Chapter 12 (© Lars Johansson/Fotolia) B_5389_ch12_ptg01_hr.4c.indd 2 1/21/15 9:57 AM 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. B_5389_ch12_ptg01_hr.4c.indd 3 1/21/15 9:57 AM 4 CHAPTER 12 12 K I N E M AT I C S OF A PARTICLE 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. B_5389_ch12_ptg01_hr.4c.indd 4 1/21/15 9:57 AM 12.2 5 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 12.2 Rectilinear Kinematics: Continuous 12 Motion 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, rather than in boldface s, since the direction always remains along the coordinate axis. s O s Position (a) 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 s s¿ s = s - s Displacement (b) 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. B_5389_ch12_ptg01_hr.4c.indd 5 Fig. 12–1 1/21/15 9:57 AM 6 CHAPTER 12 K I N E M AT I C S OF A PARTICLE Velocity. If the particle moves through a displacement s during the time interval t, the average velocity of the particle during this time interval is 12 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 (s> t), or S t 0 + ) (S 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., (vsp)avg = 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 (vsp)avg = sT > t, but its average velocity is vavg = - s> t. s P¿ P O s sT Average velocity and Average speed (d) Fig. 12–1 (cont.) B_5389_ch12_ptg01_hr.4c.indd 6 1/21/15 9:57 AM 12.2 7 RECTILINEAR KINEMATICS: CONTINUOUS MOTION Acceleration. Provided the velocity of the particle is known at two points, the average acceleration of the particle during the time interval t is defined as v t aavg = 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 (v> t), or S t 12 0 a s O v + ) (S dv a = dt v¿ Acceleration (12–2) (e) Substituting Eq. 12–1 into this result, we can also write + ) (S 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, notice that if the particle is originally at rest, then it can have an acceleration if a moment later it has a velocity v; and, if the velocity is constant, then 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. We have dt = a P P¿ O v s v¿ Deceleration (f) Fig. 12–1 (cont.) ds dv = v a or + ) (S a ds = v dv (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. B_5389_ch12_ptg01_hr.4c.indd 7 1/21/15 9:57 AM 8 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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 ac , v, s, and t. 12 Velocity as a Function of Time. Integrate ac = dv>dt, assuming that initially v = v0 when t = 0. v Lv0 When the ball is released, it has zero velocity but an acceleration of 9.81 m>s2. + ) (S (© R.C. Hibbeler) L0 dv = t ac dt v = v0 + ac t Constant Acceleration (12–4) Position as a Function of Time. Integrate v = ds>dt = v0 + act, assuming that initially s = s0 when t = 0. s Ls0 + ) (S ds = L0 t (v0 + act) dt s = s0 + v0t + 12 ac t2 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 + ) (S v dv = s Ls0 ac ds v2 = v20 + 2ac(s - s0) 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. B_5389_ch12_ptg01_hr.4c.indd 8 1/21/15 9:57 AM 12.2 9 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 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. During the time this vvvvket undergoes rectilinear motion, its altitude as a function of time can be measured and expressed as s = s(t). Its velocity can then be found using v = ds>dt, and its acceleration can be determined from a = dv>dt. (© NASA) 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. B_5389_ch12_ptg01_hr.4c.indd 9 1/23/15 2:23 PM 10 12 CHAPTER 12 EXAMPLE K I N E M AT I C S OF A PARTICLE 12.1 The car on the left in the photo and in Fig. 12–2 moves in a straight line such that for a short time its velocity is defined by v = (3t2 + 2t) 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 (© R.C. Hibbeler) SOLUTION Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right. Position. Since v = f(t), 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* + ) (S v = L0 s ds = s` s 0 ds = (3t2 + 2t) dt L0 t (3t2 + 2t)dt = t3 + t2 ` t 0 s = t3 + t2 When t = 3 s, s = (3)3 + (3)2 = 36 ft Ans. Acceleration. Since v = f(t), the acceleration is determined from a = dv>dt, since this equation relates a, v, and t. + ) (S dv d = (3t2 + 2t) dt dt = 6t + 2 a= When t = 3 s, a = 6(3) + 2 = 20 ft>s2 S Ans. NOTE: The formulas for constant acceleration cannot be used to solve this problem, because the acceleration is a function of time. *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 = (3t2 + 2t)dt yields s = t3 + t2 + C. Using the condition that at t = 0, s = 0, then C = 0. B_5389_ch12_ptg01_hr.4c.indd 10 1/21/15 9:57 AM 12.2 EXAMPLE RECTILINEAR KINEMATICS: CONTINUOUS MOTION 12.2 11 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 = (-0.4v3) 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 = f(v) 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 = -0.4v3 dt v t dv = dt 3 L60 m>s -0.4v L0 (+ T ) a = O s Fig. 12–3 1 1 1 v a b 2` = t - 0 -0.4 -2 v 60 1 1 1 c d = t 0.8 v2 (60)2 -1>2 1 v = ec + 0.8t d f m>s (60)2 Here the positive root is taken, since the projectile will continue to move downward. When t = 4 s, v = 0.559 m>s T Ans. Position. Knowing v = f(t), 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 (+ T ) v = L0 s ds = t -1>2 1 + 0.8t d dt 2 L0 (60) c 1>2 t 1 2 c + 0.8t d ` 0.8 (60)2 0 1>2 1 1 1 s = ec + 0.8t d fm 0.4 (60)2 60 s = When t = 4 s, -1>2 ds 1 = c + 0.8t d dt (60)2 s = 4.43 m B_5389_ch12_ptg01_hr.4c.indd 11 Ans. 1/21/15 9:57 AM 12 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.3 EXAMPLE 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 = +75 m>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 (+ c ) v2B = v2A + 2ac(sB - sA) 0 = (75 m>s)2 + 2(-9.81 m>s2)(sB - 40 m) 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 (+ c ) A v2C = v2B + 2ac(sC - sB) = 0 + 2(-9.81 m>s2)(0 - 327 m) sA 40 m vC = -80.1 m>s = 80.1 m>s T C Fig. 12–4 s 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., (+ c ) v2C = v2A + 2ac(sC - sA) = (75 m>s)2 + 2(-9.81 m>s2)(0 - 40 m) 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 (vB = 0) the acceleration at B is still 9.81 m>s2 downward! B_5389_ch12_ptg01_hr.4c.indd 12 1/21/15 9:57 AM 12.2 EXAMPLE 13 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 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 = (4s) 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 = f(s), 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 (+ T ) v 100 mm v dv = a ds s s L0 L0.1 m 1 2 v 4 2 s v ` = s ` 2 2 0.1 m 0 (1) vB = 0.346 m>s = 346 mm>s T Ans. v dv = A C 4s ds v = 2(s2 - 0.01)1>2 m>s 200 mm B At s = 200 mm = 0.2 m, Fig. 12–5 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, (+ T ) ds = v dt = 2(s2 - 0.01)1>2dt t ds = 2 dt L0.1 (s2 - 0.01)1>2 L0 s At s = 0.2 m, ln 1 2s2 - 0.01 + s 2 ` s 0.1 = 2t ` t 0 ln 1 2s2 - 0.01 + s 2 + 2.303 = 2t ln 1 2(0.2)2 - 0.01 + 0.2 2 + 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 = B_5389_ch12_ptg01_hr.4c.indd 13 1/21/15 9:57 AM 14 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.5 EXAMPLE A particle moves along a horizontal path with a velocity of v = (3t2 - 6t) 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 s 6.125 m O t2s t0s t 3.5 s (a) SOLUTION Coordinate System. Here positive motion is to the right, measured from the origin O, Fig. 12–6a. Distance Traveled. Since v = f(t), the position as a function of time may be found by integrating v = ds>dt with t = 0, s = 0. + ) (S ds = v dt L0 v (m/s) v 3t2 6t (0, 0) (2 s, 0) (1 s, 3 m/s) (b) t (s) s = (3t2 - 6t) dt t (3t2 - 6t) dt L0 s = (t3 - 3t2) m ds = (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 be determined from Eq. 1. This yields s t = 0 = 0 s t = 2 s = -4.0 m s t = 3.5 s = 6.125 m Fig. 12–6 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 Velocity. Ans. The displacement from t = 0 to t = 3.5 s is s = s t = 3.5 s - s t = 0 = 6.125 m - 0 = 6.125 m and so the average velocity is s 6.125 m = = 1.75 m>s S Ans. t 3.5 s - 0 The average speed is defined in terms of the distance traveled sT . This positive scalar is vavg = (vsp)avg = sT 14.125 m = = 4.04 m>s t 3.5 s - 0 Ans. NOTE: In this problem, the acceleration is a = dv>dt = (6t - 6) m>s2, which is not constant. B_5389_ch12_ptg01_hr.4c.indd 14 1/21/15 9:57 AM 12.2 15 RECTILINEAR KINEMATICS: CONTINUOUS MOTION It is highly suggested that you test yourself on the solutions to these examples, by covering them over and then trying to think about which equations of kinematics must be used and how they are applied in order to determine the unknowns. Then before solving any of the problems, try and solve some of the Preliminary and Fundamental Problems which follow. The solutions and answers to all these problems are given in the back of the book. Doing this throughout the book will help immensely in understanding how to apply the theory, and thereby develop your problem-solving skills. 12 PRELIMINARY PROBLEM P12–1. a) If s = (2t3) m, where t is in seconds, determine v when t = 2 s. g) If a = 4 m > s2, determine s when t = 3 s if v = 2 m > s and s = 2 m when t = 0. b) If v = (5s) m > s, where s is in meters, determine a at s = 1 m. h) If a = (8t2) m > s2, determine v when t = 1 s if v = 0 at t = 0. c) If v = (4t + 5) m > s, where t is in seconds, determine a when t = 2 s. i) If s = (3t2 + 2) m, determine v when t = 2 s. d) If a = 2 m > s2, determine v when t = 2 s if v = 0 when t = 0. j) When t = 0 the particle is at A. In four seconds it travels to B, then in another six seconds it travels to C. Determine the average velocity and the average speed. The origin of the coordinate is at O. e) If a = 2 m > s2, determine v at s = 4 m if v = 3 m > s at s = 0. 7m 1m A O B C s 14 m m > s2, f) If a = (s) where s is in meters, determine v when s = 5 m if v = 0 at s = 4 m. B_5389_ch12_ptg01_hr.4c.indd 15 Prob. P12–1 1/21/15 9:57 AM 16 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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. 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 Prob. F12–1 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. Prob. F12–5 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 Prob. F12–6 Prob. F12–2 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. 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. B_5389_ch12_ptg01_hr.4c.indd 16 1/21/15 9:57 AM 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 17 PROBLEMS 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m>s2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s? 12–2. 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. 12–3. 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. *12–4. A particle travels along a straight line with a constant acceleration. When s = 4 ft, v = 3 ft>s and when s = 10 ft, v = 8 ft>s. Determine the velocity as a function of position. 12–5. The velocity of a particle traveling in a straight line is given by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-s time interval, and what is its average speed? 12–6. The position of a particle along a straight line is given by s = (1.5t 3 - 13.5t 2 + 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. 12–7. A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. *12–8. A particle is moving along a straight line such that its position is defined by s = (10t2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s. B_5389_ch12_ptg01_hr.4c.indd 17 12 12–9. The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) 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. 12–10. A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>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 a numerical method to evaluate the integral. 12–11. 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. *12–12. Traveling with an initial speed of 70 km>h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km>h? Also, through what distance does the car travel during this time? 12–13. 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, please don’t drive! v1 44 ft/s d Prob. 12–13 1/21/15 9:57 AM 18 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–14. The position of a particle along a straight-line path 12 is defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds. Determine the total distance traveled when t = 10 s. What are the particle’s average velocity, average speed, and the instantaneous velocity and acceleration at this time? 12–18. The acceleration of a rocket traveling upward is given by a = (6 + 0.02s) m>s2, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0. 12–15. A particle is moving with a velocity of v0 when s = 0 and t = 0. If it is subjected to a deceleration of a = -kv3, where k is a constant, determine its velocity and position as functions of time. s *12–16. A particle is moving along a straight line with an initial velocity of 6 m>s when it is subjected to a deceleration of a = (-1.5v1> 2) m>s2, where v is in m>s. Determine how far it travels before it stops. How much time does this take? Prob. 12–18 12–19. A train starts from rest at station A and accelerates at 0.5 m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations. 12–17. Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 ft>s when the driver of B suddenly applies the brakes, causing his car to decelerate at 12 ft>s2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 15 ft>s2. Determine the minimum distance d be tween the cars so as to avoid a collision. A B *12–20. The velocity of a particle traveling along a straight line is v = (3t2 - 6t) ft>s, where t is in seconds. If s = 4 ft when t = 0, determine the position of the particle when t = 4 s. What is the total distance traveled during the time interval t = 0 to t = 4 s? Also, what is the acceleration when t = 2 s? 12–21. A freight train travels at v = 60(1- e - t ) ft>s, where t is the elapsed time in seconds. Determine the distance traveled in three seconds, and the acceleration at this time. v s d Prob. 12–17 B_5389_ch12_ptg01_hr.4c.indd 18 Prob. 12–21 1/21/15 9:57 AM 12.2 12–22. A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m>s. If the bag is released with the same upward velocity of 6 m>s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant. 