Kinematics of a Particle

Although each of these planes is rather large, from a distance their motion can be modeled as if each plane were a particle. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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. The mechanics of rigid bodies is divided into two areas: statics and dynamics. Statics is concerned with the equilibrium of a body that is either at rest or moves with constant velocity. Here we are concerned with dynamics, which deal 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.

3 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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C H A P T E R 12

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Historically, the principles of dynamics developed when it was possible to make an accurate measurement of time. Galileo Galilei (1564–1642) was one of the first major contributors to this field. His work consisted of experiments using pendulums and falling bodies. The most significant contributions in dynamics, however, were made by Isaac Newton (1642–1727), who is noted for his formulation of the three fundamental laws of motion and the law of universal gravitational attraction. Shortly after these laws were postulated, important techniques for their application were developed by Euler, D’Alembert, Lagrange, and others. There are many problems in engineering whose solutions require application of the principles of dynamics. Typically the structural design of any vehicle, such as an automobile or airplane, requires consideration of the motion to which it is subjected. This is also true for many mechanical devices, such as motors, pumps, movable tools, industrial manipulators, and machinery. Furthermore, predictions of the motions of artificial satellites, projectiles, and spacecraft are based on the theory of dynamics. With further advances in technology, there will be an even greater need for knowing how to apply the principles of this subject.

Problem Solving. Dynamics is considered to be more involved than

statics since both the forces applied to a body and its motion must be taken into account. Also, many applications require using calculus, rather than just algebra and trigonometry. In any case, the most effective way of learning the principles of dynamics is to solve problems. To be successful at this, it is necessary to present the work in a logical and orderly manner as suggested by the following sequence of steps: 1. Read the problem carefully and try to correlate the actual physical situation with the theory you have studied. 2. Draw any necessary diagrams and tabulate the problem data. 3. Establish a coordinate system and apply the relevant principles, generally in mathematical form. 4. Solve the necessary equations algebraically as far as practical; then, use a consistent set of units and complete the solution numerically. Report the answer with no more significant figures than the accuracy of the given data. 5. Study the answer using technical judgment and common sense to determine whether or not it seems reasonable. 6. Once the solution has been completed, review the problem. Try to think of other ways of obtaining the same solution. In applying this general procedure, do the work as neatly as possible. Being neat generally stimulates clear and orderly thinking, and vice versa.

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION

12.2 Rectilinear Kinematics: Continuous 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. Such objects can be considered as particles, as long as the motion of the body is characterized by 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 vector r is used to specify the location of the particle P at any given instant. Notice that r is always along the s axis, and so its direction never changes. What will change is its magnitude and its sense or arrowhead direction. For analytical work it is therefore convenient to represent r by an algebraic scalar s, representing the position coordinate of the particle, Fig. 12–1a. The magnitude of s (and r) is the distance from O to P, usually measured in meters (m) or feet (ft), and the sense (or arrowhead direction of r) 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.

r P O

¢s = s¿ - s Here ¢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. Since the displacement of a particle is a vector quantity, it should be distinguished from the distance the particle travels. Specifically, the distance traveled is a positive scalar which represents the total length of path over which the particle travels.

s Position (a)

Fig. 12–1a

r¿

Displacement. The displacement of the particle is defined as the change in its position. For example, if the particle moves from P to P¿, Fig. 12–1b, the displacement is ¢r = r¿ - r. Using algebraic scalars to represent ¢r, we also have

s

r

'r P

P¿

O s

's s¿ Displacement (b)

Fig. 12–1b


s

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Velocity. If the particle moves through a displacement ¢r from P to

P¿ during the time interval ¢t, Fig. 12–1b, the average velocity of the particle during this time interval is vavg =

¢r ¢t

If we take smaller and smaller values of ¢t, the magnitude of ¢r becomes smaller and smaller. Consequently, the instantaneous velocity is defined as v = lim 1¢r>¢t2, or ¢t : 0

v =

dr dt

Representing v as an algebraic scalar, Fig. 12–1c, we can also write

v P

P¿

O 's

s

+ 2 1:

v =

ds dt

(12–1)

Velocity (c)

Fig. 12–1c

Since ¢t or dt is always positive, the sign used to define the sense of the velocity is the same as that of ¢s or ds. For example, if the particle is moving to the right, Fig. 12–1c, the velocity is positive; whereas if it is moving to the left, the velocity is negative. (This is emphasized here by the arrow written at the left of Eq. 12–1.) The magnitude of the velocity is known as the speed, and it is generally expressed in units of m>s or ft>s. Occasionally, the term “average speed” is used. The average speed is always a positive scalar and is defined as the total distance traveled by a particle, sT , divided by the elapsed time ¢t; i.e., 1vsp2avg =

sT ¢t

For example, the particle in Fig. 12–1d travels along the path of length sT in time ¢t, so its average speed is 1vsp2avg = sT>¢t, but its average velocity is vavg = - ¢s>¢t. #'s P¿

P

O

s

sT Average velocity and Average speed (d)

Fig. 12–1 (cont.) Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

7


Acceleration. Provided the velocity of the particle is known at the

two points P and P¿, the average acceleration of the particle during the time interval ¢t is defined as

a P

P¿

O v

aavg =

¢v ¢t

s

v¿

Acceleration (e)

Fig. 12–1e

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 found by taking smaller and smaller values of ¢t and corresponding smaller and smaller values of ¢v, so that a = lim 1¢v>¢t2 or, using algebraic scalars, ¢t : 0

+ 2 1:

a =

dv dt

(12–2)

