100-slide Summary

100-slide Summary
PHY131H1F
Introduction to Physics I
Review of the first half
Chapters 1-8 + Error Analysis
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Position, Velocity, Acceleration
Significant Figures, Measurements, Errors
Equations of constant acceleration
Vectors, Relative Motion
Forces and Newton’s 3 Laws
Free Body Diagrams
Equilibrium and Non-equilibrium Problems
Circular Motion, Centripetal Force
1
The Particle Model
 Often motion of the object as a whole is not influenced by
details of the object’s size and shape
 We only need to keep track of a single point on the object
 So we can treat the object as if all its mass were
concentrated into a single point
 A mass at a single point in space is called a particle
 Particles have no size, no shape and no top, bottom, front
or back
 Below us a motion diagram of a car stopping, using the
particle model
2
Position-versus-Time Graphs
• Below is a motion diagram, made at 1 frame per minute, of a
student walking to school.
 A motion diagram is one way to represent the student’s motion.
 Another way is to make a graph of x versus t for the student:
3
Acceleration
 Sometimes an object’s velocity
is constant as it moves
 More often, an object’s velocity
changes as it moves
 Acceleration describes a change
in velocity
 Consider an object whose velocity changes from v1 to v2
during the time interval Δt
 The quantity Δv = v2 – v1 is the change in velocity
 The rate of change of velocity is called the average
acceleration:
4
Acceleration (a.k.a. “instantaneous
acceleration”)



 v  dv
a  lim 

t 0 t

 dt
 
v0 v
1

v

a

Units of v
are m/s.

Units of a
are m/s2.
5
A ball rolls up a ramp, and then down the ramp. We keep
track of the position of the ball at 6 instants as it climbs up
the ramp. At instant 6, it stops momentarily as it turns
around. Then it rolls back down. Shown below is the
motion diagram for the final 6 instants as it rolls down the
ramp.
At which instant is the speed of the ball the greatest?
A.
B.
C.
D.
E.
6
9
11
The speed is zero at point 6, but the same at points 7 to 11
The speed is the same at points 6 through 11
6
A ball rolls up a ramp, and then down the ramp. We keep
track of the position of the ball at 6 instants as it climbs up
the ramp. At instant 6, it stops momentarily as it turns
around. Then it rolls back down. Shown below is the
motion diagram for the final 6 instants as it rolls down the
ramp.
At which instant is the speed of the ball the greatest?
A.
B.
C.
D.
E.
6
9
11
The speed is zero at point 6, but the same at points 7 to 11
The speed is the same at points 6 through 11
7
A ball rolls up a ramp, and then down the ramp. We keep
track of the position of the ball at 6 instants as it climbs up
the ramp. At instant 6, it stops momentarily as it turns
around. Then it rolls back down. Shown below is the motion
diagram for the final 6 instants as it rolls down the ramp.
At which instant is the acceleration of the ball the greatest?
A.
B.
C.
D.
6
9
11
The acceleration is zero at point 6, but about the same at
points 7 to 11
E. The acceleration is about the same at points 6 through 11
8
A ball rolls up a ramp, and then down the ramp. We keep
track of the position of the ball at 6 instants as it climbs up
the ramp. At instant 6, it stops momentarily as it turns
around. Then it rolls back down. Shown below is the motion
diagram for the final 6 instants as it rolls down the ramp.
At which instant is the acceleration of the ball the greatest?
A.
B.
C.
D.
6
9
11
The acceleration is zero at point 6, but about the same at
points 7 to 11
E. The acceleration is about the same at points 6 through 11
9
Unit Conversions
 It is important to be able to
convert back and forth between
SI units and other units
 One effective method is to write
the conversion factor as a ratio
equal to one
 Because multiplying by 1 does not change a value, these
ratios are easily used for unit conversions
 For example, to convert the length 2.00 feet to meters, use
the ratio:
 So that:
10
Significant Figures
 It’s important in science and engineering to state clearly
what you know about a situation – no less, and no more
 For example, if you report a length as 6.2 m, you imply
that the actual value is between 6.15 m and 6.25 m and
has been rounded to 6.2
 The number 6.2 has two significant figures
 More precise measurement could give more significant
figures
 The appropriate number of significant figures is
determined by the data provided
 Calculations follow the “weakest link” rule: the input
value with the smallest number of significant figures
determines the number of significant figures to use in
11
reporting the output value
12
When do I round?
• The final answer of a problem should be
displayed to the correct number of significant
figures
• Numbers in intermediate calculations should not
be rounded off
• It’s best to keep lots of digits in the calculations
to avoid round-off error, which can compound if
there are several steps
13
Suggested Problem Solving Strategy
• MODEL
Think about and simplify the situation,
guess at what the right answer might be.
• VISUALIZE
• SOLVE
• ASSESS
Draw a diagram. It doesn’t have to be
artistic: stick figures and blobs are okay!
Set up the equations, solve for what
you want to find. (This takes time..)
Check your units, significant figures, do a
“sanity check”: does my answer make sense?
This is just a suggested strategy. Whatever method
works for you is fine, as long as you don’t make a mistake,
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and you show how you got to the correct answer, it’s 100%!
Error Analysis
 Almost every time you make a measurement, the
result will not be an exact number, but it will be a
range of possible values.
 The range of values associated with a measurement
is described by the uncertainty, or error.
1600 ± 100 apples:
1600 is the value
100 is the error
Exactly 3 apples (no error)
15
Errors
 Errors eliminate the need to report measurements with
vague terms like “approximately” or “≈”.
 Errors give a quantitative way of stating your confidence
level in your measurement.
 Saying the answer is 10 ± 2 means you are 68% confident
that the actual number is between 8 and 12.
 It also implies that and 14 (the 2-σ range).
N
A histogram of many, many
measurements of the same
thing:
±1 σ
±2 σ
16
(x x ) 2
The Gaussian:
N(x)
N(x)  Ae
2 2
A