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. *12–24. The acceleration of a particle traveling along a 1 straight line is a = s1> 2 m>s2, where s is in meters. If v = 0, 4 s = 1 m when t = 0, determine the particle’s velocity at s = 2 m. 12–25. If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation a = 9.81[1 - v 2 (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 S ). 12–26. The acceleration of a particle along a straight line is defined by a = (2t - 9) m>s2, where t is in seconds. At t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity. 12–27. 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 = (g>v2f) (v2f - v2), determine the time needed for the velocity to become v = vf >2. Initially the particle falls from rest. B_5389_ch12_ptg01_hr.4c.indd 19 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 19 *12–28. 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 = (12t 2 - 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. 12–29. A 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. 12–30. 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. 12–31. 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. *12–32. 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 < 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. 12–33. 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[R 2 >(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 S . 12–34. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–36), 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 y 0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y 0 = 500 km? Use the numerical data in Prob. 12–33. 1/21/15 9:57 AM 20 CHAPTER 12 K I N E M AT I C S s 12 ds ds v0 dt t 0 v2 dt t 2 ds ds v1 dt t v3 dt t 1 3 s2 s1 O t1 s3 t2 t t3 (a) v OF A PARTICLE 12.3 Rectilinear Kinematics: Erratic 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. The s–t, v–t, and a–t Graphs. To construct the v9t 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 v1 v0 v2 O t1 t2 t3 v3 t (b) ds = v dt slope of = velocity s9t graph Fig. 12–7 v a dv a0 dv dt t 0 2 dt t2 dv a3 dt t a1 dv 3 dt t1 dv = a dt v3 v2 v1 v0 O t2 t1 t t3 (a) a a1 O a0 0 t1 a2 t2 (b) Fig. 12–8 B_5389_ch12_ptg01_hr.4c.indd 20 a3 t3 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 v9t graph can be constructed by plotting this and other values at each instant. The a–t graph can be constructed from the v9t graph in a similar manner, Fig. 12–8, since t slope of = acceleration v9t 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 v9t 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 v9t 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). 1/21/15 9:57 AM 12.3 If the a–t graph is given, Fig. 12–9a, the v9t graph may be constructed using a = dv>dt, written as a 12 a0 t1 v a dt 0 L change in area under = velocity a9t graph v = 21 RECTILINEAR KINEMATICS: ERRATIC MOTION a dt Hence, to construct the v9t graph, we begin with the particle’s initial velocity v0 and then add to this small increments of area (v) determined from the a–t graph. In this manner successive points, v1 = v0 + v, etc., for the v9t 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 v9t graph is given, Fig. 12–10a, it is possible to determine the s–t graph using v = ds>dt, written as t t1 (a) v v v1 v0 t t1 (b) Fig. 12–9 v v dt L area under displacement = v9t graph s = v0 t1 s v dt 0 t t1 (a) 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 v9t 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 v9t graph. In a similar manner, the s–t graph can be obtained by integrating the equations which describe the segments of the v9t graph. As a result, if the a–t graph is linear (a first-degree curve), integration will yield a v9t graph that is parabolic (a second-degree curve) and an s–t graph that is cubic (third-degree curve). B_5389_ch12_ptg01_hr.4c.indd 21 s s1 s s0 t t1 (b) Fig. 12–10 1/21/15 9:57 AM 22 CHAPTER 12 K I N E M AT I C S OF A PARTICLE The v–s and a–s Graphs. If the a–s graph can be constructed, a 12 then points on the v9s 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, s1 a0 0 a ds —12 (v12 v02) s s1 (a) v 1 2 2 (v1 s1 a ds Ls0 area under a9s graph - v20) = v1 v0 s s1 Therefore, if the red area in Fig. 12–11a is determined, and the initial velocity v0 at s0 = 0 is known, then v1 = 1 2 10s1a ds + v20 2 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 (b) Fig. 12–11 v dv ds v0 v s s (a) a dv b ds velocity times acceleration = slope of v9s graph a = va a0 a v(dv/ds) s s (b) Fig. 12–12 B_5389_ch12_ptg01_hr.4c.indd 22 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. 1/21/15 9:57 AM 12.3 EXAMPLE 23 RECTILINEAR KINEMATICS: ERRATIC MOTION 12.6 12 A bicycle moves along a straight road such that its position is described by the graph shown in Fig. 12–13a. Construct the v9t 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 v9t 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 = (t 2) ft v = = (2t) ft>s dt ds 10 s 6 t … 30 s; s = (20t - 100) ft v = = 20 ft>s dt 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., t = 20 s; v = NOTE: Show that a = 2 ft>s2 when t = 5 s by measuring the slope of B_5389_ch12_ptg01_hr.4c.indd 23 v 2t v 20 20 10 30 t (s) (b) s 500 ft - 100 ft = = 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 v9t 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. the v9t graph. v (ft/s) a (ft/s2) 2 10 30 t (s) (c) Fig. 12–13 1/21/15 9:58 AM 24 12 CHAPTER 12 EXAMPLE K I N E M AT I C S PARTICLE 12.7 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 v9t and s–t graphs and determine the time t needed to stop the car. How far has the car traveled? a (m/s2) 10 A1 t¿ A2 10 2 OF A t (s) SOLUTION v–t Graph. Since dv = a dt, the v9t 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 a = (10) m>s2; 10 s 6 t … t; a = (-2) m>s2; dv = t v L100 m>s L10 s dv = -2 dt, v = ( -2t + 120) m>s When t = t we require v = 0. This yields, Fig. 12–14b, t = 60 s Ans. 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 = A1 + A2 , Fig. 12–14a. Thus 0 = 10 m>s2(10 s) + (-2 m>s2)(t - 10 s) v (m/s) v 10t 100 v 2t 120 10 t v 10 dt, v = 10t L0 L0 When t = 10 s, v = 10(10) = 100 m>s. Using this as the initial condition for the next time period, we have 0 … t 6 10 s; (a) t¿ 60 t (s) (b) t = 60 s Ans. s–t Graph. Since ds = v dt, integrating the equations of the v9t graph yields the corresponding equations of the s–t graph. Using the initial condition s = 0 when t = 0, we have s t L0 L0 When t = 10 s, s = 5(10)2 = 500 m. Using this initial condition, 0 … t … 10 s; v = (10t) m>s; ds = s 3000 s 5t2 t 10 60 (c) Fig. 12–14 ds = s = -(60)2 + 120(60) - 600 = 3000 m The s–t graph is shown in Fig. 12–14c. s t2 120t 600 B_5389_ch12_ptg01_hr.4c.indd 24 s = (5t2) m (-2t + 120) dt L500 m L10 s s - 500 = -t2 + 120t - [-(10)2 + 120(10)] s = (-t2 + 120t - 600) m When t = 60 s, the position is 10 s … t … 60 s; v = (-2t + 120) m>s; s (m) 500 10t dt, t (s) Ans. NOTE: A direct solution for s is possible when t = 60 s, since the triangular area under the v9t graph would yield the displacement s = s - 0 from t = 0 to t = 60 s. Hence, s = 12(60 s)(100 m>s) = 3000 m Ans. 1/21/15 9:58 AM 12.3 EXAMPLE RECTILINEAR KINEMATICS: ERRATIC MOTION 25 12.8 12 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. 0 … s 6 200 ft; v = (0.2s + 10) ft>s dv d = (0.2s + 10) (0.2s + 10) = 0.04s + 2 ds ds 200 ft 6 s … 400 ft; v = 50 ft>s a = v v (ft/s) 50 v 0.2s 10 v 50 10 200 400 s (ft) (a) dv d a = v = (50) (50) = 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 t s ds dt = 0.2s + 10 L0 L0 t = (5 ln(0.2s + 10) - 5 ln 10) s a (ft/s2) a 0.04s 2 10 2 a0 200 400 s (ft) (b) Fig. 12–15 At s = 200 ft, t = 5 ln[0.2(200) + 10] - 5 ln 10 = 8.05 s. Therefore, using these initial conditions for the second segment of motion, ds ds 200 ft 6 s … 400 ft; v = 50 ft>s; dt = = v 50 t s ds dt = ; L8.05 s L200 m 50 s s t - 8.05 = - 4; t = a + 4.05b s 50 50 Therefore, at s = 400 ft, 400 t = + 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 = v(dv>ds) = 10(50 - 10)>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 (a = 0). B_5389_ch12_ptg01_hr.4c.indd 25 1/21/15 9:58 AM 26 CHAPTER 12 K I N E M AT I C S OF A PARTICLE PRELIMINARY PROBLEM 12 P12–2. e) Draw the v–t graph if v = 0 when t = 0. Find the equation v = f(t) for each segment. a) Draw the s–t and a–t graphs if s = 0 when t = 0. v (m/s) v 2t 4 a (m/s2) 2 4 t (s) 2 2 t (s) 2 b) Draw the a–t and v–t graphs. s (m) 2 f) Determine v at s = 2 m if v = 1 m>s at s = 0. s 2t 2 a (m/s) t (s) 1 4 c) Draw the v–t and s–t graphs if v = 0, s = 0 when t = 0. a (m/s2) s (m) 2 2 t (s) 2 g) Determine a at s = 1 m. d) Determine s and a when t = 3 s if s = 0 when t = 0. v (m/s) v (m/s) 4 2 2 4 t (s) Prob. P12–2 B_5389_ch12_ptg01_hr.4c.indd 26 2 s (m) 1/21/15 9:58 AM 12.3 27 RECTILINEAR KINEMATICS: ERRATIC MOTION FUNDAMENTAL PROBLEMS F12–9. The particle travels along a straight track such that its position is described by the s9t graph. Construct the v9t graph for the same time interval. s (m) 12 F12–12. The sports car travels along a straight road such that its acceleration is described by the graph. Construct the v9s graph for the same interval and specify the velocity of the car when s = 10 m and s = 15 m. a (m/s2) 10 s 108 108 s 0.5 t3 0 6 8 10 5 t (s) Prob. F12–9 F12–10. A van travels along a straight road with a velocity described by the graph. Construct the s9t and a9t graphs during the same period. Take s = 0 when t = 0. v (ft/s) 10 s (m) 15 Prob. F12–12 F12–13. The dragster starts from rest and has an acceleration described by the graph. Construct the v9t 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) Prob. F12–10 F12–11. A bicycle travels along a straight road where its velocity is described by the v9s graph. Construct the a9s graph for the same interval. v (m/s) t (s) 5 10 Prob. F12–13 F12–14. The dragster starts from rest and has a velocity described by the graph. Construct the s9t 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 0.25 s v 15 t 225 40 Prob. F12–11 B_5389_ch12_ptg01_hr.4c.indd 27 s (m) 5 15 t (s) Prob. F12–14 1/21/15 9:58 AM 28 CHAPTER 12 K I N E M AT I C S OF A PARTICLE PROBLEMS 12 12–35. 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. 12–41. The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft>s2 and then decelerate at 2 ft>s2. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a–t, v–t, and s–t graphs for the motion. *12–36. The s–t graph for a train has been experimentally determined. From the data, construct the v–t and a–t graphs for the motion; 0 … t … 40 s. For 0 … t … 30 s, the curve is s = (0.4t2) m, and then it becomes straight for t Ú 30 s. s (m) 600 40 ft 360 30 40 t (s) Prob. 12–41 Prob. 12–36 12–37. Two rockets start from rest at the same elevation. Rocket A accelerates vertically at 20 m>s2 for 12 s and then maintains a constant speed. Rocket B accelerates at 15 m>s2 until reaching a constant speed of 150 m>s. Construct the a–t, v–t, and s–t graphs for each rocket until t = 20 s. What is the distance between the rockets when t = 20 s? 12–38. A particle starts from s = 0 and travels along a straight line with a velocity v = (t2 - 4t + 3) m>s, where t is in seconds. Construct the v–t and a–t graphs for the time interval 0 … t … 4 s. 12–39. If the position of a particle is defined by s = [2 sin (p>5)t + 4] m, where t is in seconds, construct the s9t, v9t, and a9t graphs for 0 … t … 10 s. *12–40. An airplane starts from rest, travels 5000 ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi>h. It then climbs in a straight line with a uniform acceleration of 3 ft>s2 until it reaches a constant speed of 220 mi>h. Draw the s–t, v–t, and a–t graphs that describe the motion. B_5389_ch12_ptg01_hr.4c.indd 28 12–42. The velocity of a car is plotted as shown. Determine the total distance the car moves until it stops (t = 80 s). Construct the a–t graph. v (m/s) 10 40 80 t (s) Prob. 12–42 1/21/15 9:58 AM 12.3 12–43. The motion of a jet plane just after landing on a runway is described by the a–t graph. Determine the time t when the jet plane stops. Construct the v–t and s–t graphs for the motion. Here s = 0, and v = 300 ft>s when t = 0. a (m/s2) 29 RECTILINEAR KINEMATICS: ERRATIC MOTION 12–46. The a–s graph for a rocket moving along a straight track has been experimentally determined. If the rocket 12 starts at s = 0 when v = 0, determine its speed when it is at s = 75 ft, and 125 ft, respectively. Use Simpson’s rule with n = 100 to evaluate v at s = 125 ft. a (ft/s2) 10 20 t¿ t (s) 10 a 5 6(s 10)5/3 20 5 Prob. 12–43 s (ft) 100 *12–44. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude of 4 m>s2. If the plates are spaced 200 mm apart, determine the maximum velocity vmax and the time t for the particle to travel from one plate to the other. Also draw the s–t graph. When t = t>2 the particle is at s = 100 mm. 12–45. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where t = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs for the particle. When t = t>2 the particle is at s = 0.5 m. Prob. 12–46 12–47. A two-stage rocket is fired vertically from rest at s = 0 with the acceleration as shown. After 30 s the first stage, A, burns out and the second stage, B, ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 60 s. a (m/s2) B smax 24 A v s vmax t¿/2 Probs. 12–44/45 B_5389_ch12_ptg01_hr.4c.indd 29 12 t¿ t 30 60 t (s) Prob. 12–47 1/21/15 9:58 AM 30 CHAPTER 12 K I N E M AT I C S OF A PARTICLE *12–48. The race car starts from rest and travels along a 12 straight road until it reaches a speed of 26 m>s in 8 s as shown on the v–t graph. The flat part of the graph is caused by shifting gears. Draw the a–t graph and determine the maximum acceleration of the car. 12–50. The car starts from rest at s = 0 and is subjected to an acceleration shown by the a–s graph. Draw the v–s graph and determine the time needed to travel 200 ft. a (ft/s2) v (m/s) 6 12 26 v 4t 6 a 0.04s 24 6 14 v 3.5t 300 4 5 t (s) 8 s (ft) 450 Prob. 12–50 Prob. 12–48 12–49. The jet car is originally traveling at a velocity of 10 m>s when it is subjected to the acceleration shown. Determine the car’s maximum velocity and the time t when it stops. When t = 0, s = 0. 