Substituting Eq. 12–1 into this result, we can also write

+ 2 1:

a =

d2s dt2 "a

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, it is said to be decelerating. In this case, v¿ in Fig. 12–1f is less than v, and so ¢v = v¿ - v will be negative. Consequently, a will also be negative, and therefore it will act to the left, in the opposite sense to v. Also, note that when the velocity is constant, the acceleration is zero since ¢v = v - v = 0. Units commonly used to express the magnitude of acceleration are m>s2 or ft>s2. A differential relation involving the displacement, velocity, and acceleration along the path may be obtained by eliminating the time differential dt between Eqs. 12–1 and 12–2, which gives + 2 1:

a ds = v dv

P

P¿

O v

v¿

Deceleration (f)

Fig. 12–1f

(12–3)

Although we have now produced three important kinematic equations, realize that Eq. 12–3 is not independent of Eqs. 12–1 and 12-2. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

s

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Constant Acceleration, a = ac . When the acceleration is constant, each of the three kinematic equations ac = dv>dt, v = ds>dt, and ac ds = v dv may be integrated to obtain formulas that relate ac , v, s, and t.

Velocity as a Function of Time. Integrate ac = dv>dt, assuming that initially v = v0 when t = 0. v

Lv0 + 2 1:

t

dv =

L0

ac dt

v = v0 + act Constant Acceleration

(12–4)

Position as a Function of Time. Integrate v = ds>dt = v0 + act, assuming that initially s = s0 when t = 0. s

Ls0 + 2 1:

t

ds =

L0

1v0 + act2 dt

s = s0 + v0t + 12 act2 Constant Acceleration

(12–5)

Velocity as a Function of Position. Either solve for t in

Eq. 12–4 and substitute into Eq. 12–5, or integrate v dv = ac ds, assuming that initially v = v0 at s = s0 . v

Lv0 + 2 1:

s

v dv =

Ls0

ac ds

v2 = v20 + 2ac1s - s02 Constant Acceleration

(12–6)

This equation is not independent of Eqs. 12–4 and 12–5 since it can be obtained by eliminating t between these equations. The magnitudes and signs of s0 , v0 , and ac , used in the above three equations are determined from the chosen origin and positive direction of the s axis as indicated by the arrow written at the left of each equation. Also, it is important to remember that these equations are useful only when the acceleration is constant and when t = 0, s = s0 , v = v0 . A common 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. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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Important Points • • • • • •

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.

s

• The acceleration, a = dv>dt, is negative when the particle is slowing down or 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 rocket undergoes rectilinear motion, its altitude as a function of time can be measured and expressed as s = s1t2. Its velocity can then be found using v = ds>dt, and its acceleration can be determined from a = dv>dt.

Procedure for Analysis The equations of rectilinear kinematics should be applied using the following procedure. 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 particle’s position, velocity, and acceleration can be represented as algebraic scalars. For analytical work the sense of s, v, and a is then determined from their algebraic signs.

• The positive sense for each scalar can be indicated by an arrow shown alongside each kinematic equation as it is applied.

Kinematic Equations.

• If a relationship 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, which 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 a limited use. Never apply these equations unless it is absolutely certain that the acceleration is constant.

*Some standard differentiation and integration formulas are given in Appendix A. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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EXAMPLE 12.1 The car in Fig. 12–2 moves in a straight line such that for a short time its velocity is defined by v = 13t2 + 2t2 ft>s, where t is in seconds. Determine its position and acceleration when t = 3 s. When t = 0, s = 0. a, v

s

O

Fig. 12–2

SOLUTION Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right. Position. Since v = f1t2, the car’s position can be determined from v = ds>dt, since this equation relates v, s, and t. Noting that s = 0 when t = 0, we have* + 2 1:

v = s

L0

ds = 13t2 + 2t2 dt t

ds = s

L0

13t2 + 2t2 dt

s ` = t3 + t2 ` 0

3

s = t + t When t = 3 s,

t 0

2

s = 1323 + 1322 = 36 ft

Ans.

Acceleration. Knowing v = f1t2, the acceleration is determined from a = dv>dt, since this equation relates a, v, and t. + 2 1:

a =

dv d = 13t2 + 2t2 dt dt

= 6t + 2 When t = 3 s,

a = 6132 + 2 = 20 ft>s2 :

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 = 13t2 + 2t2 dt yields s = t3 + t2 + C. Using the condition that at t = 0, s = 0, then C = 0. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.


EXAMPLE 12.2 A small projectile is fired vertically downward into a fluid medium with an initial velocity of 60 m>s. Due to the resistance of the fluid the projectile experiences a deceleration equal to a = 1-0.4v32 m>s2, where v is in m>s. Determine the projectile’s velocity and position 4 s after it is fired. SOLUTION Coordinate System. Since the motion is downward, the position coordinate is positive downward, with origin located at O, Fig. 12–3. Velocity. Here a = f1v2 and so we must determine the velocity as a function of time using a = dv>dt, since this equation relates v, a, and t. (Why not use v = v0 + act?) Separating the variables and integrating, with v0 = 60 m>s when t = 0, yields dv = -0.4v3 dt v t dv = dt 3 L60 m>s -0.4v L0

1+ T2

a =

O s

1 1 1 v a b 2` = t - 0 -0.4 -2 v 60 1 1 1 c d = t 0.8 v2 16022 -1>2 1 + 0.8t d f m>s 16022 Here the positive root is taken, since the projectile is moving downward. When t = 4 s, v = 0.559 m>s T Ans.

v = ec

Fig. 12–3

Position. Knowing v = f1t2, we can obtain the projectile’s position from v = ds>dt, since this equation relates s, v, and t. Using the initial condition s = 0, when t = 0, we have 1+ T2

-1>2 ds 1 = c + 0.8t d dt 16022

v = s

L0

s = s = When t = 4 s,

t

ds =

-1>2 1 + 0.8t d dt 2 L0 1602

c

1>2 t 2 1 c + 0.8t d ` 0.8 16022 0

1>2 1 1 1 ec + 0.8t d fm 2 0.4 1602 60

s = 4.43 m

Ans.