x 
x
x 
x
 A is the maximum amplitude.
 x is the mean or average.
 σ is the standard deviation of the distribution.
17
Normal Distribution
 σ is the standard deviation of the distribution
 Statisticians often call the square of the standard
deviation, σ2, the variance
 σ is a measure of the width of the curve: a larger σ means
a wider curve
 68% of the area under the curve of a Gaussian lies
between the mean minus the standard deviation and the
mean plus the standard deviation
 95% of the area under the curve is between the mean
minus twice the standard deviation and the mean plus
twice the standard deviation
18
Estimating the Mean from a Sample
 Suppose you make N measurements of a quantity x, and
you expect these measurements to be normally distributed
 Each measurement, or trial, you label with a number i,
where i = 1, 2, 3, etc
 You do not know what the true mean of the distribution
is, and you cannot know this
 However, you can estimate the mean by adding up all the
individual measurements and dividing by N:
N
1
xest   xi
N i1
19
Estimating the Standard Deviation from a
Sample
 Suppose you make N measurements of a quantity x, and
you expect these measurements to be normally distributed
 It is impossible to know the true standard deviation of the
distribution
 The best estimate of the standard deviation is:
N
1
2
 est 
(xi  xest )

N 1 i1
 The quantity N – 1 is called the number of degrees of
freedom
 In this case, it is the number of measurements minus one
because you used one number from a previous calculation
20
(mean) in order to find the standard deviation.
Reading Error (Digital)
 For a measurement with an instrument with a
digital readout, the reading error is usually “±
one-half of the last digit.”
 This means one-half of the power of ten
represented in the last digit.
 With the digital thermometer shown, the last
digit represents values of a tenth of a degree,
so the reading error is ½ × 0.1 = 0.05°C
 You should write the temperature as 12.80 ± 0.05 °C.
21
Significant Figures when Error are Involved
 There are two general rules for significant figures
used in experimental sciences:
1. Errors should be specified to one, or at most two,
significant figures.
2. The most precise column in the number for the
error should also be the most precise column in the
number for the value.
 So if the error is specified to the 1/100th column, the
quantity itself should also be specified to the 1/100th
column.
22
Propagation of Errors of Precision
 When you have two or more quantities with known errors
you may sometimes want to combine them to compute a
derived number
 You can use the rules of Error Propagation to infer the error
in the derived quantity
 We assume that the two directly measured quantities are x
and y, with errors Δx and Δy respectively
 The measurements x and y must be independent of each
other.
 The fractional error is the value of the error divided by the
value of the quantity: Δx / x
 To use these rules for quantities which cannot be negative,
23
the fractional error should be much less than one
Propagation of Errors
• Rule #1 (sum or difference rule):
• If z = x + y
• or z = x – y
• then z  x  y
2
2
• Rule #2 (product or division rule):
 • If z = xy
• or z = x/y
•
x  y 
z
     
 x   y 
z
2
then
2
24
Propagation of Errors
• Rule #2.1 (multiply by exact constant rule):
• If z = xy or z = x/y
• and x is an exact number, so that Δx=0
• then z  x y
• Rule #3 (exponent rule):
n
•
If
z
=
x