12–51. The v–t graph for a train has been experimentally determined. From the data, construct the s–t and a–t graphs for the motion for 0 … t … 180 s. When t = 0, s = 0. v (m/s) a (m/s2) 10 6 6 t¿ 15 t (s) 4 60 Prob. 12–49 B_5389_ch12_ptg01_hr.4c.indd 30 120 180 t (s) Prob. 12–51 1/21/15 9:58 AM 12.3 *12–52. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the total distance the motorcycle travels until it stops when t = 15 s. Also plot the a–t and s–t graphs. 31 RECTILINEAR KINEMATICS: ERRATIC MOTION 12–55. An airplane lands on the straight runway, originally traveling at 110 ft>s when s = 0. If it is subjected to the 12 decelerations shown, determine the time t needed to stop the plane and construct the s–t graph for the motion. 12–53. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the motorcycle’s acceleration and position when t = 8 s and t = 12 s. a (ft/s2) 5 v (m/s) 5 v 1.25t 15 20 t¿ t (s) 3 v5 8 v t 15 Prob. 12–55 4 10 15 t (s) Probs. 12–52/53 12–54. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the s–t and a–t graphs. Also determine the average speed and the distance traveled for the 15-s time interval. When t = 0, s = 0. *12–56. Starting from rest at s = 0, a boat travels in a straight line with the acceleration shown by the a–s graph. Determine the boat’s speed when s = 50 ft, 100 ft, and 150 ft. 12–57. Starting from rest at s = 0, a boat travels in a straight line with the acceleration shown by the a–s graph. Construct the v–s graph. a (ft/s2) v (m/s) 8 15 6 v 0.6t 2 5 15 Prob. 12–54 B_5389_ch12_ptg01_hr.4c.indd 31 t (s) 100 150 s (ft) Probs. 12–56/57 1/21/15 9:58 AM 32 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–58. A two-stage rocket is fired vertically from rest with 12 the acceleration shown. After 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 motion of the second stage for 0 … t … 40 s. 12–62. If the position of a particle is defined as s = (5t - 3t2) ft, where t is in seconds, construct the s–t, v–t, and a–t graphs for 0 … t … 10 s. 12–63. From experimental data, the motion of a jet plane while traveling along a runway is defined by the v–t graph. Construct the s–t and a–t graphs for the motion. When t = 0, s = 0. a (m/s2) B A 20 15 v (m/s) 60 15 40 t (s) Prob. 12–58 20 5 20 30 t (s) Prob. 12–63 12–59. The speed of a train during the first minute has been recorded as follows: t 1s2 v 1m>s2 0 20 40 60 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. *12–60. A man riding upward in a freight elevator accidentally drops a package off the elevator when it is 100 ft from the ground. If the elevator maintains a constant upward speed of 4 ft>s, determine how high the elevator is from the ground the instant the package hits the ground. Draw the v–t curve for the package during the time it is in motion. Assume that the package was released with the same upward speed as the elevator. 12–61. Two cars start from rest side by side and travel along a straight road. Car A accelerates at 4 m>s2 for 10 s and then maintains a constant speed. Car B accelerates at 5 m>s 2 until reaching a constant speed of 25 m>s and then maintains this speed. Construct the a–t, v–t, and s–t graphs for each car until t = 15 s. What is the distance between the two cars when t = 15 s? B_5389_ch12_ptg01_hr.4c.indd 32 *12–64. The motion of a train is described by the a–s graph shown. Draw the v–s graph if v = 0 at s = 0. a (m/s2) 3 300 600 s (m) Prob. 12–64 1/21/15 9:58 AM 12.3 12–65. The jet plane starts from rest at s = 0 and is subjected to the acceleration shown. Determine the speed of the plane when it has traveled 1000 ft. Also, how much time is required for it to travel 1000 ft? 33 RECTILINEAR KINEMATICS: ERRATIC MOTION 12–67. The v–s graph of a cyclist traveling along a straight road is shown. Construct the a–s graph. 12 v (ft/s) a (ft/s2) 75 a 75 0.025s 15 50 v 0.04 s 19 v 0.1s 5 5 s (ft) 1000 s (ft) 100 Prob. 12–65 350 Prob. 12–67 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. *12–68. The v–s graph for a test vehicle is shown. Determine its acceleration when s = 100 m and when s = 175 m. v (m/s) v (m/s) 80 v 0.2s 50 2 v 4s 20 s (m) 100 400 Prob. 12–66 B_5389_ch12_ptg01_hr.4c.indd 33 150 200 s (m) Prob. 12–68 1/23/15 2:23 PM 34 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.4 12 General Curvilinear Motion 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. s s r O Position Path (a) Position. Consider a particle located at a point on a space curve 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 = r(t). 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. s s Velocity. During the time t, the average velocity of the particle is r r¿ vavg = r O Displacement (b) The instantaneous velocity is determined from this equation by letting t S 0, and consequently the direction of r approaches the tangent to the curve. Hence, v = lim (r> t) or S 0 t v = v s r O 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 S 0, we have v = lim (r> t) = lim (s> t), or S S 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. B_5389_ch12_ptg01_hr.4c.indd 34 1/21/15 9:58 AM 12.4 35 GENERAL CURVILINEAR MOTION Acceleration. If the particle has a velocity v at time t and a velocity v 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 S 0 in the above equation. In the limit v will approach the tangent to the hodograph, and so a = lim (v> t), or S t v v v¿ O¿ (e) 0 a = dv dt (12–9) Hodograph v a O¿ Substituting Eq. 12–7 into this result, we can also write a = d2r 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. B_5389_ch12_ptg01_hr.4c.indd 35 (f) a Acceleration path (g) Fig. 12–16 1/21/15 9:58 AM 36 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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. 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 z 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 = x(t), y = y(t), z = z(t), so that r = r(t). At any instant the magnitude of r is defined from Eq. B–3 in Appendix B as (a) r = 2x2 + y2 + z2 And the direction of r is specified by the unit vector ur = r>r. z Velocity. The first time derivative of r yields the velocity of the particle. Hence, s v = v vxi vyj vzk y x Velocity dr d d d = (xi) + (yj) + (zk) dt dt dt dt 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 (b) d dx di (xi) = i + x dt dt dt Fig. 12–17 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 B_5389_ch12_ptg01_hr.4c.indd 36 (12–12) 1/21/15 9:58 AM 12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 37 # # # The “dot” notation x, y, z represents the first time derivatives of x = x(t), y = y(t), z = z(t), respectively. The velocity has a magnitude that is found from 12 v = 2v2x + 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 y x dv a = = ax i + ay j + az k dt (12–13) # $ ax = vx = x # $ ay = vy = y # $ az = vz = z (12–14) Acceleration (c) where Here ax , ay , az represent, respectively, the first time derivatives of vx = vx(t), vy = vy(t), vz = vz(t), or the second time derivatives of the functions x = x(t), y = y(t), z = z(t). The acceleration has a magnitude a = 2a2x + 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. B_5389_ch12_ptg01_hr.4c.indd 37 1/21/15 9:58 AM 38 CHAPTER 12 12 K I N E M AT I C S OF A PARTICLE 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. B_5389_ch12_ptg01_hr.4c.indd 38 1/21/15 9:58 AM 12.5 EXAMPLE 12.9 12 At any instant the horizontal position of the weather balloon in Fig. 12–18a is defined by x = (8t) 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. 39 CURVILINEAR MOTION: RECTANGULAR COMPONENTS x2 10 The velocity component in the x direction is x A d # vx = x = (8t) = 8 ft>s S dt 16 ft To find the relationship between the velocity components we will use the chain rule of calculus. When t = 2 s, x = 8122 = 16 ft, Fig. 12–18a, and so (a) d # # vy = y = (x2 >10) = 2xx >10 = 2(16)(8)>10 = 25.6 ft>s c dt When t = 2 s, the magnitude of velocity is therefore v = 2(8 ft>s)2 + (25.6 ft>s)2 = 26.8 ft>s Ans. The direction is tangent to the path, Fig. 12–18b, where uv = tan -1 vy vx = tan -1 25.6 8 v 26.8 ft/s = 72.6 Ans. uv 72.6 B Acceleration. The relationship between the acceleration components is determined using the chain rule. (See Appendix C.) We have (b) d # ax = vx = (8) = 0 dt d # # # # $ ay = vy = (2xx >10) = 2(x)x >10 + 2x(x)>10 dt Thus, = 2(8)2 >10 + 2(16)(0)>10 = 12.8 ft>s2 c a = 2(0)2 + (12.8)2 = 12.8 ft>s2 Ans. a 12.8 ft/s2 ua 90 B The direction of a, as shown in Fig. 12–18c, is (c) 12.8 ua = tan-1 = 90 0 Ans. Fig. 12–18 NOTE: It is also possible to obtain vy and ay by first expressing y = f (t) = (8t)2 >10 = 6.4t2 and then taking successive time derivatives. B_5389_ch12_ptg01_hr.4c.indd 39 1/21/15 9:58 AM 40 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.10 EXAMPLE 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 upward velocity of 10 m>s, determine the magnitudes of the velocity and acceleration of the plane when it reaches an altitude of y = 100 m. y x SOLUTION When y = 100 m, then 100 = 0.001x2 or x = 316.2 m. Also, due to constant velocity vy = 10 m>s, so (© R.C. Hibbeler) y = vy t; 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 y = 0.001x2 d # # vy = y = (0.001x 2) = (0.002x)x = 0.002 xvx dt y (1) Thus y 0.001x2 10 m>s = 0.002(316.2 m)(vx) vx = 15.81 m>s 100 m x The magnitude of the velocity is therefore v = 2v2x + v2y = 2(15.81 m>s)2 + (10 m>s)2 = 18.7 m>s (a) Ans. Acceleration. Using the chain rule, the time derivative of Eq. (1) gives the relation between the acceleration components. # # # $ ay = vy = (0.002x)x + 0.002x(x) = 0.002(vx2 + xax) y # When x = 316.2 m, vx = 15.81 m>s , vy = ay = 0, vy a 100 m 0 = 0.002 3 (15.81 m>s)2 + 316.2 m(ax)4 ax = -0.791 m>s2 v vx x (b) Fig. 12–19 The magnitude of the plane’s acceleration is therefore a = 2a2x + a2y = 2(-0.791 m>s2)2 + (0 m>s2)2 = 0.791 m>s2 Ans. These results are shown in Fig. 12–19b. B_5389_ch12_ptg01_hr.4c.indd 40 1/21/15 9:58 AM 12.6 12.6 MOTION OF A PROJECTILE 41 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 (v0)x and (v0)y , 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 ag 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 + ) (S + ) (S + ) (S v = v0 + act; x = x0 + v0t + 12 act2; v2 = v20 + 2ac(x - x0); vx = (v0)x x = x0 + (v0)xt vx = (v0)x 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 (+ c ) (+ c ) (+ c ) v = v0 + act; 1 2 2 act ; y = y0 + v0t + v2 = v20 + 2ac(y - y0); vy = (v0)y - gt y = y0 + (v0)yt - 12 gt2 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. (© R.C. Hibbeler) v2y = (v0)2y - 2g(y - y0) 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. B_5389_ch12_ptg01_hr.4c.indd 41 1/21/15 9:58 AM 42 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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 Once thrown, the basketball follows a parabolic trajectory. (© R.C. Hibbeler) 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 = (v0)x , and x = x0 + (v0)x t Vertical Motion. • In the vertical or y direction only two of the following three equations can be used for solution. vy = (v0)y + ac t y = y0 + (v0)y t + 12 ac t2 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. (© R.C. Hibbeler) B_5389_ch12_ptg01_hr.4c.indd 42 v2y = (v0)2y + 2ac( y - y0) For example, if the particle’s final velocity vy is not needed, then the first and third of these equations will not be useful. 1/21/15 9:58 AM 12.6 EXAMPLE 12.11 MOTION OF A PROJECTILE 43 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 ag 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 (vA)x = 12 m>s and (vA)y = 0. Also, between points A and B the acceleration is ay = -9.81 m>s2. Since (vB)x = (vA)x = 12 m>s, the three unknowns are (vB)y , R, and the time of flight tAB . Here we do not need to determine (vB)y . 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 (+ c ) yB = yA + (vA)ytAB + 12 act2AB -6 m = 0 + 0 + 12(-9.81 m>s2)t2AB tAB = 1.11 s Horizontal Motion. as follows: + ) (S Ans. Since tAB has been calculated, R is determined xB = xA + (vA)xtAB R = 0 + 12 m>s (1.11 s) 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. B_5389_ch12_ptg01_hr.4c.indd 43 1/21/15 9:58 AM 44 12 CHAPTER 12 EXAMPLE K I N E M AT I C S OF A PARTICLE 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 (vA)y . [Note that (vA)x = (vO)x .] With the origin of coordinates at O, Fig. 12–22, the initial velocity of a chip has components of (vO)x = (25 cos 30) ft>s = 21.65 ft>s S (vO)y = (25 sin 30) ft>s = 12.5 ft>s c Also, (vA)x = (vO)x = 21.65 ft>s and ay = -32.2 ft>s2. Since we do not need to determine (vA)y , we have Horizontal Motion. + ) (S xA = xO + (vO)xtOA 20 ft = 0 + (21.65 ft>s)tOA tOA = 0.9238 s Vertical Motion. Relating tOA to the initial and final elevations of a chip, we have (+ c ) yA = yO + (vO)ytOA + 12 ac t2OA (h - 4 ft) = 0 + (12.5 ft>s)(0.9238 s) + 12(-32.2 ft>s2)(0.9238 s)2 h = 1.81 ft Ans. NOTE: We can determine (vA)y by using (vA)y = (vO)y + actOA . B_5389_ch12_ptg01_hr.4c.indd 44 1/21/15 9:58 AM 12.6 EXAMPLE 45 MOTION OF A PROJECTILE 12.13 12 (© R.C. Hibbeler) 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 (vB)y . y 30 A C Vertical Motion. Since the time of flight and the vertical distance between the ends of the path are known, we can determine vA . yB = yA + (vA)ytAB + 12 act2AB (+ c ) -1 m = 0 + vA sin 30(1.5 s) + 12(-9.81 m>s2)(1.5 s)2 vA = 13.38 m>s = 13.4 m>s Ans. Horizontal Motion. The range R can now be determined. + ) (S xB = xA + (vA)xtAB x h 1m B R (b) Fig. 12–23 R = 0 + 13.38 cos 30 m>s (1.5 s) = 17.4 m Ans. 