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EXAMPLE 12.3 During a test a rocket is traveling 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.

vB $ 0 B

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, sB

1+ c 2

v2B = v2A + 2ac1sB - sA2 0 = 175 m>s22 + 21-9.81 m>s221sB - 40 m2

vA $ 75 m/s

sB = 327 m

A

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.

sA $ 40 m C

s

Ans.

O

1+ c 2

Fig. 12–4

v2C = v2B + 2ac1sC - sB2 = 0 + 21-9.81 m>s2210 - 327 m2 vC = -80.1 m>s = 80.1 m>s T

Ans.

The negative root was chosen since the rocket is moving downward. Similarly, Eq. 12–6 may also be applied between points A and C, i.e., 1+ c 2

v2C = v2A + 2ac1sC - sA2 = 175 m>s22 + 21-9.81 m>s2210 - 40 m2 vC = -80.1 m>s = 80.1 m>s T

NOTE: It should be realized that the rocket is subjected to a

deceleration from A to B of 9.81 m>s2, and then from B to C it is accelerated at this rate. Furthermore, even though the rocket momentarily comes to rest at B 1vB = 02 the acceleration at B is 9.81 m>s2 downward! Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

13


EXAMPLE 12.4 A metallic particle is subjected to the influence of a magnetic field as it travels downward through a fluid that extends from plate A to plate B, Fig. 12–5. If the particle is released from rest at the midpoint C, s = 100 mm, and the acceleration is a = 14s2 m>s2, where s is in meters, determine the velocity of the particle when it reaches plate B, s = 200 mm, and the time it needs to travel from C to B. SOLUTION Coordinate System. As shown in Fig. 12–5, s is taken positive downward, measured from plate A. Velocity. Since a = f1s2, the velocity as a function of position can be obtained by using v dv = a ds. Realizing that v = 0 at s = 0.1 m, we have 1+ T2

A

v dv = a ds v

L0

s

v dv = 1 2 2v

v

4s ds

4 2

` = s2 ` 0

s

L0.1

s

C

200 mm

0.1

v = 21s2 - 0.0121>2

(1)

At s = 200 mm = 0.2 m, vB = 0.346 m>s = 346 mm>s T Ans. The positive root is chosen since the particle is traveling downward, i.e., in the +s direction. Time. The time for the particle to travel from C to B can be obtained using v = ds>dt and Eq. 1, where s = 0.1 m when t = 0. From Appendix A, 1+ T2

100 mm

B

Fig. 12–5

ds = v dt = 21s2 - 0.0121>2 dt s

t

ds = 2 dt 2 1>2 L0.1 1s - 0.012 L0 ln A 4s2 - 0.01 + s B `

s 0.1

= 2t `

t 0

ln A 4s2 - 0.01 + s B + 2.303 = 2t At s = 0.2 m, t =

ln A 410.222 - 0.01 + 0.2 B + 2.303 2

= 0.658 s

Ans.

NOTE: The formulas for constant acceleration cannot be used here,

because the acceleration changes with position, i.e., a = 4s.


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EXAMPLE 12.5 A particle moves along a horizontal path with a velocity of v = 13t2 - 6t2 m>s, where t is the time in seconds. If it is initially located at the origin O, determine the distance traveled in 3.5 s, and the particle’s average velocity and average speed during the time interval. s $ #4.0 m

SOLUTION Coordinate System. Here we will assume positive motion to the right, measured from the origin O, Fig. 12–6a.

s $ 6.125 m O

t$2s

t$0s

t $ 3.5 s

Distance Traveled. Since v = f1t2, the position as a function of time may be found by integrating v = ds>dt with t = 0, s = 0. + 2 1:

(a)

Fig. 12–6a

ds = v dt = 13t2 - 6t2 dt s

L0

t

ds = 3

L0

t

2

t dt - 6

L0

t dt

s = 1t3 - 3t22 m v (m/s) 2

v $ 3t # 6t (0, 0)

(2 s, 0)

(1 s, #3 m/s) (b)

Fig. 12–6b

t (s)

(1)

In order to determine the distance traveled in 3.5 s, it is necessary to investigate the path of motion. The graph of the velocity function, Fig. 12–6b, reveals that for 0 … 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, 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 ƒ t=0 = 0

s ƒ t = 2 s = -4.0 m

s ƒ t = 3.5 s = 6.125 m

The path is shown in Fig. 12–6a. Hence, the distance traveled in 3.5 s is sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m

Ans.

Velocity. The displacement from t = 0 to t = 3.5 s is ¢s = s ƒ t = 3.5 s - s ƒ t = 0 = 6.12 - 0 = 6.12 m and so the average velocity is ¢s 6.12 = = 1.75 m>s : Ans. ¢t 3.5 - 0 The average speed is defined in terms of the distance traveled sT . This positive scalar is sT 14.125 1vsp2avg = = = 4.04 m>s Ans. ¢t 3.5 - 0 vavg =

NOTE: In this problem, the acceleration is a = dv>dt = 16t - 62 m>s2,

which is not constant.