• then z
x
n
z
x
25
The Error in the Mean
 Many individual, independent measurements
are repeated N times
 Each individual measurement has the same
error Δx
 Using error propagation you can show that the
error in the estimated mean is:
xest
x

N
26
The 4 Equations of Constant Acceleration:
1. vf  vi  at
Does not contain position!
2. s  s  v t  1 a(t) 2
f
i
i
2
3. v 2  v 2  2a(s  s )
f
i
f
i
v i  v f 
4. sf  si  
t
 2 
Does not contain vf !
Does not contain t !
Does not contain a ! (but
you know it’s constant)
Strategy: When a = constant, you can use one of these
equations. Figure out which variable you don’t know and
don’t care about, and use the equation which doesn’t
27
contain it.
Free Fall
 The motion of an object moving
under the influence of gravity only,
and no other forces, is called free
fall
 Two objects dropped from the
same height will, if air resistance
can be neglected, hit the ground at
the same time and with the same
speed
 Consequently, any two objects in
free fall, regardless of their mass,
have the same acceleration:
The apple and feather seen
28
here are falling in a vacuum.
Free Fall
 Figure (a) shows the motion
diagram of an object that was
released from rest and falls freely
 Figure (b) shows the object’s
velocity graph
 The velocity graph is a straight
line with a slope:
 where g is a positive number which
is equal to 9.80 m/s2 on the surface
of the earth
 Other planets have different values
of g
29
Two-Dimensional Kinematics
 If the velocity vector’s
angle θ is measured from
the positive x-direction, the
velocity components are
where the particle’s speed
is
 Conversely, if we know the velocity components, we can
determine the direction of motion:
30
Projectile Motion
 The start of a
projectile’s motion is
called the launch
 The angle θ of the
initial velocity v0 above
the x-axis is called the
launch angle
 The initial velocity vector can be broken into
components
where v0 is the initial speed
31
Projectile Motion
 Gravity acts downward
 Therefore, a projectile
has no horizontal
acceleration
 Thus
 The vertical component of acceleration ay is −g of free fall
 The horizontal component of ax is zero
 Projectiles are in free fall
32
Reasoning About Projectile Motion
A heavy ball is launched exactly horizontally at height h above
a horizontal field. At the exact instant that the ball is launched, a
second ball is simply dropped from height h. Which ball hits the
ground first?
 If air resistance is neglected,
the balls hit the ground
simultaneously
 The initial horizontal velocity of
the first ball has no influence
over its vertical motion
 Neither ball has any initial
vertical motion, so both fall
distance h in the same amount
of time
33
Range of a Projectile
A projectile with initial speed v0 has a launch angle of θ above
the horizontal. How far does it travel over level ground before
it returns to the same elevation from which it was launched?
 This distance is
sometimes called the
range of a projectile
 Example 4.5 from your
textbook shows:
 The maximum distance
occurs for θ = 45°
Trajectories of a projectile launched at
different angles with a speed of 99 m/s.
34
Reference Frames
Relative Velocity
• Relative velocities are found as the time derivative of
the relative positions.
• CA is the velocity of C relative to A.
• CB is the velocity of C relative to B.
• AB is the velocity of reference frame A relative to
reference frame B.
• This is known as the Galilean transformation of
velocity.
35
Slide 4-69
Relative Motion
• Note the “cancellation”
• vTG = velocity of the
Train relative to the
Ground
• vPT = velocity of the
Passenger relative to the
Train
• vPG = velocity of the
Passenger relative to the
Ground
vPG = vPT + vTG
Inner subscripts
disappear
36
You are running toward the right at 5 m/s
toward an elevator that is moving up at 2 m/s.
Relative to you, the direction and magnitude
of the elevator’s velocity are
A.
B.
C.
D.
E.
down and to the right, less than 2 m/s.
up and to the left, less than 2 m/s.
up and to the left, more than 2 m/s.
up and to the right, less than 2 m/s.
up and to the right, more than 2 m/s.
37
You are running toward the right at 5 m/s
toward an elevator that is moving up at 2 m/s.
Relative to you, the direction and magnitude
of the elevator’s velocity are
A.
B.
C.
D.
E.
down and to the right, less than 2 m/s.
up and to the left, less than 2 m/s.
up and to the left, more than 2 m/s.
up and to the right, less than 2 m/s.
up and to the right, more than 2 m/s.
38
What is a force?
 A force is a push or a
pull
A force acts on an object
 Pushes and pulls are
applied to something
 From the object’s
perspective, it has a force
exerted on it
• The S.I. unit of force is
the Newton (N)
• 1 N = 1 kg m s–2
39
Tactics: Drawing force vectors
40
41
N1
Newton’s First Law
The natural state of an object with no net
external force on it is to either remain at rest
or continue to move in a straight line with a
constant velocity.
42
Inertial Reference Frames
 If a car stops
suddenly, you may be
“thrown” forward
 You do have a
forward acceleration
relative to the car
 However, there is no
force pushing you
forward
This guy thinks there’s a force hurling him
into the windshield. What a dummy!
 We define an inertial reference frame as one in which
Newton’s laws are valid
 The interior of a crashing car is not an inertial reference
43
frame!
Thinking About Force
 Every force has an agent which causes the force
 Forces exist at the point of contact between the agent and
the object (except for the few special cases of long-range
forces)
 Forces exist due to interactions happening now, not due to
what happened in the past
 Consider a flying arrow
 A pushing force was required to
accelerate the arrow as it was
shot
 However, no force is needed to
keep the arrow moving forward
as it flies
 It continues to move because of inertia
44
What is Mass?
• Mass is a scalar quantity that describes an
object’s inertia.
• It describes the amount of matter in an object.
• Mass is an intrinsic property of an object.
• It tells us something
about the object,
regardless of where the
object is, what it’s
doing, or whatever
forces may be acting on
it.
45
What Do Forces Do? A Virtual Experiment
 Attach a stretched rubber band to a 1 kg block
 Use the rubber band to pull the block across a horizontal,
frictionless table
 Keep the rubber band stretched by a fixed amount
 We find that the block moves with a constant acceleration
46
N2
Newton’s Second Law
The acceleration of an object is directly
proportional to the net force acting on it, and
inversely proportional to its mass.