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 (vC)y = 0, and since vA is known, we can determine h directly without considering tAC using the following equation. (vC)2y = (vA)2y + 2ac[ yC - yA] 02 = (13.38 sin 30 m>s)2 + 2(-9.81 m>s2)[(h - 1 m) - 0] h = 3.28 m Ans. NOTE: Show that the bike will strike the ground at B with a velocity having components of (vB)x = 11.6 m>s S , (vB)y = 8.02 m>s T B_5389_ch12_ptg01_hr.4c.indd 45 1/21/15 9:58 AM 46 CHAPTER 12 K I N E M AT I C S OF A PARTICLE PRELIMINARY PROBLEMS 12 P12–3. Use the chain-rule and find y· and ÿ in terms of x, x· and ẍ if a) The particle travels from A to B. Identify the three unknowns, and write the three equations needed to solve for them. P12–5. y = 4x2 y 10 m/s b) y = 3ex 30 A 8m c) y = 6 sin x B x Prob. P12–5 The particle travels from A to B. Identify the three unknowns, and write the three equations needed to solve for them. P12–4. The particle travels from A to B. Identify the three unknowns, and write the three equations needed to solve for them. P12–6. y y 40 m/s A B 60 m/s B 20 m Prob. P12–4 B_5389_ch12_ptg01_hr.4c.indd 46 20 x A x tAB 5 s Prob. P12–6 1/21/15 9:58 AM 12.6 47 MOTION OF A PROJECTILE 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), if 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. 12 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 Prob. F12–18 3m x x 8t F12–19. A particle is traveling along the parabolic path y = 0.25x2. If x = 8 m, vx = 8 m>s, and ax = 4 m>s2 when t = 2 s, determine the magnitude of the particle’s velocity and acceleration at this instant. y 4m Prob. F12–16 y 0.25x2 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 x Prob. F12–19 F12–20. 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. y2 4x y x y 0.05 x2 x x (4t4) m Prob. F12–17 B_5389_ch12_ptg01_hr.4c.indd 47 Prob. F12–20 1/21/15 9:58 AM 48 CHAPTER 12 K I N E M AT I C S OF A PARTICLE F12–21. The ball is kicked from point A with the initial 12 velocity vA = 10 m>s. Determine the maximum height h it reaches. F12–25. A ball is thrown from A. If it is required to clear the wall at B, determine the minimum magnitude of its 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 8 ft x x 3 ft x Prob. 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 basket at B. y B vA 30 A Prob. 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 Prob. 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 Prob. F12–24 B_5389_ch12_ptg01_hr.4c.indd 48 Prob. F12–26 1/21/15 9:58 AM 12.6 49 MOTION OF A PROJECTILE PROBLEMS 12 12–69. If the velocity of a particle is defined as v(t) = {0.8t2i + 12t1 > 2j + 5k} m >s, determine the magnitude and coordinate direction angles a, b, g of the particle’s acceleration when t = 2 s. 12–70. The velocity of a particle is v = 5 3i + (6 - 2t)j 6 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. 12–74. A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6t i + 12t2 k} ft>s2. Determine the particle’s position (x, y, z) when t = 2 s. 12–75. A particle travels along the curve from A to B in 2 s. It takes 4 s for it to go from B to C and then 3 s to go from C to D. Determine its average speed when it goes from A to D. y D B 12–71. A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = 5 6t i + 12 t 2k 6 ft>s2. Determine the particle’s position (x, y, z) at t = 1 s. 5m 15 m C 10 m x A *12–72. The velocity of a particle is given by v = 5 16t 2 i + 4t 3j + (5t + 2)k 6 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? 12–73. The water sprinkler, positioned at the base of a hill, releases a stream of water with a velocity of 15 ft>s as shown. Determine the point B(x, y) where the water strikes the ground on the hill. Assume that the hill is defined by the equation y = (0.05x2) ft and neglect the size of the sprinkler. y Prob. 12–75 *12–76. A particle travels along the curve from A to B in 5 s. It takes 8 s for it to go from B to C and then 10 s to go from C to A. Determine its average speed when it goes around the closed path. y B y (0.05x2) ft 15 ft/s B 20 m 60 A x Prob. 12–73 B_5389_ch12_ptg01_hr.4c.indd 49 30 m C x Prob. 12–76 1/21/15 9:58 AM 50 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–77. The position of a crate sliding down a ramp is given > 12 by x = (0.25t3) m, y = (1.5t2) m, z = (6 − 0.75t5 2) m, where t is in seconds. Determine the magnitude of the crate’s velocity and acceleration when t = 2 s. 12–81. A particle travels along the curve 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. 12–78. A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y2 = [120(103)x] m. 1 If the x component of acceleration is ax = a t2 b m>s2, 4 where t is in seconds, determine the magnitude of the rocket’s velocity and acceleration when t = 10 s. y 12–79. The particle travels along the path defined by the parabola y = 0.5x 2. If the component of velocity along the x axis is vx = (5t) ft>s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0. 30 C 45 30 m B x A y Prob. 12–81 y 0.5x2 12–82. 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. x O Prob. 12–79 z *12–80. 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. y v0 p x) y c sin ( –– L x c L Prob. 12–80 B_5389_ch12_ptg01_hr.4c.indd 50 y c L x Prob. 12–82 1/21/15 9:58 AM 12.6 12–83. 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. 51 MOTION OF A PROJECTILE 12–86. Determine the minimum initial velocity v0 and the corresponding angle u0 at which the ball must be kicked in 12 order for it to just cross over the 3-m high fence. y v0 A C B 3m u0 D x 6m v 10 m/s Prob. 12–86 x2 v2 1 4 Prob. 12–83 *12–84. The van travels over the hill described by y = (- 1.5(10 - 3) x 2 + 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. 12–87. The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity vA at which it was launched, the angle of release u, and the height h. y B y (1.5 (103) x2 15) ft 15 ft x h vA 100 ft A Prob. 12–84 u 3.5 ft 12–85. The flight path of the helicopter as it takes off from A is defined by the parametric equations x = (2t2) m and y = (0.04t3) m, where t is the time in seconds. Determine the distance the helicopter is from point A and the magnitudes of its velocity and acceleration when t = 10 s. y 18 ft Prob. 12–87 *12–88. Neglecting the size of the ball, determine the magnitude vA of the basketball’s initial velocity and its velocity when it passes through the basket. B 30 A A x Prob. 12–85 B_5389_ch12_ptg01_hr.4c.indd 51 vA 3m P 10 m Prob. 12–88 1/21/15 9:58 AM 52 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–89. The girl at A can throw a ball at vA = 10 m>s. 12 Calculate the maximum possible range R = Rmax and the associated angle u at which it should be thrown. Assume the ball is caught at B at the same elevation from which it is thrown. 12–93. A golf ball is struck with a velocity of 80 ft>s as shown. Determine the distance d to where it will land. 12–94. A golf ball is struck with a velocity of 80 ft>s as shown. Determine the speed at which it strikes the ground at B and the time of flight from A to B. 12–90. Show that the girl at A can throw the ball to the boy at B by launching it at equal angles measured up or down from a 45° inclination. If vA = 10 m >s, determine the range R if this value is 15°, i.e., u1 = 45° − 15° = 30° and u2 = 45° + 15° = 60°. Assume the ball is caught at the same elevation from which it is thrown. vA 80 ft/s vA 10 m/s B u A B A 45 10 d R Probs. 12–93/94 Probs. 12–89/90 12–91. The ball at A is kicked with a speed vA = 80 ft>s and at an angle uA = 30°. Determine the point (x, –y) where it strikes the ground. Assume the ground has the shape of a parabola as shown. *12–92. The ball at A is kicked such that uA = 30. If it strikes the ground at B having coordinates x = 15 ft, y = - 9 ft, determine the speed at which it is kicked and the speed at which it strikes the ground. 12–95. The basketball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Neglecting the size of the ball, determine the magnitude vA of its initial velocity and the height h of the ball when it passes over player B. y vA A uA x C 30 vA y A B x Probs. 12–91/92 B_5389_ch12_ptg01_hr.4c.indd 52 B h 7 ft 10 ft y 0.04x2 25 ft 5 ft Prob. 12–95 1/21/15 9:59 AM 12.6 *12–96. 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 . 12–97. 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 speed at which he strikes the ground. 53 MOTION OF A PROJECTILE 12–99. The missile at A takes off from rest and rises vertically to B, where its fuel runs out in 8 s. If the 12 acceleration varies with time as shown, determine the missile’s height hB and speed vB. If by internal controls the missile is then suddenly pointed 45° as shown, and allowed to travel in free flight, determine the maximum height attained, hC, and the range R to where it crashes at D. 45 vB C B vA uA A hC hB 4m D A 3 5 R 4 a (m/s2) 100 m 40 B t (s) 8 Prob. 12–99 Probs. 12–96/97 12–98. 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. *12–100. The projectile is launched with a velocity v0. Determine the range R, the maximum height h attained, and the time of flight. Express the results in terms of the angle u and v0. The acceleration due to gravity is g. y vA B 7.5 ft 3 ft C s v0 u 21 ft Prob. 12–98 B_5389_ch12_ptg01_hr.4c.indd 53 A h x R Prob. 12–100 1/21/15 9:59 AM 54 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–101. The drinking fountain is designed such that the 12 nozzle is located from the edge of the basin as shown. Determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C. *12–104. The man at A wishes to throw two darts at the target at B so that they arrive at the same time. If each dart is thrown with a speed of 10 m>s, determine the angles uC and uD at which they should be thrown and the time between each throw. Note that the first dart must be thrown at uC (7 uD), then the second dart is thrown at uD. vA 40 A 50 mm B 5m C C uC 250 mm A 100 mm uD D B Prob. 12–101 Prob. 12–104 12–102. If the dart is thrown with a speed of 10 m>s, determine the shortest possible time before it strikes the target. Also, what is the corresponding angle uA at which it should be thrown, and what is the velocity of the dart when it strikes the target? 12–103. If the dart is thrown with a speed of 10 m>s, determine the longest possible time when it strikes the target. Also, what is the corresponding angle uA at which it should be thrown, and what is the velocity of the dart when it strikes the target? 12–105. 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. 4m A vA 2m uA A vA B 1.5 m B x Probs. 12–102/103 B_5389_ch12_ptg01_hr.4c.indd 54 Prob. 12–105 1/21/15 9:59 AM 12.6 12–106. The snowmobile is traveling at 10 m>s when it leaves the embankment at A. Determine the time of flight from A to B and the range R of the trajectory. A 55 MOTION OF A PROJECTILE *12–108. The baseball player A hits the baseball at vA = 40 ft>s and uA = 60 from the horizontal. When the 12 ball is directly overhead of player B he begins to run under it. Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the ball was hit. 40 vA 40 ft/s uA A 3 B C vA 5 4 B d 15 ft R Prob. 12–108 Prob. 12–106 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. The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity vA at which it was launched, the angle of release u, and the height h. A B u vA 20 ft h vA B A u 3.5 ft 35 ft Prob. 12–107 B_5389_ch12_ptg01_hr.4c.indd 55 18 ft Prob. 12–109 1/21/15 9:59 AM 56 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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. ut t Position (a) O¿ O¿ r r ds r r ds r r O¿ ds Radius of curvature (b) Velocity. Since the particle moves, s is a function of time. As 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 = s(t), i.e., v = ds>dt (Eq. 12–8). Hence O¿ r r v Velocity (c) 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 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.* v = vut (12–15) # v = s (12–16) where 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. B_5389_ch12_ptg01_hr.4c.indd 56 1/21/15 9:59 AM 12.7 57 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS Acceleration. The acceleration of the particle is the time rate of change of the velocity. Thus, # # # a = v = vut + vut 12 O¿ du r (12–17) r un # 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 ut= , Fig. 12–24d. As shown in Fig. 12–24e, we require ut= = ut + dut . Here dut stretches between the arrowheads of ut and ut= , which lie on an infinitesimal arc of radius ut = 1. Hence, dut has a magnitude of dut = (1) 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 ds u¿t ut (d) # # s v # u t = uu n = u n = u n r r Substituting into Eq. 12–17, a can be written as the sum of its two components, a = atut + anun un du u¿ t (12–18) dut ut (e) where # at = v or at ds = v dv (12–19) and O¿ an = v2 r (12–20) an P These two mutually perpendicular components are shown in Fig. 12–24f. Therefore, the magnitude of acceleration is the positive value of a at Acceleration (f) a = 2a2t + a2n B_5389_ch12_ptg01_hr.4c.indd 57 (12–21) Fig. 12–24 (cont.) 1/21/15 9:59 AM 58 CHAPTER 12 K I N E M AT I C S OF A PARTICLE To better understand these results, consider the following two special cases of motion. 12 n t v an at As the boy swings upward with a velocity v, his motion can be analyzed using n–t coordinates. As he rises, the magnitude of his velocity (speed) is decreasing, and so at will be negative. The rate at which the direction of his velocity changes is an, which is always positive, that is, towards the center of rotation. (© R.C. Hibbeler) 1. If the particle moves along a straight line, then r S 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 an 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 b osculating plane O n s O¿ ub un ut Fig. 12–26 B_5389_ch12_ptg01_hr.4c.indd 58 t 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, no motion occurs in the ub direction, and if 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. 1/21/15 9:59 AM 12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 59 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. Once the rotation is constant, the riders will then have only a normal component of acceleration. (© R.C. Hibbeler) # 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 = (at)c , the above equations, when integrated, yield s = s0 + v0t + 12(at)ct2 v = v0 + (at)ct v2 = v20 + 2(at)c(s - s0) 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 = f(x), the radius of curvature r at any point on the path is determined from the equation r = [1 + (dy>dx)2]3>2 d2y>dx2 The derivation of this result is given in any standard calculus text. B_5389_ch12_ptg01_hr.4c.indd 59 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. (© R.C. Hibbeler) 1/21/15 9:59 AM 60 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.14 EXAMPLE 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 1 2 1 the 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. 