PROBLEMS

15

PROBLEMS 12–1. A truck, traveling along a straight road at 20 km>h, increases its speed to 120 km>h in 15 s. If its acceleration is constant, determine the distance traveled. 12–2. A car starts from rest and reaches a speed of 80 ft>s after traveling 500 ft along a straight road. Determine its constant acceleration and the time of travel. 12–3. A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft>s. Determine the speed at which it hits the ground and the time of travel. *12–4. Starting from rest, a particle moving in a straight line has an acceleration of a = 12t - 62 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? 12–5. 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? 12–6. A freight train travels at v = 6011 - e -t2 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

*12–8. From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed of 80.7 ft>s 155 mi>h2 when it hits the ground? Each floor is 12 ft higher than the one below it. (Note: You may want to remember this when traveling 55 mi>h.)

12–9. A particle moves along a straight line such that its position is defined by s = 1t3 - 3t2 + 22 m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 4 s.

12–10. A particle is moving along a straight line such that its acceleration is defined as a = 1-2v2 m>s2, where v is in meters per second. If v = 20 m>s when s = 0 and t = 0, determine the particle’s velocity as a function of position and the distance the particle moves before it stops.

12–11. The acceleration of a particle as it moves along a straight line is given by a = 12t - 12 m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.

*12–12. A particle, initially at the origin, moves along a straight line through a fluid medium such that its velocity is defined as v = 1.811 - e -0.3t2 m>s, where t is in seconds. Determine the displacement of the particle during the first 3 s.

Prob. 12–6

12–7. The position of a particle along a straight line is given by s = 1t3 - 9t2 + 15t2 ft, where t is in seconds. Determine its maximum acceleration and maximum velocity during the time interval 0 … t … 10 s.

12–13. The velocity of a particle traveling in a straight line is given by v = 16t - 3t22 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?


16

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–14. A particle moves along a straight line such that its position is defined by s = 1t2 - 6t + 52 m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. 12–15. A particle is moving along a straight line such that when it is at the origin it has a velocity of 4 m>s. If it begins to decelerate at the rate of a = 1- 1.5v1>22 m>s2, where v is in m>s, determine the particle’s position and velocity when t = 2 s.

12–21. A particle has an initial speed of 27 m>s. If it experiences a deceleration of a = 1-6t2 m>s2, where t is in seconds, determine the distance traveled before it stops. 12–22. The acceleration of a rocket traveling upward is given by a = 16 + 0.02s2 m>s2, where s is in meters. Determine the rocket’s velocity when s = 2 km and the time needed to reach this altitude. Initially, v = 0 and s = 0 when t = 0.

*12–16. A particle travels to the right along a straight line with a velocity v = [5>14 + s2] m>s, where s is in meters. Determine its deceleration when s = 2 m. 12–17. Two particles A and B start from rest at the origin s = 0 and move along a straight line such that aA = 16t - 32 ft>s2 and a B = 112t2 - 82 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. 12–18. A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at 0.6 ft>s2, decelerate at 0.3 ft>s2, and reach a maximum speed of 8 ft>s, determine the shortest time to make the lift, starting from rest and ending at rest. 12–19. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later. *12–20. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the time interval between the instant A strikes the water and the instant B strikes the water. Also, at what speed do they strike the water?

s

Prob. 12–22 12–23. The acceleration of a rocket traveling upward is given by a = 16 + 0.02s2 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.

B A 80 ft

s

Probs. 12–19/20

Prob. 12–23


17

PROBLEMS *12–24. 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.

"12–25. A particle has an initial speed of 27 m>s. If it experiences a deceleration of a = 1-6t2 m>s2, where t is in seconds, determine its velocity when it travels 10 m. How much time does this take?

12–26. Ball A is released from rest at a height of 40 ft at the same time that a second ball B is thrown upward 5 ft from the ground. If the balls pass one another at a height of 20 ft, determine the speed at which ball B was thrown upward.

*12–28. The acceleration of a particle along a straight line is defined by a = 12t - 92 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.

12–29. A particle is moving along a straight line such that its acceleration is defined as a = 14s22 m>s2, where s is in meters. If v = -100 m>s when s = 10 m and t = 0, determine the particle’s velocity as a function of position.

12–30. A car can have an acceleration and a deceleration of 5 m>s2. If it starts from rest, and can have a maximum speed of 60 m>s, determine the shortest time it can travel a distance of 1200 m when it stops.

12–31. Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m>s2 and decelerate at 2 m>s2.

A

40 ft

B 5 ft

*12–32. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB , respectively. If B maintains its constant speed, while A begins to decelerate at aA , determine the distance d between the cars at the instant A stops.

Prob. 12–26

A

B d

12–27. A car starts from rest and moves along a straight line with an acceleration of a = 13s -1>32 m>s2, where s is in meters. Determine the car’s velocity and position when t = 6 s.

Prob. 12–32


18

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–33. If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation a = 9.81[1 - v2110 -42] m>s2, where v is in m>s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as t : q ).

12–35. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–34), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y0 = 500 km? Use the numerical data in Prob. 12–34.

12–34. 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>1R + y22], where g0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g0 = 9.81 m>s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that v = 0 as y : q .

*12–36. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf . If this variation of the acceleration can be expressed as a = 1g>v2f21v2f - v22, determine the time needed for the velocity to become v = vf . Initially the particle falls from rest.