 Fnet
a
m
47
Equilibrium
 An object on which the net force is zero is in equilibrium
 If the object is at rest, it is in static equilibrium
 If the object is moving along a straight line with a constant
velocity it is in dynamic equilibrium
 The requirement for either
type of equilibrium is:
The concept of equilibrium is
essential for the engineering
analysis of stationary objects such
48
as bridges.
Non-Equilibrium
 Suppose the x- and y-components of acceleration are
independent of each other
 That is, ax does not depend on y or vy, and ay does not
depend on x or vx
 Your problem-solving strategy is to:
1. Draw a free-body diagram
2. Use Newton’s second law in component form:
The force components (including proper signs) are
found from the free-body diagram
49
Universal Law of Gravitation
Gravity is an attractive, long-range force between any
two objects.
When two objects with masses m1 and m2 are separated
by distance r, each object pulls on the other with a force
given by Newton’s law of gravity, as follows:
(Sometimes called “Newton’s 4th Law”, or
“Newton’s Law of Universal Gravitation”)
50
Gravity for Earthlings
If you happen to live on the surface of a large planet with
radius R and mass M, you can write the gravitational force
even more simply as
where the quantity g is defined to be:
At sea level, g = 9.83 m/s2.
At 39 km altitude, g = 9.71 m/s2.
51
Gravity: FG = mg is just a short form!
and
are the same equation, with different notation!
The only difference is that in the second equation
we have assumed that m2 = M (mass of the
earth) and r ≈ R (radius of the earth).
52
Weight: A Measurement
 You weigh apples in the grocery
store by placing them in a spring
scale and stretching a spring
 The reading of the spring scale is
the magnitude of Fsp
 We define the weight of an object
as the reading Fsp of a calibrated
spring scale on which the object is
stationary
 Because Fsp is a force, weight is
measured in newtons
53
Weight: A Measurement
 The figure shows a man weighing
himself in an accelerating elevator
 Looking at the free-body diagram,
the y-component of Newton’s
second law is:
 The man’s weight as he
accelerates vertically is:
 You weigh more as an elevator
accelerates upward!
54
Normal Force
 When an object sits on a table,
the table surface exerts an
upward contact force on the
object
 This pushing force is directed
perpendicular to the surface,
and thus is called the normal
force
 A table is made of atoms
joined together by molecular
bonds which can be modeled as
springs
 Normal force is a result of
many molecular springs being
compressed ever so slightly
55
Tension Force
 When a string or rope or wire
pulls on an object, it exerts a
contact force called the tension
force
 The tension force is in the
direction of the string or rope
 A rope is made of atoms joined
together by molecular bonds
 Molecular bonds can be
modeled as tiny springs holding
the atoms together
 Tension is a result of many
molecular springs stretching
ever so slightly
Why does
friction exist?
Because at the
microscopic
level, nothing
is smooth!
Static Friction
 The figure shows a person
pushing on a box that, due to static
friction, isn’t moving
 Looking at the free-body diagram,
the x-component of Newton’s first
law requires that the static friction
force must exactly balance the
pushing force:
 fs points in the direction opposite
to the way the object would move if
there were no static friction
58
Static friction acts in response to an applied force.
59
Maximum Static Friction
There’s a limit to how big fs can get. If you push
hard enough, the object slips and starts to move. In
other words, the static friction force has a maximum
possible size fs max.
• The two surfaces don’t slip against each other as
long as fs ≤ fs max.
•A static friction force fs > fs max is not
physically possible. Many experiments have
shown the following approximate relation usually
holds:
where n is the magnitude of the normal force, and the
proportionality constant μs is called the “coefficient of
60
static friction”.
“Kinetic Friction”