12–27b. Therefore, vA = 6 m>s 45 d Ans. # The acceleration is determined from a = vut + (v2 >r)un . However, it is first necessary to determine the radius of curvature of the path at A 1 (10 m, 5 m). Since d2y>dx2 = 10 , then y 1 x2 20 y n r = u vA A t x (a) d2y>dx2 The acceleration becomes 5m 10 m [1 + (dy>dx)2]3>2 = 31 + 1 101 x 2 2 4 3>2 1 10 ` x = 10 m = 28.28 m v2 # aA = vut + u r n (6 m>s)2 = 2ut + u 28.28 m n = 5 2ut + 1.273un 6 m>s2 As shown in Fig. 12–27b, n 1.273 m/s2 90 45 f a 2 m/s2 t a = 2(2 m>s2)2 + (1.273 m>s2)2 = 2.37 m>s2 2 f = tan-1 = 57.5 1.273 Thus, 45 + 90 + 57.5 - 180 = 12.5 so that, a = 2.37 m>s2 12.5 d (b) Fig. 12–27 B_5389_ch12_ptg01_hr.4c.indd 60 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. 1/21/15 9:59 AM 12.7 EXAMPLE CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 12.15 61 12 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 = 2a2t + a2n . Here at = 7 ft>s2. Since an = v2 >r, the velocity as a function of time must be determined first. v = v0 + (at)ct v = 0 + 7t Thus (7t)2 v2 = 0.163t2 ft>s2 an = = r 300 The time needed for the acceleration to reach 8 ft>s2 is therefore a = 2a2t + a2n 8 ft>s2 = 2(7 ft>s2)2 + (0.163t2)2 Solving for the positive value of t yields 0.163t2 = 2(8 ft>s2)2 - (7 ft>s2)2 t = 4.87 s Ans. Velocity. The speed at time t = 4.87 s is v = 7t = 7(4.87) = 34.1 ft>s Ans. NOTE: Remember the velocity will always be tangent to the path, whereas the acceleration will be directed within the curvature of the path. B_5389_ch12_ptg01_hr.4c.indd 61 1/21/15 9:59 AM 62 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.16 EXAMPLE (© R.C. Hibbeler) 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 = (0.2t) m>s2, where t is in seconds, determine the magnitude of its acceleration when it arrives at point B. (a) A 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 # at = v = 0.2t (1) s L0 3m v dv = L0 t 0.2t dt v = 0.1t2 2m n t (2) The time needed for the box to reach point B can be determined by realizing that the position of B is sB = 3 + 2p(2)>4 = 6.142 m, Fig. 12–29b, and since sA = 0 when t = 0 we have ds v = = 0.1t2 dt L0 B (b) 6.142 m tB 0.1t2dt L0 6.142 m = 0.0333t3B ds = tB = 5.690s Substituting into Eqs. 1 and 2 yields # (aB)t = vB = 0.2(5.690) = 1.138 m>s2 n 2 5.242 m/s vB = 0.1(5.69)2 = 3.238 m>s aB At B, rB = 2 m, so that t B 1.138 m/s2 (c) Fig. 12–29 B_5389_ch12_ptg01_hr.4c.indd 62 (aB)n = (3.238 m>s)2 v2B = = 5.242 m>s2 rB 2m The magnitude of aB , Fig. 12–29c, is therefore aB = 2(1.138 m>s2)2 + (5.242 m>s2)2 = 5.36 m>s2 Ans. 1/21/15 9:59 AM 12.7 63 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS PRELIMINARY PROBLEM 12 P12–7. d) Determine the normal and tangential components of acceleration at s = 0 if v = (4s + 1) m > s, where s is in meters. a) Determine the acceleration at the instant shown. v 2 m/s v 3 m/s2 s 2m 1m # b) Determine the increase in speed and the normal component of acceleration at s = 2 m. At s = 0, v = 0. e) Determine the acceleration at s = 2 m if v = (2 s) m > s2, where s is in meters. At s = 0, v = 1 m > s. s s2m v 4 m/s2 3m 2m c) Determine the acceleration at the instant shown. The particle has a constant speed of 2 m > s. f. Determine the acceleration when t = 1 s if v = (4t2 + 2) m > s, where t is in seconds. v (4 t2 + 2) m/s y y 2 x2 6m x 2 m/s Prob. P12–7 B_5389_ch12_ptg01_hr.4c.indd 63 1/21/15 9:59 AM 64 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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. y v 0.0625t2 40 m y 1 x2 24 n 20 ft/s O Prob. F12–27 x F12–28. The car is traveling along the road with a speed of v = (2 s) m>s, where s is in meters. Determine the magnitude of its acceleration when s = 10 m. 10 ft Prob. F12–30 v (2s) 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 50 m n A 90 s O 300 m n Prob. F12–28 B 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. t Prob. 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. A 250 m y rB 300 m n 50 m B C A s 50 m t x O Prob. F12–29 B_5389_ch12_ptg01_hr.4c.indd 64 Prob. F12–32 1/21/15 9:59 AM 12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 65 PROBLEMS 12 12–110. An automobile is traveling on a 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. 12–111. 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. *12–112. A boat has an initial speed of 16 ft>s. If it then increases its speed along a circular path of radius r = 80 ft at # the rate of v = (1.5s) ft>s, where s is in feet, determine the time needed for the boat to travel s = 50 ft. 12–113. The position of a particle is defined by r = {4(t - sin t)i + (2t2 - 3)j} m, where t is in seconds and the argument for the sine is in radians. Determine the speed of the particle and its normal and tangential components of acceleration when t = 1 s. 12–114. The automobile has a speed of 80 ft>s at point A and an acceleration 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 12–115. The automobile is originally at rest at s = 0. If its # speed is increased by v = (0.05t 2) ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration when t = 18 s. *12–116. The automobile is originally at rest s = 0. If it # then starts to increase its speed at v = (0.05t 2) ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration at s = 550 ft. 300 ft s 240 ft Probs. 12–115/116 12–117. The two cars A and B travel along the circular path at constant speeds vA = 80 ft>s and vB = 100 ft>s, respectively. If they are at the positions shown when t = 0, determine the time when the cars are side by side, and the time when they are 90° apart. 12–118. Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 60 ft>s and is increasing its speed at the rate of 15 ft>s2 until it travels for a distance of 100p ft, after which it maintains a constant speed. Car B has a speed of 120 ft>s and is decreasing its speed at 15 ft>s2 until it travels a distance of 65p ft, after which it maintains a constant speed. Determine the time when they come side by side. vA A A u 30 a rA 400 ft rB 390 ft B n Prob. 12–114 B_5389_ch12_ptg01_hr.4c.indd 65 vB Probs. 12–117/118 1/23/15 2:23 PM 66 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–119. The satellite S travels around the earth in a 12 circular path with a constant speed of 20 Mm>h. If the acceleration is 2.5 m>s2, determine the altitude h. Assume the earth’s diameter to be 12 713 km. 12–121. 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 and x = 50 m. 12–122. 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 = 101.68 m and x = 0. S h y y 16 1 x 625 B s A Probs. 12–121/122 Prob. 12–119 *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. x 12–123. The motorcycle is traveling at 1 m>s when it is . at A. If the speed is then increased at v = 0.1 m>s2, determine its speed and acceleration at the instant t = 5 s. s 18 m y y 0.5x2 s ρ 30 m x A Prob. 12–120 B_5389_ch12_ptg01_hr.4c.indd 66 Prob. 12–123 1/21/15 9:59 AM 12.7 67 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS *12–124. 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 v = 8 m>s and the increase in speed is dv>dt = 4 m>s2. Determine the magnitude of the acceleration of the box at this instant. 12–127. At a given instant the train engine at E has a speed of 20 m>s and an acceleration of 14 m>s2 acting in the 12 direction shown. Determine the rate of increase in the train’s speed and the radius of curvature r of the path. y v 20 m/s 75 a 14 m/s2 A E y 0.4x2 x r 2m Prob. 12–124 12–125. The car travels around the circular track having a radius of r = 300 m such that when it is at point A it has a velocity of 5 m>s, which is increasing at the rate of . v = (0.06t) m>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration when it has traveled one-third the way around the track. 12–126. The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a velocity of 2 ft>s, which is increasing at the # rate of v = (0.002t) ft>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration when it has traveled three-fourths the way around the track. Prob. 12–127 *12–128. The car has an initial speed v0 = 20 m>s. If it increases its speed along the circular track at s = 0, at = (0.8s) m>s2, where s is in meters, determine the time needed for the car to travel s = 25 m. 12–129. The car starts from rest at s = 0 and increases its speed at at = 4 m>s2. Determine the time when the magnitude of acceleration becomes 20 m>s2. At what position s does this occur? y s r A Probs. 12–125/126 B_5389_ch12_ptg01_hr.4c.indd 67 x r 40 m Probs. 12–128/129 1/23/15 2:24 PM 68 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–130. A boat is traveling along a circular curve having a 12 radius of 100 ft. If its speed at t = 0 is 15 ft>s and is # increasing at v = (0.8t) ft>s2, determine the magnitude of its acceleration at the instant t = 5 s. *12–136. At a given instant the jet plane has a speed of 550 m>s and an acceleration of 50 m>s2 acting in the direction shown. Determine the rate of increase in the plane’s speed, and also the radius of curvature r of the path. 550 m/s 12–131. A boat is traveling along a circular path having a radius of 20 m. Determine the magnitude of the boat’s acceleration when the speed is v = 5 m>s and the rate of # increase in the speed is v = 2 m>s2. 70 a 50 m/s2 *12–132. Starting from rest, a bicyclist travels around a horizontal circular path, r = 10 m, at a speed of v = (0.09t2 + 0.1t) m>s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m. 12–133. A particle travels around a circular path having a radius of 50 m. If it is initially traveling with a speed of 10 m>s and its speed then increases at a rate of # v = (0.05 v) m>s2, determine the magnitude of the particle’s acceleration four seconds later. 12–134. The motorcycle is traveling at a constant speed of 60 km>h. Determine the magnitude of its acceleration when it is at point A. r Prob. 12–136 12–137. The ball is ejected horizontally from the tube with a speed of 8 m>s. Find the equation of the path, y = f(x), and then find the ball’s velocity and the normal and tangential components of acceleration when t = 0.25 s. y vA 8 m/s y x A 2 y 2x A x Prob. 12–137 25 m Prob. 12–134 12–135. When t = 0, the train has a speed of 8 m>s, which is increasing at 0.5 m>s2. Determine the magnitude of the acceleration of the engine when it reaches point A, at t = 20 s. Here the radius of curvature of the tracks is rA = 400 m. A B_5389_ch12_ptg01_hr.4c.indd 68 60 150 m 150 m vt 8 m/s Prob. 12–135 12–138. The motorcycle is traveling at 40 m>s when it is at # A. If the speed is then decreased at v = - (0.05 s) m>s2, where s is in meters measured from A, determine its speed and acceleration when it reaches B. B A Prob. 12–138 1/21/15 9:59 AM 12.7 69 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 12–139. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the minimum acceleration experienced by the passengers. 12–142. The race car has an initial speed vA = 15 m>s at A. If it increases its speed along the circular track at the rate 12 at = (0.4s) m>s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m. *12–140. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the maximum acceleration experienced by the passengers. y 2 x2 y 1 (60)2 (40) 0)2 r 40 m s x A 60 m Probs. 12–139/140 Prob. 12–142 12–141. A package is dropped from the plane which is flying with a constant horizontal velocity of vA = 150 ft>s. Determine the normal and tangential components of acceleration and the radius of curvature of the path of motion (a) at the moment the package is released at A, where it has a horizontal velocity of vA = 150 ft>s, and (b) just before it strikes the ground at B. 12–143. The motorcycle travels along the elliptical track at a constant speed v. Determine its greatest acceleration if a 7 b. *12–144. The motorcycle travels along the elliptical track at a constant speed v. Determine its smallest acceleration if a 7 b. A vA 1500 ft y b 2 y a x b 2 B Prob. 12–141 B_5389_ch12_ptg01_hr.4c.indd 69 a 2 x 2 1 Probs. 12–143/144 1/21/15 9:59 AM 70 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–145. Particles A and B are traveling counter-clockwise 12 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? 12–146. 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? 12–149. The train passes point B with a speed of 20 m>s which is decreasing at at = - 0.5 m>s2. Determine the magnitude of acceleration of the train at this point. 12–150. 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. y x y 200 e 1000 A sA B A u 120 sB x B 400 m r5m Probs. 12–149/150 Probs. 12–145/146 12–147. 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. *12–148. 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). 12–151. The particle travels with a constant speed of 300 mm>s along the curve. Determine the particle’s acceleration when it is located at point (200 mm, 100 mm) and sketch this vector on the curve. y (mm) y y 15 lnQ x R 80 y 20(103) x 80 m A x v P x (mm) Probs. 12–147/148 B_5389_ch12_ptg01_hr.4c.indd 70 Prob. 12–151 1/21/15 9:59 AM 12.8 *12–152. A particle P travels along an elliptical spiral path such that its position vector r is defined by r = 5 2 cos(0.1t)i + 1.5 sin(0.1t)j + (2t)k 6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. When t = 8 s, determine the coordinate direction angles a, b, and g, which the binormal axis to the osculating plane makes with the x, y, and z axes. Hint: Solve for the velocity vP and acceleration aP of the particle in terms of their i, j, k components. The binormal is parallel to vP * aP. Why? 71 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 12–153. 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. 