12.3 Rectilinear Kinematics: Erratic Motion When a particle’s motion during a time period is erratic or there is a discontinuity in the motion, then it may be difficult to obtain a continuous mathematical function to describe its position, velocity, or acceleration. Instead, the motion may best be described graphically using a series of curves that can be generated experimentally. If the resulting graph describes the relationship between any two of the variables, a, v, s, t, a graph describing the relationship between the other variables can be established by using the kinematic equations a = dv>dt, v = ds>dt, a ds = v dv. Several situations occur frequently. s ds ds v0 $ dt t $ 0 v2 $ dt t 2 ds ds v1 $ dt t v3 $ dt t 1 3

s1 O

t1

s2

t2 (a)

Fig. 12–7a

s3 t3

t

Given the s –t Graph, Construct the v–t Graph. If the position of a particle can be determined experimentally during a time period t, the s – t graph for the particle can be plotted, Fig. 12–7a. To determine the particle’s velocity as a function of time, i.e., the v–t graph, we must use v = ds>dt since this equation relates v, s, and t. Therefore, the velocity at any instant is determined by measuring the slope of the s–t graph, i.e., ds = v dt slope of s–t graph = velocity

(12–36)


19

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION

For example, measurement of the slopes v0 , v1 , v2 , v3 at the intermediate points (0, 0), (t1 , s1), (t2 , s2), (t3 , s3) on the s–t graph, Fig. 12–7a, gives the corresponding points on the v–t graph shown in Fig. 12–7b. It may also be possible to establish the v–t graph mathematically, provided the segments of the s–t graph can be expressed in the form of equations s = f1t2. Corresponding equations describing the segments of the v– t graph are then determined by time differentiation, since v = ds>dt.

Given the v – t Graph, Construct the a–t Graph. When the

particle’s v–t graph is known, as in Fig. 12–8a, the acceleration as a function of time, i.e., the a–t graph, can be determined using a = dv>dt. Hence, the acceleration at any instant is determined by measuring the slope of the v–t graph, i.e.,

v

v0

v1 v2

O

t1

t2

t3 v3

t

(b)

Fig. 12–7b

dv = a dt slope of v–t graph = acceleration

For example, measurement of the slopes a0 , a1 , a2 , a3 at the intermediate points (0, v0), (t1 , v1), (t2 , v2), (t3 , v3) on the v–t graph, Fig. 12–8a, yields the corresponding points on the a –t graph shown in Fig. 12–8b. Any segments of the a–t graph can also be determined mathematically, provided the equations of the corresponding segments of the v–t graph are known, v = g1t2. This is done by simply taking the time derivative of v = g1t2, since a = dv>dt. Since differentiation reduces a polynomial of degree n to that of degree n – 1, then if the s–t graph is parabolic (a second-degree curve), the v–t graph will be a sloping line (a first-degree curve), and the a–t graph will be a constant or a horizontal line (a zero-degree curve). Fig. 12–8b

v — a2 $ dv dt

— $0 a0 $ dv dt t$0

— a1 $ dv dt

v0 O

v1 t1

t1

v2

t2 (a)

t2

— a3 $ dv dt

a

t3

v3

t3

t

O

a0 $ 0

a1 t1

a2 t2

a3 t3

t

(b)

Fig. 12–8a


20

C H A P T E R 12

K I N E M AT I C S

EXAMPLE 12.6

OF A

PA R T I C L E

A bicycle moves along a straight road such that its position is described by the graph shown in Fig. 12–9a. Construct the v–t and a–t graphs for 0 … t … 30 s. s (ft)

500

s $ 20t # 100

100

s $ t2

10

30

t (s)

(a)

Fig. 12–9a SOLUTION v–t Graph. Since v = ds>dt, the v–t graph can be determined by differentiating the equations defining the s –t graph, Fig. 12–9a.We have

v (ft/s) v $ 2t

v $ 20

20

s = t2

0 … t 6 10 s;

30

t (s)

(b)

Fig. 12–9b

t = 20 s;

a (ft/s2)

ds = 2t dt

ds = 20 dt The results are plotted in Fig. 12–9b. 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., 10 s 6 t … 30 s;

10

v =

s = 20t - 100

v =

v =

¢s 500 - 100 = = 20 ft>s ¢t 30 - 10

a–t Graph. Since a = dv>dt, the a–t graph can be determined by differentiating the equations defining the lines of the v–t graph.This yields

2

30

10 (c)

Fig. 12–9c

t (s)

0 … t 6 10 s;

v = 2t

a =

dv = 2 dt

10 6 t … 30 s;

v = 20

a =

dv = 0 dt

The results are plotted in Fig. 12–9c. NOTE: Show that a = 2 ft>s2 when t = 5 s by measuring the slope of

the v–t graph.


21


Given the a–t Graph, Construct the v–t Graph. If the a–t

a

graph is given, Fig. 12–10a, the v–t graph may be constructed using a = dv>dt, written in integrated form as

a0

¢v =

t

'v $ "01 a dt t1

a dt

L change in area under = velocity a–t graph

t

(a) v

Hence, to construct the v–t graph, we begin with the particle’s initial velocity v0 and then add to this small increments of area 1¢v2 determined from the a–t graph. In this manner successive points, v1 = v0 + ¢v, etc., for the v–t graph are determined, Fig. 12–10b. Notice that an algebraic addition of the area increments 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). If segments of the a–t graph can be described by a series of equations, then each of these equations may be integrated to yield equations describing the corresponding segments of the v–t graph. For example, if the a–t graph is linear (a first-degree curve), integration will yield a v–t graph that is parabolic (a second-degree curve), etc.

