
Provided r0 represents a minimum height for launching, in which
frictional resistance from the atmosphere is neglected, speeds at launch
which are less than vc will cause the satellite to reenter the earth’s
atmosphere and either burn up or crash, Fig. 13–25.
Hyperbolic trajectory
e&1
Parabolic trajectory
e!1
Elliptical trajectory
0%e%1
v0
v0 % vc
Crash
trajectory
Circular
trajectory
e!0
r0
Fig. 13–25
159
13
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b
13
rp O
ra
b
a
a
Fig. 13–26
Elliptical Orbit. All the trajectories attained by planets and most
satellites are elliptical, Fig. 13–26. For a satellite’s orbit about the earth,
the minimum distance from the orbit to the center of the earth O (which
is located at one of the foci of the ellipse) is rp and can be found using
Eq. 13–22 with u = 0°. Therefore;
rp = r0
(13–26)
This minimum distance is called the perigee of the orbit. The apogee or
maximum distance ra can be found using Eq. 13–22 with u = 180°. * Thus,
ra =
r0
12GMe>r0v202 - 1
(13–27)
With reference to Fig. 13–26, the half length of the major axis of the ellipse is
a =
rp + ra
2
(13–28)
Using analytical geometry, it can be shown that the half length of the
minor axis is determined from the equation
b = 2rpra
(13–29)
*Actually, the terminology perigee and apogee pertains only to orbits about the earth.
If any other heavenly body is located at the focus of an elliptical orbit, the minimum and
maximum distances are referred to respectively as the periapsis and apoapsis of the orbit.
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161
Furthermore, by direct integration, the area of an ellipse is
A = pab =
p
1r + ra22rpra
2 p
(13–30)
13
The areal velocity has been defined by Eq. 13–13, dA>dt = h>2.
Integrating yields A = hT>2, where T is the period of time required to
make one orbital revolution. From Eq. 13–30, the period is
T =
p
1r + ra22rpra
h p
(13–31)
In addition to predicting the orbital trajectory of earth satellites, the
theory developed in this section is valid, to a surprisingly close
approximation, at predicting the actual motion of the planets traveling
around the sun. In this case the mass of the sun, Ms , should be substituted
for Me when the appropriate formulas are used.
The fact that the planets do indeed follow elliptic orbits about the sun
was discovered by the German astronomer Johannes Kepler in the early
developed the laws of motion and the law of gravitation, and so at the
time it provided important proof as to the validity of these laws. Kepler’s
laws, developed after 20 years of planetary observation, are summarized
as follows:
1. Every planet travels in its orbit such that the line joining it to the
center of the sun sweeps over equal areas in equal intervals of time,
whatever the line’s length.
2. The orbit of every planet is an ellipse with the sun placed at one of
its foci.
3. The square of the period of any planet is directly proportional to
the cube of the major axis of its orbit.
A mathematical statement of the first and second laws is given by
Eqs. 13–13 and 13–22, respectively. The third law can be shown from
Eq. 13–31 using Eqs. 13–19, 13–28, and 13–29. (See Prob. 13–116.)
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EXAMPLE 13.13
A satellite is launched 600 km from the surface of the earth, with an
initial velocity of 30 Mm>h acting parallel to the tangent at the surface
of the earth, Fig. 13–27.Assuming that the radius of the earth is 6378 km
and that its mass is 5.976110242 kg, determine (a) the eccentricity of the
orbital path, and (b) the velocity of the satellite at apogee.
13
SOLUTION
Part (a). The eccentricity of the orbit is obtained using Eq. 13–18.
The constants h and C are first determined from Eqs. 13–20 and
13–21. Since
v0 ! 30 Mm/h
rp = r0 = 6378 km + 600 km = 6.97811062 m
v0 = 30 Mm>h = 8333.3 m>s
ra
rp
A
O
vA
600 km
Fig. 13–27
then
h = rpv0 = 6.9781106218333.32 = 58.1511092 m2>s
GMe
1
C =
a1 b
rp
rpv20
66.73110-122[5.976110242]
1
e
1
f = 25.4110-92 m-1
6.97811062
6.9781106218333.322
Hence,
2.54110-82[58.1511092]2
Ch2
e =
=
= 0.215 6 1 Ans.
GMe
66.73110-122[5.976110242]
From Eq. 13–23, observe that the orbit is an ellipse.
Part (b). If the satellite were launched at the apogee A shown in
Fig. 13–27, with a velocity vA , the same orbit would be maintained
provided
h = rpv0 = ravA = 58.1511092 m2>s
Using Eq. 13–27, we have
rp
6.97811062
= 10.80411062
ra =
=
2GMe
2[66.73110-122][5.976110242]
-1
-1
rpv20
6.9781106218333.322
Thus,
58.1511092
vA =
Ans.