12–154. If the speed of the crate at A is 15 ft>s, which is # increasing at a rate v = 3 ft>s2 , determine the magnitude of the acceleration of the crate at this instant. z y y 1 x2 16 P A r y x 10 ft x Prob. 12–152 Prob. 12–154 12.8 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. u uu 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. B_5389_ch12_ptg01_hr.4c.indd 71 r ur r u O Position (a) Fig. 12–30 1/21/15 9:59 AM 72 CHAPTER 12 K I N E M AT I C S uu r ur Position u¿r u (12–22) # 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 ur= , where ur= = ur + ur , Fig. 12–30b. The time change in ur is then ur . For small angles u this vector has a magnitude ur 1(u) and acts in the uu direction. Therefore, ur = uuu , and so (a) uu r = r 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 # # # v = r = r ur + r ur r u O PARTICLE Position. At any instant the position of the particle, Fig. 12–30a, is defined by the position vector u 12 OF A ur ur (b) ur u # ur = lim = a lim b uu S t 0 t t S 0 t # # 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.) B_5389_ch12_ptg01_hr.4c.indd 72 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 = 2(r)2 + (ru)2 (12–26) and the direction of v is, of course, tangent to the path, Fig. 12–30c. 1/21/15 9:59 AM 12.8 73 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS Acceleration. Taking the time derivatives of Eq. 12–24, using Eqs. 12–25, we obtain the particle’s instantaneous acceleration, 12 $ ## # $ ## # # a = v = rur + rur + ruuu + ru uu + 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 uu= , where uu= = uu + uu , Fig. 12–30d. The time change in uu is thus uu . For small angles this vector has a magnitude uu 1(u) and acts in the -ur direction; i.e., uu = - uur . Thus, uu u # uu = lim = - a lim bur t S 0 t t S 0 t # # uu = -uur uu uu u¿u ur u (d) (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 = a rur + a uuu (12–28) # $ ar = r - ru2 $ # # au = ru + 2ru (12–29) where $ The term u = d2u>dt2 = d>dt(du>dt) 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 ar and au are always perpendicular, the magnitude of acceleration is simply the positive value of # $ $ # # a = 2(r - r u 2)2 + (ru + 2r u)2 ar r u (12–30) 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. B_5389_ch12_ptg01_hr.4c.indd 73 a au O Acceleration (e) 1/21/15 9:59 AM 74 CHAPTER 12 K I N E M AT I C S uz 12 uu ur rP z O u r Fig. 12–31 OF A PARTICLE 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 = (r - ru2)ur + (ru + 2ru)uu + 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: 1. If the polar coordinates are specified as time parametric equations, r = r(t) and u = u(t), then the time derivatives can be found directly. 2. If the time-parametric equations are not given, then the path r = f(u) 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 Coordinate System. • Polar coordinates are a suitable choice for solving problems when The spiral motion of this girl can be followed by using cylindrical components. Here the radial coordinate r is constant, the transverse coordinate u will increase with time as the girl rotates about the vertical, and her altitude z will decrease with time. (© R.C. Hibbeler) 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 line to the radial line. 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 = f(u), 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. B_5389_ch12_ptg01_hr.4c.indd 74 1/21/15 9:59 AM 12.8 EXAMPLE CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 12.17 75 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 ru 2 u B r ·· au ru r u, t (a) (b) 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, # 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. NOTE: 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, # # (ru)2 v2 = ru2 -ar = an = = r r # $ dv d # dr # du au = at = = (ru) = u + r = 0 + ru dt dt dt dt B_5389_ch12_ptg01_hr.4c.indd 75 1/21/15 9:59 AM 76 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.18 EXAMPLE The rod OA in Fig. 12–33a rotates in the horizontal plane such that u = (t3) rad. At the same time, the collar B is sliding outward along OA so that r = (100t2) mm. If in both cases t is in seconds, determine the velocity and acceleration of the collar when t = 1 s. O SOLUTION Coordinate System. Since time-parametric equations of the path are given, it is not necessary to relate r to u. r B u Velocity and Acceleration. Determining the time derivatives and evaluating them when t = 1 s, we have 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 ` = 1 rad = 57.3 t=1 s t=1 s = 3 rad>s = 6 rad>s2. = 200ur + 100(3)uu = 5 200ur + 300uu 6 mm>s r The magnitude of v is (b) v = 2(200)2 + (300)2 = 361 mm>s d = tan-1 a 300 b = 56.3 d + 57.3 = 114 200 Ans. Ans. As shown in Fig. 12–33c, # $ $ # # a = ( r - ru2)ur + (ru + 2ru)uu a u 57.3 f = [200 - 100(3)2]ur + [100(6) + 2(200)3]uu u au 1800 mm/s2 The magnitude of a is ar 700 mm/s2 r (c) Fig. 12–33 = 5 -700ur + 1800uu 6 mm>s2 f = tan-1 a a = 2(-700)2 + (1800)2 = 1930 mm>s2 1800 b = 68.7 (180 - f) + 57.3 = 169 700 Ans. Ans. NOTE: The velocity is tangent to the path; however, the acceleration is directed within the curvature of the path, as expected. B_5389_ch12_ptg01_hr.4c.indd 76 1/21/15 9:59 AM 12.8 EXAMPLE 77 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 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 of u = 4 rad>s. u r · 100 m u 4 rad/s 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, (a) r r = 100>cos u = 100 sec u Velocity and Acceleration. Using the chain rule of calculus, noting that d(sec u) = sec u tan u du, and d(tan u) = sec2 u du, we have # # r = 100(sec u tan u)u# # # # $ 2 r = 100(sec u tan u)u(tan $ u)u + 100 sec u(sec u)u(u) + 100 sec u tan u(u # ) # $ = 100 sec u tan2 u (u)2 + 100 sec3u (u)2 + 100(sec u tan u)u $ # Since u = 4 rad>s = constant, then u = 0, and the above equations, when u = 45, become vr v u r (b) r ar As shown in Fig. 12–34b, # # v = rur + ruuu = 565.7ur + 141.4(4)uu As shown in Fig. 12–34c, # $ $ # # a = ( r - ru2)ur + (ru + 2ru)uu = [6788.2 - 141.4(4)2]ur + [141.4(0) + 2(565.7)4]uu = 5 4525.5ur + 4525.5uu 6 m>s2 a = 2a2r + a2u = 2(4525.5)2 + (4525.5)2 = 6400 m>s2 a u r u 100 m au Ans. Ans. $ NOTE: It is also possible to find a without having to calculate r (or ar). As shown in Fig. 12–34d, since au = 4525.5 m>s2, then by vector resolution, a = 4525.5>cos 45 = 6400 m>s2. B_5389_ch12_ptg01_hr.4c.indd 77 100 m u r = 100 sec 45 = 141.4 # r = 400 sec 45 tan 45 = 565.7 $ r = 1600 (sec 45 tan2 45 + sec3 45) = 6788.2 = 5 565.7ur + 565.7uu 6 m>s v = 2v2r + v2u = 2(565.7)2 + (565.7)2 = 800 m>s u vu u (c) a u 45 ar au 4525.5 m/s2 (d) Fig. 12–34 1/21/15 9:59 AM 78 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12.20 EXAMPLE r 0.5 (1 cos u) ft u r · ·· u, u (a) 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 = 0.5(1 - cos u) 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. Velocity and Acceleration. The time derivatives of r and u can be determined using the chain rule. r = 0.5(1 - cos u) # # r = 0.5(sin u)u # # $ $ r = 0.5(cos u) u(u) + 0.5(sin u)u 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 = 2(r)2 + (ru)2 # 4 = 2(0)2 + (1u)2 # u = 4 rad>s Ans. $ In a similar manner, u can be found using Eq. 12–30. # $ $ # # a = 2(r - ru2)2 + (ru + 2ru)2 r v 4 ft/s a 30 ft/s2 $ 30 = 2[-0.5(4)2 - 1(4)2]2 + [1u + 2(0)(4)]2 $ (30)2 = ( -24)2 + u 2 $ u = 18 rad>s2 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. B_5389_ch12_ptg01_hr.4c.indd 78 1/21/15 9:59 AM 12.8 79 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 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 = e u, where r is in meters. When u = p4 rad, the link has an angular # $ velocity and angular acceleration of u = 2 rad>s and u = 4 rad>s2. Determine the radial and transverse components of the peg’s acceleration at this instant. A r eu A r 400 ft P r u Prob. F12–33 O 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, u O Prob. 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 magnitude of the velocity of the collars at this point. A u, u r 0.2(l + cos u) m B r u r Prob. 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 u = p>4 rad, the # angular velocity $ and angular acceleration of the link are u = 3 rad>s and u = 1 rad>s2. Determine the magnitude of the peg’s acceleration at this instant. u O C u 3 rad/s Prob. 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 v A r P u, u u u O A Prob. F12–35 B_5389_ch12_ptg01_hr.4c.indd 79 30 m r u Prob. F12–38 1/21/15 10:00 AM 80 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE PROBLEMS 12–155. A particle is moving 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, where u is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when u = 30. *12–156. For a short time a rocket travels up and to the right at a constant speed of 800 m>s along the parabolic path y = 600 - 35x2. Determine the radial and transverse components of velocity of the rocket at the instant u = 60°, where u is measured counterclockwise from the x axis. *12–160. A# radar gun at O rotates with the angular $ velocity of u = 0.1 rad>s and angular acceleration of u = 2 0.025 rad>s , at the instant u = 45°, as it follows the motion of the car traveling along the circular road having a radius of r = 200 m. Determine the magnitudes of velocity and acceleration of the car at this instant. 12–157. A particle moves along a path defined by polar coordinates r = (2et) ft and u = (8t2) rad, where t is in seconds. Determine the components of its velocity and acceleration when t = 1 s. 12–158. 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 a constant angular rate of 120 rad>s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller. 12–159. The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m>s and its acceleration is 7 m>s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical components. r 200 m u O Prob. 12–160 12–161. If a particle moves along a path such that r = (2 cos t) ft and u = (t>2) rad, where t is in seconds, plot the path r = f(u) and determine the particle’s radial and transverse components of velocity and acceleration. 12–162. If a particle moves along a path such that r = (eat) m and u = t, where t is in seconds, plot the path r = f(u), and determine the particle’s radial and transverse components of velocity and acceleration. 12–163. The car travels along the circular curve having a radius r = 400 ft. At the instant shown, its angular rate of # rotation is u = 0.025 rad>s, which is decreasing at the rate $ u = - 0.008 rad>s2. Determine the radial and transverse components of the car’s velocity and acceleration at this instant and sketch these components on the curve. *12–164. 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 6m r 400 ft O x 3m Prob. 12–159 B_5389_ch12_ptg01_hr.4c.indd 80 2m y . u Probs. 12–163/164 1/21/15 10:00 AM 12.8 12–165. The time rate of change of acceleration is referred to as the jerk, which is often used as a means of measuring # passenger discomfort. Calculate this vector, a, in terms of its cylindrical components, using Eq. 12–32. 12–166. A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by u = sin 3t, where u and the argument for the sine are in radians, and t is in seconds. Determine the magnitude of the acceleration of the particle at u = 30. The particle starts from rest at u = 0. 12–167. The slotted link is pinned at O, and as a result of # the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = (0.4 u) m, where u is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant u = p>3 rad. CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 81 12–169. The slotted link is pinned at O, and as a result of # the constant angular velocity u = 3 rad>s it drives the peg P 12 for a short distance along the spiral guide r = (0.4 u) m, where u is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = 0.5 m. 0.5 m P r r 0.4u · u 3 rad/s u O Prob. 12–169 0.5 m P r r 0.4u · u 3 rad/s u O Prob. 12–167 *12–168. For a short time the bucket of the backhoe traces the path of the cardioid r = 25(1 − cos u) ft. Determine the magnitudes of the velocity and acceleration of the bucket when #u = 120° if the boom is rotating with an angular velocity of u = 2 rad>s and an angular acceleration of $ u = 0.2 rad>s2 at the instant shown. 12–170. A particle moves in the x -y plane such that its position is defined by r = {2ti + 4t2j} ft, where t is in seconds. Determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s. 12–171. At the instant shown, the man# is twirling a hose over his head with an$ angular velocity u = 2 rad>s and an angular acceleration u = 3 rad>s2. If it is assumed that the hose lies in a horizontal plane, and water is flowing through it at a constant rate of 3 m>s, determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r = 1.5 m. · u 2 rad/s ·· u 3 rad/s2 r 1.5 m u r Prob. 12–168 B_5389_ch12_ptg01_hr.4c.indd 81 u 120 Prob. 12–171 1/23/15 2:24 PM 82 CHAPTER 12 K I N E M AT I C S OF A PARTICLE *12–172. The rod OA rotates clockwise with a constant 12 angular velocity of 6 rad>s. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 200(2 − cos u) mm. Determine the speed of the slider blocks at the instant u = 150°. 12–173. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–172 when u = 150°. A *12–176. The car travels around the circular track with a constant speed of 20 m>s. Determine the car’s radial and transverse components of velocity and acceleration at the instant u = p>4 rad. 12–177. The car travels around the circular track such that its transverse component is u = (0.006t2) rad, where t is in seconds. Determine the car’s radial and transverse components of velocity and acceleration at the instant t = 4 s. B r 6 rad/s u O 400 mm r (400 cos u) m 600 mm r 200 mm u Probs. 12–172/173 12–174. A double collar C is pin connected together such that one collar slides over a fixed rod and the other slides over a rotating rod. If the geometry of the fixed rod for a short distance can be defined by a lemniscate, r2 = (4 cos 2u) ft2, determine the collar’s radial and transverse components of velocity and acceleration at the instant u = 0° as shown. Rod # OA is rotating at a constant rate of u = 6 rad>s. Probs. 