Given the v–t Graph, Construct the s–t Graph. When the v –t graph is given, Fig. 12–11a, it is possible to determine the s–t graph using v = ds>dt, written in integrated form

'v

v1 v0

t

t1 (b)

Fig. 12–10

v

v0 t

's $ "01 v dt t1

t

(a)

¢s =

v dt L displacement = area under v–t graph

In the same manner as stated above, we begin with the particle’s initial position s0 and add (algebraically) to this small area increments ¢s determined from the v–t graph, Fig. 12–11b. If it is possible to describe segments of the v–t graph by a series of equations, then each of these equations may be integrated to yield equations that describe corresponding segments of the s–t graph.

s

s1

's

s0 t

t1 (b)

Fig. 12–11


22

C H A P T E R 12

K I N E M AT I C S

OF A

EXAMPLE 12.7

The test car in Fig. 12–12a starts from rest and travels along a straight track such that it accelerates at a constant rate for 10 s and then decelerates at a constant rate. Draw the v–t 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 #2

t¿

A2

10

PA R T I C L E

t (s)

SOLUTION v–t Graph. Since dv = a dt, the v–t graph is determined by integrating the straight-line segments of the a–t graph. Using the initial condition v = 0 when t = 0, we have v

(a)

a = 10;

10 s 6 t … t¿;

a = -2;

t

10 dt, v = 10t L0 L0 When t = 10 s, v = 101102 = 100 m>s. Using this as the initial condition for the next time period, we have

Fig. 12–12a

0 … t 6 10 s;

dv =

v

t

-2 dt, v = -2t + 120 L100 L10 When t = t¿ we require v = 0. This yields, Fig. 12–12b, t¿ = 60 s Ans.

v (m/s) v $ 10t 100

A more direct solution for t¿ is possible by realizing that the area under the a–t graph is equal to the change in the car’s velocity. We require ¢v = 0 = A 1 + A 2 , Fig. 12–12a. Thus

v $ #2t " 120

10

dv =

t¿ $ 60

t (s)

(b)

Fig. 12–12b

0 = 10 m>s2110 s2 + 1-2 m>s221t¿ - 10 s2 t¿ = 60 s Ans. s–t Graph. Since ds = v dt, integrating the equations of the v –t graph yields the corresponding equations of the s–t graph. Using the initial condition s = 0 when t = 0, we have s

3000

0 … t … 10 s;

v = 10t;

ds =

10 s … t … 60 s;

v = -2t + 120;

s

s $ 5t2 500

t

10t dt, s = 5t2 L0 L0 When t = 10 s, s = 511022 = 500 m. Using this initial condition,

s (m)

2

s $ #t " 120t # 600 10

60 (c)

Fig. 12–12c

t (s)

t

L500

ds =

L10

1-2t + 1202 dt

s - 500 = -t2 + 120t - [-11022 + 1201102] s = -t2 + 120t - 600 When t¿ = 60 s, the position is s = -16022 + 1201602 - 600 = 3000 m

Ans.

The s–t graph is shown in Fig. 12–12c. NOTE: A direct solution for s is possible when t¿ = 60 s, since the

triangular area under the v –t graph would yield the displacement ¢s = s - 0 from t = 0 to t¿ = 60 s. Hence, ¢s = 12160211002 = 3000 m

Ans.


23


Given the a–s Graph, Construct the v–s Graph. In some

cases an a–s graph for the particle can be constructed, so that points on the v –s graph can be determined by using v dv = a ds. Integrating this equation between the limits v = v0 at s = s0 and v = v1 at s = s1 , we have,

a

a0

s

"01a ds $ —12 (v12 # v02) s

s1 1 2 2 1v1

- v202 =

(a)

s1

a ds Ls0 area under a–s graph

Fig. 12–13a

v

v1 s

Thus, the initial small segment of area under the a–s graph, 1s01a ds, shown colored in Fig. 12–13a, equals one-half the difference in the squares of the speed, 12 1v21 - v202. Therefore, if the area is determined and the initial value s of v0 at s0 = 0 is known, then v1 = A 2 1s01a ds + v20 B 1>2, Fig. 12–13b. Successive points on the v–s graph can be constructed in this manner starting from the initial velocity v0 . Another way to construct the v–s graph is to first determine the equations which define the segments of the a–s graph. Then the corresponding equations defining the segments of the v–s graph can be obtained directly from integration, using v dv = a ds.

Given the v–s Graph, Construct the a–s Graph. If the v–s graph is known, the acceleration a at any position s can be determined using a ds = v dv, written as

v0 s

s1 (b)

Fig. 12–13b v dv ds

v0

v

s

s

dv b ds acceleration = velocity times slope of v–s graph a = va

(a)

a

Fig. 12–14a

a0

Thus, at any point (s, v) in Fig. 12–14a, the slope dv>ds of the v–s graph a $ v(dv/ds) is measured. Then since v and dv>ds are known, the value of a can be calculated, Fig. 12–14b. We can also determine the segments describing the a –s graph s analytically, provided the equations of the corresponding segments of (b) the v–s graph are known. As above, this requires integration using Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, a ds = v dv. Fig. 12–14b

New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

s

24

C H A P T E R 12

K I N E M AT I C S

EXAMPLE 12.8

PA R T I C L E

The v–s graph describing the motion of a motorcycle is shown in Fig. 12–15a. Construct the a–s graph of the motion and determine the time needed for the motorcycle to reach the position s = 400 ft. SOLUTION a–s Graph. Since the equations for segments of the v–s graph are given, the a–s graph can be determined using a ds = v dv.