fk
• Also called “sliding friction”
• When two flat surfaces are in contact and sliding relative to one
another, heat is created, so it slows down the motion (kinetic
energy is being converted to thermal energy).
• Many experiments have shown the following approximate relation
usually holds for the magnitude of fk:
f k  k n

fk

where n is the magnitude of
the normal force.
The direction of fk is opposite
the direction of motion. 61
A Model of Friction
The friction force response to an increasing applied force.
62
A wooden block weighs 100 N, and is sitting stationary
on a smooth horizontal concrete surface. The
coefficient of static friction between wood and
concrete is 0.2.
A 5 N horizontal force is applied to the block, pushing
toward the right, but the block does not move. What
is the force of static friction of the concrete on the
block?
A. 100 N, to the left
B. 20 N, to the left

F
C. 5 N, to the left
D. 20 N, to the right
E. 5 N, to the right
63
A wooden block weighs 100 N, and is sitting stationary
on a smooth horizontal concrete surface. The
coefficient of static friction between wood and
concrete is 0.2.
A 5 N horizontal force is applied to the block, pushing
toward the right, but the block does not move. What
is the force of static friction of the concrete on the
block?
A. 100 N, to the left
B. 20 N, to the left

F
C. 5 N, to the left
D. 20 N, to the right
E. 5 N, to the right
64
A wooden block weighs 100 N, and is sitting stationary
on a smooth horizontal concrete surface. The
coefficient of static friction between wood and
concrete is 0.2.
A horizontal force is applied to the block, pushing
toward the right. What is the magnitude of the
maximum pushing force you can apply and have the
block remain stationary?

F
A.
B.
C.
D.
E.
200 N
100 N
20 N
10 N
5N
65
A wooden block weighs 100 N, and is sitting stationary
on a smooth horizontal concrete surface. The
coefficient of static friction between wood and
concrete is 0.2.
A horizontal force is applied to the block, pushing
toward the right. What is the magnitude of the
maximum pushing force you can apply and have the
block remain stationary?