12–176/177 r2 4 cos 2 u · u 6 rad/s O r A C Prob. 12–174 12–178. The 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 u = p>3 rad. 12–175. A block moves outward along the slot in the # platform with a speed of r = (4t) m>s, where t is in seconds. The platform rotates at a constant rate of 6 rad>s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = 1 s. u · u 6 rad/s r r u Prob. 12–175 B_5389_ch12_ptg01_hr.4c.indd 82 Prob. 12–178 1/21/15 10:00 AM 12.8 83 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 12–179. A horse on the merry-go-round moves according to the equations r = 8 ft, u = (0.6t) rad, and z = (1.5 sin u) ft, where t is in seconds. Determine the cylindrical components of the velocity and acceleration of the horse when t = 4 s. 12–183. A truck is traveling along the horizontal circular curve of radius r = 60 m with a constant# speed v = 20 m>s. 12 Determine the angular rate of rotation u of the radial line r and the magnitude of the truck’s acceleration. *12–180. A horse on the merry-go-round moves according # to the equations r = 8 ft, u = 2 rad>s and z = (1.5 sin u) ft, where t is in seconds. Determine the maximum and minimum magnitudes of the velocity and acceleration of the horse during the motion. *12–184. A truck is traveling along the horizontal circular curve of radius r = 60 m with a speed of 20 m>s which is increasing at 3 m>s2, Determine the truck’s radial and transverse components of acceleration. z u r 60 m u z u r Probs. 12–183/184 Probs. 12–179/180 12–181. If the slotted arm AB 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 is constrained to move in the slots of the fixed bar CD and rotating bar AB. 12–182. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When u = 30°, the # angular velocity acceleration of arm AB are u = 2 rad>s $ and angular and u = 3 rad>s2, respectively. Determine the magnitudes of the velocity and acceleration of the peg P at this instant. 12–185. The rod OA rotates counterclockwise with a # constant angular velocity of u = 5 rad>s. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 100(2 − cos u) mm. Determine the speed of the slider blocks at the instant u = 120°. 12–186. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–185 when u = 120°. · u 5 rad/s y A B r D u B x O P r (4 sec u) ft u A C r 100 (2 cos u) mm 4 ft Probs. 12–181/182 B_5389_ch12_ptg01_hr.4c.indd 83 Probs. 12–185/186 1/21/15 10:00 AM 84 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–187. The searchlight on the boat anchored 2000 ft from 12 shore is turned on the automobile, which is traveling along the straight road at a constant speed 80 ft>s, Determine the angular rate of rotation of the light when the automobile is r = 3000 ft from the boat. 12–191. The arm of the robot moves so that r = 3 ft is constant, and its grip A moves along the path z = (3 sin 4u) ft, where u is in radians. If u = (0.5t) rad, where t is in seconds, determine the magnitudes of the grip’s velocity and acceleration when t = 3 s. *12–188. If the car in Prob. 12–187 is accelerating at 15 ft>s2 at the instant$ r = 3000 ft determine the required angular acceleration u of the light at this instant. *12–192. For a short time the arm of the robot is extending # such that r = 1.5 ft>s when r = 3 ft, z = (4t2) ft, and u = 0.5t rad, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the grip A when t = 3 s. A 80 ft/s r u z u r u Probs. 12–191/192 2000 ft Probs. 12–187/188 12–189. 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. 12–190. Solve $ Prob. 12–189 if the # particle has an angular acceleration u = 5 rad>s2 when u = 4 rad>s at u = p>2 rad. 12–193. The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the #rotating rod AB. If the angular velocity of AB is 2 given as u = (e0.5 t ) rad>s, where t is in seconds, and the path defined by the fixed rod is r = |(0.4 sin u + 0.2)| m, determine the radial and transverse components of the collar’s velocity and acceleration when t = 1 s. When t = 0, u = 0. Use Simpson’s rule with n = 50 to determine u at t = 1 s. 12–194. The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod AB. If the mechanism is to be designed so that the largest speed given to the collar # is 6 m>s, determine the required constant angular velocity u of rod AB. The path defined by the fixed rod is r = (0.4 sin u + 0.2) m. B y 0.6 m r (8 u) ft A r C r u 0.2 m u x Probs. 12–189/190 B_5389_ch12_ptg01_hr.4c.indd 84 0.2 m 0.2 m Probs. 12–193/194 1/21/15 10:00 AM 12.9 85 ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES 12.9 Absolute Dependent Motion 12 Analysis of Two Particles 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 the fixed datums to A and 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 dt dt or vB = -vA 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., aB = -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) from the fixed datums 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 2sB + h + sA = l Datum sB B h A Since l and h are constant during the motion, the two time derivatives yield 2vB = -vA 2aB = -aA Hence, when B moves downward (+sB), A moves to the left (-sA) with twice the motion. B_5389_ch12_ptg01_hr.4c.indd 85 Datum sA (a) Fig. 12–37 1/21/15 10:00 AM 86 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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 2(h - sB) + h + sA = l Time differentiation yields sB 2vB = vA B h Datum Here the signs are the same. Why? A Datum 2aB = aA sA (b) Fig. 12–37 (cont.) Procedure for Analysis 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. • 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 The cable is wrapped around the pulleys on this crane in order to reduce the required force needed to hoist a load. (© R.C. Hibbeler) B_5389_ch12_ptg01_hr.4c.indd 86 those that specify the positive and negative sense of the position coordinates. 1/21/15 10:00 AM 12.9 EXAMPLE ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES 12.21 87 12 Determine the speed of block A in Fig. 12–38 if block B has an upward speed of 6 ft>s. C D Datum 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>s T B_5389_ch12_ptg01_hr.4c.indd 87 Ans. 1/21/15 10:00 AM 88 12 CHAPTER 12 EXAMPLE K I N E M AT I C S OF A PARTICLE 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 + (sB - sC) = 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 B_5389_ch12_ptg01_hr.4c.indd 88 Ans. 1/21/15 10:00 AM 12.9 EXAMPLE ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES 12.23 89 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 (sA - sC) + (sB - sC) + 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 B_5389_ch12_ptg01_hr.4c.indd 89 Ans. 1/21/15 10:00 AM 90 12 CHAPTER 12 EXAMPLE K I N E M AT I C S OF A PARTICLE 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. D E 15 m 10 m C S y A x Fig. 12–41 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 C 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 = 2(15)2 + x2; also, lCD = 15 - y. Hence, l = lDA + lCD 30 = 2(15)2 + x2 + (15 - y) vA 0.5 m/s y = 2225 + x2 - 15 (1) Time Derivatives. Taking the time derivative, using the chain rule (see Appendix C), where vS = dy>dt and vA = dx>dt, yields dy 1 2x dx = J R 2 dt 2 2225 + x dt x = (2) vA 2225 + x2 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, vS = 20 Ans. (0.5) = 0.4 m>s = 400 mm>s c 2225 + (20)2 The acceleration is determined by taking the time derivative of Eq. 2. Since vA is constant, then aA = dvA >dt = 0, and we have 2 -x(dx>dt) dy dvA 225v2A 1 dx 1 aS = 2 = c d xv + c d a b v + c d x = A A dt dt (225 + x2)3>2 (225 + x2)3>2 2225 + x2 dt 2225 + x2 vS = At x = 20 m, with vA = 0.5 m>s, the acceleration becomes aS = 225(0.5 m>s)2 [225 + (20 m)2]3>2 = 0.00360 m>s2 = 3.60 mm>s2 c Ans. 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. B_5389_ch12_ptg01_hr.4c.indd 90 1/21/15 10:00 AM 12.10 91 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 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. 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 rB = rA + rB>A z¿ z A Translating observer rB/A rA Fixed observer B y O y¿ rB x¿ (12–33) x Fig. 12–42 Velocity. An equation that relates the velocities of the particles is determined by taking the time derivative of the above equation; i.e., 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. B_5389_ch12_ptg01_hr.4c.indd 91 1/21/15 10:00 AM 92 CHAPTER 12 12 K I N E M AT I C S OF A PARTICLE Acceleration. The time derivative of Eq. 12–34 yields a similar vector relation between the absolute and relative accelerations of particles A and B. aB = 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 close-flying planes must be aware of their relative positions and velocities at all times in order to avoid a collision. (© R.C. Hibbeler) * 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 . B_5389_ch12_ptg01_hr.4c.indd 92 1/21/15 10:00 AM 12.10 EXAMPLE 93 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 12.25 12 A train travels at a constant speed of 60 mi>h and 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 = (45 cos 45i + 45 sin 45j) + vT>A vT>A = 5 28.2i - 31.8j 6 mi>h The magnitude of vT>A is thus vT>A = 2(28.2)2 + (-31.8)2 = 42.5 mi>h Ans. 28.2 mi/h From the direction of each component, Fig. 12–43b, the direction of vT>A is (vT>A)y 31.8 tan u = = (vT>A)x 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 (v ) (v ) d = c a 45 d + c T>A x d + c T>A y d S S c Resolving each vector into its x and y components yields + ) (S 60 = 45 cos 45 + (vT>A)x + 0 (+ c ) 0 = 45 sin 45 + 0 + (vT>A)y Solving, we obtain the previous results, (vT>A)x = 28.2 mi>h = 28.2 mi>h S (vT>A)y = -31.8 mi>h = 31.8 mi>h T B_5389_ch12_ptg01_hr.4c.indd 93 vT/A vA 45 mi/h 45 u vT 60 mi/h (c) Fig. 12–43 1/21/15 10:00 AM 94 12 CHAPTER 12 K I N E M AT I C S PARTICLE 12.26 EXAMPLE y¿ y 700 km/h 600 km/h A B x¿ 50 km/h2 OF A 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 (+ c ) vB = vA + vB>A 600 km>h = 700 km>h + vB>A vB>A = -100 km>h = 100 km>h T vB/A 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) (600 km>h)2 v2B = 900 km>h2 = r 400 km Applying the relative-acceleration equation gives (aB)n = aB = aA + aB>A 900i - 100j = 50j + aB>A Thus, aB>A = 5 900i - 150j 6 km>h2 From Fig. 12–44c, the magnitude and direction of aB>A are therefore aB>A = 912 km>h2 u = tan-1 900 km/h2 u aB/A 150 km/h2 (c) Fig. 12–44 B_5389_ch12_ptg01_hr.4c.indd 94 150 = 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. 1/21/15 10:00 AM 12.10 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¿ 3 m/s2 18 m/s r 100 m 12 m/s B y 60 -12j = (-18 cos 60i - 18 sin 60j) + vB>A vB>A = 5 9i + 3.588j 6 m>s 2 m/s2 60 x¿ A vB = vA + vB>A Thus, 95 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES x (a) vB>A = 2(9)2 + (3.588)2 = 9.69 m>s Ans. Noting that vB>A has +i and +j components, Fig. 12–45b, its direction is tan u = (vB>A)y (vB>A)x = 3.588 9 u = 21.7 a Ans. 3.588 m/s Acceleration. Car B has both tangential and normal components of acceleration. Why? The magnitude of the normal component is u 9 m/s (12 m>s)2 v2B = 1.440 m>s2 = r 100 m Applying the equation for relative acceleration yields (aB)n = (b) aB = aA + aB>A (-1.440i - 3j) = (2 cos 60i + 2 sin 60j) + aB>A aB>A = 5 -2.440i - 4.732j 6 m>s2 Here aB>A has -i and -j components. Thus, from Fig. 12–45c, aB>A = 2(2.440)2 + (4.732)2 = 5.32 m>s2 tan f = (aB>A)y (aB>A)x = 2.440 m/s2 f Ans. 4.732 2.440 f = 62.7 d Ans. NOTE: Is it possible to obtain the relative acceleration of aA>B using this method? Refer to the comment made at the end of Example 12.26. B_5389_ch12_ptg01_hr.4c.indd 95 vB/A aB/A 4.732 m/s2 (c) Fig. 12–45 1/21/15 10:00 AM 96 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE FUNDAMENTAL PROBLEMS F12–39. Determine the velocity 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 velocity of block A if end F of the rope is pulled down with a speed of vF = 3 m>s. C C B A E D A vA 3 m/s F D vF 3 m/s Prob. F12–39 F12–40. Determine the velocity of block A if end B of the rope is pulled down with a speed of 6 m>s. Prob. F12–42 F12–43. Determine the velocity 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 P B A A Prob. F12–43 Prob. F12–40 F12–41. Determine the velocity of block A if end B of the rope is pulled down with a speed of 1.5 m>s. F12–44. Determine the velocity of cylinder B if cylinder A moves downward with a speed of vA = 4 ft>s. F E C 1.5 m/s D B A vA 4 ft/s A B Prob. F12–41 B_5389_ch12_ptg01_hr.4c.indd 96 Prob. F12–44 1/21/15 10:00 AM 12.10 97 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 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. 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 O 45 vA 15 m/s 30 x A 80 km/h Prob. F12–47 Prob. F12–45 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 100 m 60 20 km/h B A 65 km/h Prob. F12–46 B_5389_ch12_ptg01_hr.4c.indd 97 Prob. F12–48 1/21/15 10:00 AM 98 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE PROBLEMS 12–195. 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. D 12–198. If the end of the cable at A is pulled down with a speed of 5 m>s, determine the speed at which block B rises. C A A 2 m/s 5 m/s B B Prob. 12–195 *12–196. 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. Prob. 12–198 12–199. Determine the displacement of the log if the truck at C pulls the cable 4 ft to the right. C B D B A C Prob. 12–199 d Prob. 12–196 12–197. The pulley arrangement shown is designed for hoisting materials. If BC remains fixed while the plunger P is pushed downward with a speed of 4 ft>s, determine the speed of the load at A. *12–200. Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5 s. 12–201. Starting from rest, the cable can be wound onto the drum of the motor at a rate of vA = (3t2) m>s, where t is in seconds. Determine the time needed to lift the load 7 m. B C P 4 ft/s A Prob. 12–197 B_5389_ch12_ptg01_hr.4c.indd 98 A D C B Probs. 