v (ft/s)

50

OF A

v $ 0.2s " 10 v $ 50

0 … s 6 200 ft; a = v

10 200

400

s (ft)

v = 0.2s + 10

dv d = 10.2s + 102 10.2s + 102 = 0.04s + 2 ds ds

200 ft 6 s … 400 ft;

v = 50;

(a)

Fig. 12–15a

a = v

d dv = 1502 1502 = 0 ds ds

The results are plotted in Fig. 12–15b.

a (ft/s2) a $ 0.04s " 2 10 2

a$0 200

400 (b)

Fig. 12–15b

s (ft)

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 at t = 0, so ds ds 0 … s 6 200 ft; v = 0.2s + 10; dt = = v 0.2s + 10 t s ds dt = L0 L0 0.2s + 10 t = 5 ln10.2s + 102 - 5 ln 10 At s = 200 ft, t = 5 ln[0.212002 + 10] - 5 ln 10 = 8.05 s. Therefore, for the second segment of motion, ds ds 200 ft 6 s … 400 ft; v = 50; dt = = v 50 t s ds s dt = ; t - 8.05 = - 4 50 L8.05 L200 50 t =

s + 4.05 50

Therefore, at s = 400 ft, t =

400 + 4.05 = 12.0 s 50

Ans.

NOTE: The graphical results can be checked in part by calculating slopes.

For example, at s = 0, a = v1dv>ds2 = 10150 - 102>200 = 2 m>s2. Also, the results can be checked in part by inspection. The v–s graph indicates the initial increase in velocity (acceleration) followed by constant velocity 1a = 02.


25


PROBLEMS 12–37. 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. 12–38. 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.

12–41. 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. 12–42. 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. smax v s

vmax

40 ft

t¿/ 2

t¿

t

Probs. 12–41/42

Prob. 12–38

12–43. A car starting from rest moves along a straight track with an acceleration as shown. Determine the time –t for the car to reach a speed of 50 m>s and construct the v–t graph that describes the motion until the time t. a (m/s2)

12–39. If the position of a particle is defined as s = 15t - 3t22 ft, where t is in seconds, construct the s–t, v–t, and a–t graphs for 0 … t … 10 s. *12–40. If the position of a particle is defined by s = [2 sin1p>52t + 4] m, where t is in seconds, construct the s–t, v–t, and a–t graphs for 0 … t … 10 s.

8

10

t

t (s)

Prob. 12–43


26

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

*12–44. 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.

12–46. A car travels along a straight road with the speed shown by the v–t graph. Determine the total distance the car travels until it stops when t = 48 s. Also plot the s–t and a–t graphs.

v (m/s)

v (m/s) 5

6

v$5 v $ 1.25t

v$

1 — t 5

1

v $ #— 3 (t # 48)

v $ #t " 15

4

10

t (s)

15

30

t (s)

48

Prob. 12–44 Prob. 12–46

12–45. From experimental data, the motion of a jet plane while traveling along a runway is defined by the v–t graph shown. Construct the s–t and a–t graphs for the motion.

v (m/s)

12–47. The v–t graph for the motion of a train as it moves from station A to station B is shown. Draw the a–t graph and determine the average speed and the distance between the stations.

v (ft/s)

40 80

10

40

Prob. 12–45

t (s)

30

90

120

t (s)

Prob. 12–47


27

PROBLEMS *12–48. 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 = 10.4t22 m, and then it becomes straight for t Ú 30 s.

"12–51. The a–s graph for a boat moving along a straight path is given. If the boat 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.

s (m)

a (ft/s2)

600

360

a $ 5 " 6( s # 10)5/3 5 30

t (s)

40

100

Prob. 12–48

s (ft)

Prob. 12–51 12–49. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the a–t graph and determine the maximum acceleration during the 30-s time interval. The car starts from rest at s = 0. 12–50. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the s–t graph and determine the average speed and the distance traveled for the 30-s time interval. The car starts from rest at s = 0. v (ft/s)

60 40

*12–52. 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.

v $ t " 30 v $ 0.4t2

10

Probs. 12–49/50

30

t (s)

12–53. 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>s2 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?


28

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–54. A two-stage rocket is fired vertically from rest at s = 0 with an 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

*12–56. 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 200 ft. Also, how much time is required for it to travel 200 ft?

a (ft/s2)

A

15

75 a $ 75 # 0.15s

9 a $ 0.01t2

s (ft)

500 30

t (s)

60

Prob. 12–56

Prob. 12–54 12–55. The a–t graph for a motorcycle traveling along a straight road has been estimated as shown. Determine the time needed for the motorcycle to reach a maximum speed of 100 ft>s and the distance traveled in this time. Draw the v–t and s–t graphs. The motorcycle starts from rest at s = 0.

12–57. The jet car is originally traveling at a speed of 20 m>s when it is subjected to the acceleration shown in the graph. Determine the car’s maximum speed and the time t when it stops.

a (m/s2)

a (ft/s2)

20 a $ 0.5t " 5 2

10

a $ 10t 10

10

30

Prob. 12–55

t

t (s) 20

t (s)

Prob. 12–57


29

PROBLEMS 12–58. A motorcyclist at A is traveling at 60 ft>s when he wishes to pass the truck T which is traveling at a constant speed of 60 ft>s. To do so the motorcyclist accelerates at 6 ft>s2 until reaching a maximum speed of 85 ft>s. If he then maintains this speed, determine the time needed for him to reach a point located 100 ft in front of the truck. Draw the v–t and s–t graphs for the motorcycle during this time.