F
A.
B.
C.
D.
E.
200 N
100 N
20 N
10 N
5N
66
Rolling Motion
 If you slam on the brakes
so hard that the car tires
slide against the road
surface, this is kinetic friction
 Under normal driving
conditions, the portion of the
rolling wheel that contacts
the surface is stationary, not
sliding
 If your car is accelerating or decelerating or turning,
it is static friction of the road on the wheels that
provides the net force which accelerates the car
67
Rolling Friction
 A car with no engine or
brakes applied does not
roll forever; it gradually
slows down
 This is due to rolling
friction
 The force of rolling friction can be calculated as
where μr is called the coefficient of rolling friction.
 The rolling friction direction is opposite to the velocity of
the rolling object relative to the surface
68
Drag
 The air exerts a drag force on objects as they move
through the air
 Faster objects experience a greater drag force than
slower objects
 The drag force on a
high-speed
motorcyclist is
significant
 The drag force
direction is opposite
the object’s velocity
69
Drag
 For normal sized objects on earth traveling at a speed v which
is less than a few hundred meters per second, air resistance can
be modeled as:
 A is the cross-section area of the object
 ρ is the density of the air, which is about 1.2 kg/m3
 C is the drag coefficient, which is a dimensionless number
that depends on the shape of the object
70
Cross Sectional Area depends on size, shape, and
direction of motion.
…Consider the forces on a falling piece of paper,
crumpled and not crumpled.
71
Terminal Speed
 The drag force from the air
increases as an object falls and
gains speed
 If the object falls far enough, it
will eventually reach a speed at
which D = FG
 At this speed, the net force is
zero, so the object falls at a
constant speed, called the
terminal speed vterm
72
Terminal Speed
 The figure shows the
velocity-versus-time graph
of a falling object with and
without drag
 Without drag, the velocity
graph is a straight line with
ay = −g
 When drag is included, the
vertical component of the
velocity asymptotically
approaches −vterm
73
Propulsion
 If you try to walk across a
frictionless floor, your foot slips
and slides backward
 In order to walk, your foot
must stick to the floor as you
straighten your leg, moving
your body forward
 The force that prevents
slipping is static friction
 The static friction force points
in the forward direction
 It is static friction that propels
you forward!
What force causes this sprinter
to accelerate?
74
Interacting Objects
 If object A exerts a force on object B, then object B exerts a
force on object A.
 The pair of forces, as shown, is called an action/reaction
pair.
75
N3
Newton’s Third Law
If object 1 acts on object 2 with a force, then
object 2 acts on object 1 with an equal force
in the opposite direction.