12–200/201 1/21/15 10:00 AM 12.10 99 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 12–202. If the end A of the cable is moving at vA = 3 m>s, determine the speed of block B. 12–206. Determine the speed of the block at B. 12 6 m/s A D C vA 3 m/s A B B Prob. 12–206 Prob. 12–202 12–207. Determine the speed of block A if the end of the rope is pulled down with a speed of 4 m>s. 12–203. Determine the time needed for the load at B to attain a speed of 10 m>s, starting from rest, if the cable is drawn into the motor with an acceleration of 3 m>s2. *12–204. The cable at A is being drawn toward the motor at vA = 8 m>s. Determine the velocity of the block. A vA 4 m/s B C A B Prob. 12–207 Probs. 12–203/204 12–205. If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relative velocity of block B with respect to C. *12–208. The motor draws in the cable at C with a constant velocity of vC = 4 m>s. The motor draws in the cable at D with a constant acceleration of aD = 8 m>s2. If vD = 0 when t = 0, determine (a) the time needed for block A to rise 3 m, and (b) the relative velocity of block A with respect to block B when this occurs. C D B A C A B Prob. 12–205 B_5389_ch12_ptg01_hr.4c.indd 99 Prob. 12–208 1/21/15 10:00 AM 100 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–209. The cord is attached to the pin at C and passes 12 over the two pulleys at A and D. The pulley at A is attached to the smooth collar that travels along the vertical rod. Determine the velocity and acceleration of the end of the cord at B if at the instant sA = 4 ft the collar is moving upward at 5 ft>s, which is decreasing at 2 ft>s2. 12–210. The 16-ft-long cord is attached to the pin at C and passes over the two pulleys at A and D. The pulley at A is attached to the smooth collar that travels along the vertical rod. When sB = 6 ft, the end of the cord at B is pulled downward with a velocity of 4 ft>s and is given an acceleration of 3 ft>s2. Determine the velocity and acceleration of the collar at this instant. 3 ft *12–212. The girl at C stands near the edge of the pier and pulls in the rope horizontally at a constant speed of 6 ft>s. Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft. 6 ft/s xC C A 8 ft B xB Prob. 12–212 3 ft C D sB sA B 12–213. If the hydraulic cylinder H draws in rod BC at 2 ft > s, determine the speed of slider A. A A B C H Prob. 12–213 Probs. 12–209/210 12–211. The roller at A is moving with a velocity of vA = 4 m>s and has an acceleration of aA = 2 m>s2 when x A = 3 m. Determine the velocity and acceleration of block B at this instant. vA 4 m/s xA A 12–214. At the instant shown, the car at A is traveling at 10 m>s around the curve while increasing its speed at 5 m>s2. The car at B is traveling at 18.5 m>s along the straightaway and increasing its speed at 2 m>s2. Determine the relative velocity and relative acceleration of A with respect to B at this instant. yB 18.5 m/s 4m B A 100 m % Prob. 12–211 B_5389_ch12_ptg01_hr.4c.indd 100 yA 10 m/s 100 m 45 Prob. 12–214 1/21/15 10:00 AM 12.10 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 12–215. The motor draws in the cord at B with an acceleration of aB = 2 m>s2. When sA = 1.5 m, vB = 6 m>s. Determine the velocity and acceleration of the collar at this instant. B 101 12–218. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 500 km>h and 12 vB = 700 km>h such that the angle between their straightline courses is u = 60, determine the velocity of plane B with respect to plane A. A 2m vA 500 km/h A vB 700 km/h 60 sA B Prob. 12–215 *12–216. If block B is moving down with a velocity vB and has an acceleration aB , determine the velocity and acceleration of block A in terms of the parameters shown. sA A h Prob. 12–218 12–219. At the instant shown, cars A and B are traveling at speeds of 55 mi>h and 40 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. Car B moves along a curve having a radius of curvature of 0.5 mi. vB, aB vB 40 mi/h B B Prob. 12–216 12–217. The crate C is being lifted by moving the roller at A downward with a constant speed of vA = 2 m>s along the guide. Determine the velocity and acceleration of the crate at the instant s = 1 m. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hint: Relate the coordinates xC and xA using the problem geometry, then take the first and second time derivatives. 4m 30 A vA 55 mi/h Prob. 12–219 *12–220. The boat can travel with a speed of 16 km>h in still water. The point of destination is located along the dashed line. If the water is moving at 4 km>h, determine the bearing angle u at which the boat must travel to stay on course. B xA xC 4m vW 4 km/h u 70 A C s Prob. 12–217 B_5389_ch12_ptg01_hr.4c.indd 101 Prob. 12–220 1/21/15 10:00 AM 102 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–221. Two boats leave the pier P at the same time and 12 travel in the directions shown. If vA = 40 ft>s and vB = 30 ft>s, determine the velocity of boat A relative to boat B. How long after leaving the pier will the boats be 1500 ft apart? *12–224. At the instant shown, car A has a speed of 20 km>h, which is being increased at the rate of 300 km>h2 as the car enters the expressway. At the same instant, car B is decelerating at 250 km>h2 while traveling forward at 100 km>h. Determine the velocity and acceleration of A with respect to B. y vA 40 ft/s vB 30 ft/s A B 30 A 45 100 m x P B Prob. 12–221 Prob. 12–224 12–222. A car is traveling north along a straight road at 50 km>h. An instrument in the car indicates that the wind is coming from the east. If the car’s speed is 80 km>h, the instrument indicates that the wind is coming from the northeast. Determine the speed and direction of the wind. 12–223. Two boats leave the shore at the same time and travel in the directions shown. If vA = 10 m>s and vB = 15 m>s, determine the velocity of boat A with respect to boat B. How long after leaving the shore will the boats be 600 m apart? 12–225. Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 90 ft>s and is increasing its speed at the rate of 15 ft>s2, whereas B has a speed of 105 ft>s and is decreasing its speed at 25 ft>s2. Determine the relative velocity and relative acceleration of car A with respect to car B at this instant. vA A vA 10 m/s B A vB 15 m/s B rA 300 ft vB 60 rB 250 ft 30 O 45 Prob. 12–223 B_5389_ch12_ptg01_hr.4c.indd 102 Prob. 12–225 1/21/15 10:00 AM 12.10 103 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 12–226. 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 direction in which the drops appear to fall with respect to the man. 12–229. A passenger in an automobile observes that raindrops make an angle of 30° with the horizontal as the 12 auto travels forward with a speed of 60 km>h. Compute the terminal (constant) velocity vr of the rain if it is assumed to fall vertically. vw 20 km/h vr va 60 km/h vm 5 km/h Prob. 12–229 Prob. 12–226 12–227. At the instant shown, cars A and B are traveling at velocities of 40 m>s and 30 m>s, respectively. If B is increasing its velocity by 2 m>s2, while A maintains a constant velocity, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is rB = 200 m. *12–228. At the instant shown, cars A and B are traveling at velocities of 40 m>s and 30 m>s, respectively. If A is increasing its speed at 4 m>s2, whereas the speed of B is decreasing at 3 m>s2, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is rB = 200 m. B 12–230. A man can swim at 4 ft>s in still water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. If the river flows with a velocity of 2 ft>s, determine the speed of the man and the time needed to make the crossing. Note: While in the water he must not direct himself toward point B to reach this point. Why? 30 ft A B vA 40 m/s vB 30 m/s vr 2 ft/s 40 ft 30 A Probs. 12–227/228 B_5389_ch12_ptg01_hr.4c.indd 103 Prob. 12–230 1/21/15 10:00 AM 104 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 12–231. The ship travels at a constant speed of vs = 20 m>s 12 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 12–234. At a given instant the football player at A throws a football C with a velocity of 20 m>s in the direction shown. Determine the constant speed at which the player at B must run so that he can catch the football at the same elevation at which it was thrown. Also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made. Player B is 15 m away from A when A starts to throw the football. 45 vw 10 m/s y x C 20 m/s 60 A 15 m Prob. 12–231 *12–232. The football player at A throws the ball in the y–z plane at a speed vA = 50 ft>s and an angle uA = 60° with the horizontal. At the instant the ball is thrown, the player is at B and is running with constant speed along the line BC in order to catch it. Determine this speed, vB, so that he makes the catch at the same elevation from which the ball was thrown. 12–233. The football player at A throws the ball in the y–z plane with a speed vA = 50 ft>s and an angle uA = 60° with the horizontal. At the instant the ball is thrown, the player is at B and is running at a constant speed of vB = 23 ft>s along the line BC. Determine if he can reach point C, which has the same elevation as A, before the ball gets there. z uA Prob. 12–234 12–235. At the instant shown, car A travels along the 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. y C vA vB 15 A 20 ft 30 ft 15 r 200 m B A C 30 x Probs. 12–232/233 B_5389_ch12_ptg01_hr.4c.indd 104 B B Prob. 12–235 1/21/15 10:00 AM 12.10 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 105 CONCEPTUAL PROBLEMS PROBLEMS C12–1. 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 C12–3. The basketball was thrown at an angle measured from the horizontal to the man’s outstretched arm. If the basket is 3 m 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 Prob. C12–3 (© R.C. Hibbeler) C12–4. 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. A Prob. C12–1 (© R.C. Hibbeler) C12–2. 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. A Prob. C12–2 (© R.C. Hibbeler) B_5389_ch12_ptg01_hr.4c.indd 105 Prob. C12–4 (© R.C. Hibbeler) 1/21/15 10:00 AM 106 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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. s s t vavg = 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. (vsp)avg a = dv , dt v = sT = t ds , dt a ds = v dv v = v0 + a c t s = s0 + v0t + 12 act2 v2 = v20 + 2ac(s - s0) 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. B_5389_ch12_ptg01_hr.4c.indd 106 dv , dt ds v = , dt a ds = v dv a = 1/21/15 10:01 AM 107 CHAPTER REVIEW Curvilinear Motion, x, y, 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. 12 # vx = x # a x = vx # vy = y # a y = vy # vz = z # a z = vz z s k i v a z r xi yj zk y j x y x 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. (+ c) vy = (v0)y + act (+ c) y = y0 + (v0)yt + 12 act2 (+ c) v2y = (v0)2y + 2ac(y - y0) + ) (S x = x0 + (v0)xt y ag vx v0 (v0)y vy (v0)x v r y y0 x x0 x B_5389_ch12_ptg01_hr.4c.indd 107 1/21/15 10:01 AM 108 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE 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. 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 = f(u) is given, then the chain rule of calculus must be used to obtain time derivatives. (See Appendix C.) O¿ n O # at = v or an = an s at ds = v dv v2 r a at v t # vr = r v # vu = ru # $ ar = r - ru2 $ # # au = ru + 2ru 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 B_5389_ch12_ptg01_hr.4c.indd 108 1/21/15 10:01 AM 109 CHAPTER REVIEW 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¿ rB = rA + rB>A a z vB = vA + vB>A aB = aA + aB>A A a Translating observer rB/A rA Fixed observer x¿ b B y O y¿ rB b x B_5389_ch12_ptg01_hr.4c.indd 109 1/21/15 10:01 AM 110 12 CHAPTER 12 K I N E M AT I C S OF A PARTICLE REVIEW PROBLEMS R12–1. The position of a particle along a straight line is given by s = (t 3 - 9t 2 + 15t) ft, where t is in seconds. Determine its maximum acceleration and maximum velocity during the time interval 0 … t … 10 s. R12–2. If a particle has an initial velocity v0 = 12 ft>s to the right, and a constant acceleration of 2 ft>s2 to the left, determine the particle’s displacement in 10 s. Originally s0 = 0. R12–5. A car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points A, B, and C. If it takes 3 s to go from A to B, and then 5 s to go from B to C, determine the average acceleration between points A and B and between points A and C. y vC 40 m/s x R12–3. A projectile, initially at the origin, moves along a straight-line path through a fluid medium such that its velocity is v = 1800(1 - e-0.3t) mm>s where t is in seconds. Determine the displacement of the projectile during the first 3 s. vB 30 m/s B C 45 vA 20 m/s A Prob. R12–5 R12–4. The v–t graph of a car while traveling along a road is shown. Determine the acceleration when t = 2.5 s, 10 s, and 25 s. Also if s = 0 when t = 0, find the position when t = 5 s, 20 s, and 30 s. R12–6. From a videotape, it was observed that a player kicked a football 126 ft during a measured time of 3.6 seconds. Determine the initial speed of the ball and the angle u at which it was kicked. v (m/s) 20 v0 u 5 20 Prob. R12–4 B_5389_ch12_ptg01_hr.4c.indd 110 30 t (s) A 126 ft Prob. R12–6 1/21/15 10:01 AM 111 REVIEW PROBLEMS R12–7. The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from # s = 0, its speed is increased by v = (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. R12–9. A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by 12 u = (5t2) rad, where t is in seconds. Determine the magnitude of the particle’s acceleration when u = 30°. The particle starts from rest when u = 0°. R12–10. Determine the time needed for the load at B to attain a speed of 8 m>s, starting from rest, if the cable is drawn into the motor with an acceleration of 0.2 m>s2. . v (0.05s) m/s2 v 4 m/s A vA 50 m Prob. R12–7 B vB Prob. R12–10 R12–8. 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. R12–11. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 600 km>h and vB = 500 km>h such that the angle between their straightline courses is u = 75, determine the velocity of plane B with respect to plane A. v B A vA B u 5m A u Prob. R12–8 B_5389_ch12_ptg01_hr.4c.indd 111 vB Prob. R12–11 1/21/15 10:01 AM

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