= 5382.2 m>s = 19.4 Mm>h
10.80411062
NOTE: The farther the satellite is from the earth, the slower it
moves, which is to be expected since h is constant.
=
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CENTRAL-FORCE MOTION AND SPACE MECHANICS
PROBLEMS
In the following problems, except where otherwise
indicated, assume that the radius of the earth is 6378
km, the earth’s mass is 5.976110242 kg, the mass of the
sun is 1.99110302 kg, and the gravitational constant is
G = 66.73110-122 m3>1kg # s22.
*13–116. Prove Kepler’s third law of motion. Hint: Use
Eqs. 13–19, 13–28, 13–29, and 13–31.
•13–117. The Viking explorer approaches the planet Mars
on a parabolic trajectory as shown. When it reaches point A
its velocity is 10 Mm> h. Determine r0 and the required
velocity at A so that it can then maintain a circular orbit as
shown. The mass of Mars is 0.1074 times the mass of the
earth.
A
13–119. The satellite is moving in an elliptical orbit with
an eccentricity e = 0.25. Determine its speed when it is at
its maximum distance A and minimum distance B from
13
the earth.
2 Mm
A
B
Prob. 13–119
*13–120. The space shuttle is launched with a velocity of
17 500 mi/h parallel to the tangent of the earth’s surface at
point P and then travels around the elliptical orbit. When
it reaches point A, its engines are turned on and its
velocity is suddenly increased. Determine the required
increase in velocity so that it enters the second elliptical
orbit. Take G = 34.4(10 - 9) ft4>lb # s4, Me = 409(1021) slug,
and re = 3960 mi, where 5280 ft = mi.
1500 mi
P¿
A
P
r0
Prob. 13–117
13–118. The satellite is in an elliptical orbit around the
earth as shown. Determine its velocity at perigee P and
apogee A, and the period of the satellite.
P
4500 mi
Prob. 13–120
•13–121. Determine the increase in velocity of the space
shuttle at point P so that it travels from a circular orbit to an
elliptical orbit that passes through point A. Also, compute
the speed of the shuttle at A.
A
P
2 Mm
8 Mm
Prob. 13–118
A
8 Mm
2 Mm
Prob. 13–121
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13–122. The rocket is in free flight along an elliptical
trajectory A¿A. The planet has no atmosphere, and its
mass is 0.60 times that of the earth. If the orbit has the
apoapsis and periapsis shown, determine the rocket’s velocity
when it is at point A. Take G = 34.4110-921lb # ft22>slug 2,
Me = 409110212 slug, 1 mi = 5280 ft.
13–123. If the rocket is to land on the surface of the planet,
determine the required free-flight speed it must have at A¿
so that the landing occurs at B. How long does it take for the
rocket to land, in going from A¿ to B? The planet has no
atmosphere, and its mass is 0.6 times that of the earth.
Take G = 34.4110-921lb # ft22>slug 2, Me = 409110212 slug,
1 mi = 5280 ft.
AND
A C C E L E R AT I O N
13–127. A rocket is in a free-flight elliptical orbit about
the earth such that the eccentricity of its orbit is e and its
perigee is r0. Determine the minimum increment of speed it
should have in order to escape the earth’s gravitational field
when it is at this point along its orbit.
*13–128. A rocket is in circular orbit about the earth at an
altitude of h = 4 Mm. Determine the minimum increment
in speed it must have in order to escape the earth’s
gravitational field.
•13–129. The rocket is in free flight along an elliptical
trajectory A¿A. The planet has no atmosphere, and its mass
is 0.70 times that of the earth. If the rocket has an apoapsis
and periapsis as shown in the figure, determine the speed of
the rocket when it is at point A.
r ! 3 Mm
A
B
4000 mi
O
A¿
B
A
O
A¿
r ! 2000 mi
6 Mm
10 000 mi
Probs. 13–122/123
*13–124. A communications satellite is to be placed into
an equatorial circular orbit around the earth so that it
always remains directly over a point on the earth’s surface.
If this requires the period to be 24 hours (approximately),
determine the radius of the orbit and the satellite’s velocity.
•13–125. The speed of a satellite launched into a
circular orbit about the earth is given by Eq. 13–25.
Determine the speed of a satellite launched parallel to
the surface of the earth so that it travels in a circular orbit
800 km from the earth’s surface.