F1 on 2   F2 on 1
76
Acceleration Constraints
 If two objects A and B move together, their accelerations are
constrained to be equal: aA = aB
 This equation is called an acceleration constraint
 Consider a car being towed by a truck
 In this case, the
acceleration constraint is
aCx = aTx = ax
 Because the
accelerations of both
objects are equal, we can
drop the subscripts C and
T and call both of them
ax
77
Acceleration Constraints
 Sometimes the acceleration
of A and B may have
different signs
 Consider the blocks A and
B in the figure
 The string constrains the
two objects to accelerate
together
 But, as A moves to the right in the +x direction, B moves
down in the −y direction
 In this case, the acceleration constraint is aAx = −aBy
78
The Massless String Approximation
Often in physics problems the mass of the string or rope is
much less than the masses of the objects that it connects.
In such cases, we can adopt the following massless string
approximation:
79
In the figure to the right, is
the tension in the string
greater than, less than, or
equal to the force of
gravity on block B?
A. Equal to
B. Greater than
C. Less than
80
In the figure to the right, is
the tension in the string
greater than, less than, or
equal to the force of
gravity on block B?
A. Equal to
B. Greater than
C. Less than
81
Motion on a Circular Path
 Consider a particle at a
distance r from the origin, at
an angle θ from the positive x
axis
 The angle may be measured
in degrees, revolutions (rev) or
radians (rad), that are related
by:
1 rev = 360° = 2π rad
 If the angle is measured in radians, then there is a simple
relation between θ and the arc length s that the particle travels
along the edge of a circle of radius r:
82
Angular Velocity
 As the time interval Δt becomes very small, we arrive at the
definition of instantaneous angular velocity
83
Angular Velocity in Uniform Circular Motion
 When angular velocity ω is constant, this is uniform
circular motion
 In this case, as the particle goes around a circle one
time, its angular displacement is Δθ = 2π during one
period Δt = T
 The absolute value of the constant angular velocity is
related to the period of the motion by
84
Angular Velocity of a
Rotating Object
 The figure shows a wheel
rotating on an axle
 Points 1 and 2 turn through
the same angle as the wheel
rotates
 That is, Δθ1 = Δθ2 during
some time interval Δt
 Therefore ω1 = ω2 = ω
 All points on the wheel rotate with the same angular velocity
 We can refer to ω as the angular velocity of the wheel
85
Tangential Velocity
 The tangential velocity
component vt is the rate ds/dt at
which the particle moves around
the circle, where s is the arc
length
 The tangential velocity and the
angular velocity are related by
 In this equation, the units of vt are m/s, the units of ω are
rad/s, and the units of r are m
86
Centripetal Acceleration
 In uniform circular motion,
although the speed is constant,
there is an acceleration
because the direction of the
velocity vector is always
changing
 The acceleration of uniform
circular motion is called
centripetal acceleration
 The direction of the centripetal acceleration is toward the
center of the circle:
87
Dynamics of Uniform Circular Motion
 An object in uniform circular motion is not traveling at a
constant velocity in a straight line
 Consequently, the particle must have a net force acting on it
 Without such a force, the
object would move off in a
straight line tangent to the circle
 The car would end up in the
ditch!
Highway and racetrack curves are banked to allow the normal force
88
of the road to provide the centripetal acceleration of the
turn.
Banked Curves
 Real highway curves are banked by being tilted up at the outside edge of
the curve
 The radial component of the normal force can provide centripetal
acceleration needed to turn the car
 For a curve of radius r banked at an angle θ, the exact speed at which a
car must take the curve without assistance from friction is
v0  rg tan
89
Banked Curves
 Consider a car going around a banked curve at a speed
higher than v0  rg tan
 In this case, static friction must prevent the car from slipping
up the hill
90
Banked Curves
 Consider a car going around a banked curve at a speed
slower than v  rg tan
0
 In this case, static friction must prevent the car from slipping
down the hill
91
Circular Orbits
 An object in a low circular orbit
has acceleration:
 If the object moves in a circle of
radius r at speed vorbit the centripetal
acceleration is:
 The required speed for a circular orbit near a planet’s surface,
neglecting air resistance, is:
92
Loop-the-loop
 The figure shows a rollercoaster going around a
vertical loop-the-loop of
radius r
 Because the car is moving in a circle, there must be a net force
toward the center of the circle
93
Loop-the-loop
 The figure shows the roller-coaster
free body diagram at the bottom of the
loop
 Since the net force is toward the
center (upward at this point) n > FG
 This is why you “feel heavy” at the
bottom of the valley on a roller
coaster
 The normal force at the bottom is larger than mg
94
A car is rolling over the top of a
hill at speed v. At this instant,


FG  mg
A. n > FG.
B. n < FG.
C. n = FG.
D.We can’t tell about n without knowing v.
95
A car is rolling over the top of a
hill at speed v. At this instant,


FG  mg
A. n > FG.
B. n < FG.
C. n = FG.
D.We can’t tell about n without knowing v.
96
Angular Acceleration
 Suppose a wheel’s rotation is
speeding up or slowing down
 This is called nonuniform circular
motion
 We can define the angular
acceleration as
 The units of α are rad/s2
 The figure to the right shows a wheel
with angular acceleration α = 2 rad/s2
97
The Sign of Angular Acceleration
 If ω is counter-clockwise and |ω| is increasing, then α is positive
 If ω is counter-clockwise and |ω| is decreasing, then α is negative
 If ω is clockwise and |ω| is decreasing, then α is positive
 If ω is clockwise and |ω| is increasing, then α is negative
98
Angular Kinematics
 The same relations that hold for linear motion between ax, vx and
x apply analogously to rotational motion for α, ω and θ
 There is a graphical relationship between α and ω:
 The table shows a comparison of the rotational and linear
kinematics equations for constant α or constant as:
99
Nonuniform Circular
Motion
 The particle in the figure is
speeding up as it moves
around the circle
 The tangential acceleration
is
 The centripetal acceleration
is
ar = v2/r = ω2r
100
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