13–126. The earth has an orbit with eccentricity e = 0.0821
around the sun. Knowing that the earth’s minimum distance
from the sun is 151.3(106) km, find the speed at which a
rocket travels when it is at this distance. Determine the
equation in polar coordinates which describes the earth’s
9 Mm
Prob. 13–129
13–130. If the rocket is to land on the surface of the
planet, determine the required free-flight speed it must
have at A¿ so that it strikes the planet at B. How long does
it take for the rocket to land, going from A¿ to B along an
elliptical path? The planet has no atmosphere, and its mass
is 0.70 times that of the earth.
r ! 3 Mm
B
A
O
6 Mm
A¿
9 Mm
Prob. 13–130
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13–131. The satellite is launched parallel to the tangent of
the earth’s surface with a velocity of v0 = 30 Mm>h from an
altitude of 2 Mm above the earth as shown. Show that the
orbit is elliptical, and determine the satellite’s velocity when
it reaches point A.
v0 ! 30 Mm/h
165
CENTRAL-FORCE MOTION AND SPACE MECHANICS
13–134. A satellite is launched with an initial velocity
v0 = 4000 km>h parallel to the surface of the earth.
Determine the required altitude (or range of altitudes)
above the earth’s surface for launching if the free-flight
trajectory is to be (a) circular, (b) parabolic, (c) elliptical,
and (d) hyperbolic.
13–135. The rocket is in a free-flight elliptical orbit about
the earth such that e = 0.76 as shown. Determine its speed
when it is at point A. Also determine the sudden change in
speed the rocket must experience at B in order to travel in
free flight along the orbit indicated by the dashed path.
A
u ! 150"
P
2 Mm
Prob. 13–131
B
C
A
*13–132. The satellite is in an elliptical orbit having an
eccentricity of e = 0.15. If its velocity at perigee is
vP = 15 Mm>h, determine its velocity at apogee A and the
period of the satellite.
9 Mm
8 Mm
5 Mm
Prob. 13–135
15 Mm/h
P
*13–136. A communications satellite is in a circular orbit
above the earth such that it always remains directly over a
point on the earth’s surface. As a result, the period of the
satellite must equal the rotation of the earth, which is
approximately 24 hours. Determine the satellite’s altitude h
above the earth’s surface and its orbital speed.
A
Prob. 13–132
•13–133. The satellite is in an elliptical orbit. When it is at
perigee P, its velocity is vP = 25 Mm>h, and when it reaches
point A, its velocity is vA = 15 Mm>h and its altitude above
the earth’s surface is 18 Mm. Determine the period of the
satellite.
•13–137. Determine the constant speed of satellite S so
that it circles the earth with an orbit of radius r = 15 Mm .
Hint: Use Eq. 13–1.
vP ! 25 Mm/h
r ! 15 Mm
P
18 Mm
A
S
Prob. 13–133
Prob. 13–137
13
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CONCEPTUAL PROBLEMS
13
P13–1. If the box is released from rest at A, use
numerical values to show how you would estimate the
time for it to arrive at B. Also, list the assumptions for your
analysis.
P13–3. Determine the smallest speed of each car A and B
so that the passengers do not lose contact with the seat
while the arms turn at a constant rate. What is the largest
normal force of the seat on each passenger? Use numerical
B
A
A
B
P13–1
P13–2. The tugboat has a known mass and its propeller
provides a known maximum thrust. When the tug is fully
powered you observe the time it takes for the tug to reach a
speed of known value starting from rest. Show how you
could determine the mass of the barge. Neglect the drag
force of the water on the tug. Use numerical values to
P13–3
P13–4. Each car is pin connected at its ends to the rim of the
wheel which turns at a constant speed. Using numerical
values, show how to determine the resultant force the seat
exerts on the passenger located in the top car A. The
passengers are seated towards the center of the wheel. Also,
list the assumptions for your analysis.
A
P13–2
P13–4
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CHAPTER REVIEW
CHAPTER REVIEW
Kinetics
Kinetics is the study of the relation between
forces and the acceleration they cause.
This relation is based on Newton’s second
law of motion, expressed mathematically as
13
F2
FR = #F
Before applying the equation of motion, it
is important to first draw the particle’s
free-body diagram in order to account for
all of the forces that act on the particle.
Graphically, this diagram is equal to the
kinetic diagram, which shows the result of the
forces, that is, the ma vector.
Inertial Coordinate Systems
When applying the equation of motion, it
is important to measure the acceleration
from an inertial coordinate system. This
system has axes that do not rotate but are
either fixed or translate with a constant
velocity. Various types of inertial
coordinate systems can be used to apply
©F = ma in component form.
Rectangular x, y, z axes are used to describe
rectilinear motion along each of the axes.
Normal and tangential n, t axes are often
used when the path is known. Recall that a n
is always directed in the + n direction. It
indicates the change in the velocity
direction. Also recall that at is tangent to
the path. It indicates the change in the
velocity magnitude.
Cylindrical coordinates are useful when
angular motion of the radial line r
is specified or when the path can
conveniently be described with these
coordinates.
ma
!
F1
Free-body
diagram
Kinetic
diagram
y
a
Path of particle
vO
O
Inertial frame of reference
x
at = dv>dt or at = v dv>ds
an = v2>r where r =
[1 + 1dy>dx22]3>2 ƒ
ƒ d2y>dx2 ƒ
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