*12–60. The a–t graph for a car is shown. Construct the v–t and s–t graphs if the car starts from rest at t = 0. At what time t¿ does the car stop?

a (m/s2) (vm)1 $ 60 ft/s

(vm)2 $ 85 ft/s vt $ 60 ft/s

A T 40 ft

5

55 ft

100 ft

t¿

Prob. 12–58

10

t (s)

#2

Prob. 12–60 12–59. The v–s graph for a go-cart traveling on a straight road is shown. Determine the acceleration of the go-cart at s = 50 m and s = 150 m. Draw the a–s graph.

12–61. The a–s graph for a train traveling along a straight track is given for the first 400 m of its motion. Plot the v–s graph. v = 0 at s = 0.

a (m/s2) v (m/s)

8 2 100

200

s (m) 200

Prob. 12–59

400

s (m)

Prob. 12–61


30

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–62. The v–s graph for an airplane traveling on a straight runway is shown. Determine the acceleration of the plane at s = 100 m and s = 150 m. Draw the a–s graph.

v (m/s)

v (m/s)

50 40

*12–64. The v–s graph for a test vehicle is shown. Determine its acceleration when s = 100 m and when s = 175 m.

v $ 0.1s " 30 50 v $ 0.4s

100

200

s (m) 150

Prob. 12–62

s (m)

200

Prob. 12–64

12–63. Starting from rest at s = 0, a boat travels in a straight line with an acceleration as shown by the a–s graph. Determine the boat’s speed when s = 40, 90, and 200 ft.

12–65. The v–s graph was determined experimentally to describe the straight-line motion of a rocket sled. Determine the acceleration of the sled when s = 100 m, and when s = 200 m.

v (m/s)

a (ft/s2)

60 4 2 20 50

150

250

s (ft)

50

Prob. 12–63

300

s (m)

Prob. 12–65


31

12.4 GENERAL CURVILINEAR MOTION

12.4 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.

Position. Consider a particle located at point P on a space curve

defined by the path function s, Fig. 12–16a. The position of the particle, measured from a fixed point O, will be designated by the position vector r = r1t2. This vector is a function of time since, in general, both its magnitude and direction change as the particle moves along the curve.

P

Path

O

Position

Displacement. Suppose that during a small time interval ¢t the

Fig. 12–16a

Velocity. During the time ¢t, the average velocity of the particle is vavg

P¿

¢r = ¢t

dr dt

(b)

(12–7)

¢t : 0

ds v = dt

s

Displacement

Fig. 12–16b

Since dr will be tangent to the curve at P, 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 noting that the magnitude of the displacement ¢r is the length of the straight line segment from P to P¿, Fig. 12–16b. Realizing that this length, ¢r, approaches the arc length ¢s as ¢t : 0, we have v = lim 1¢r>¢t2 = lim 1¢s>¢t2, or ¢t : 0

P

r

O

¢t : 0

v =

's 'r

r¿

The instantaneous velocity is determined from this equation by letting ¢t : 0, and consequently the direction of ¢r approaches the tangent to the curve at point P. Hence, v = lim 1¢r>¢t2 or

s

(a)

particle moves a distance ¢s along the curve to a new position P¿, 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. defined as

s

r

v P

O

r Velocity

(12–8)

Thus, the speed can be obtained by differentiating the path function s with respect to time.

s

(c)

Fig. 12–16c

*A summary of some of the important concepts of vector analysis is given in Appendix C. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

32

C H A P T E R 12 v¿

K I N E M AT I C S

OF A

PA R T I C L E

Acceleration. If the particle has a velocity v at time t and a velocity

v

v¿ = v + ¢v at t + ¢t, Fig. 12–16d, then the average acceleration of the particle during the time interval ¢t is

P

P¿

a avg = (d)

¢v ¢t

Fig. 12–16d

'v

where ¢v = v¿ - v. To study this time rate of change, the two velocity vectors in Fig. 12–16d are plotted in Fig. 12–16e such that their tails are located at the fixed point O¿ and their arrowheads touch points on a curve. This curve is called a hodograph, and when constructed, it describes the locus of points for the arrowhead of the velocity vector in the same manner as the path s describes the locus of points for the arrowhead of the position vector, Fig. 12–16a. To obtain the instantaneous acceleration, let ¢t : 0 in the above equation. In the limit ¢v will approach the tangent to the hodograph, and so a = lim 1¢v>¢t2, or

v v¿

O¿

(e)

Fig. 12–16e

¢t : 0

a =

dv dt

(12–9)

Hodograph

Substituting Eq. 12–7 into this result, we can also write

v

a

O¿

a =

(f)

Fig. 12–16f

P a Acceleration

path

(g)

Fig. 12–16 (cont.)

d 2r dt2

By definition of the derivative, a acts tangent to the hodograph, Fig. 12–16f, and therefore, in general, a 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 P to P¿, Fig. 12–16d. Just a magnitude change increases (or decreases) the “length” of v, and this in itself would allow a to remain tangent to the path. However, in order for the particle to follow the 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.


33

12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS

12.5 Curvilinear Motion:

Rectangular Components

Occasionally the motion of a particle can best be described along a path that is represented using a fixed x, y, z frame of reference.

Position. If at a given instant the particle P is at point (x, y, z) on the

z

curved path s, Fig. 12–17a, its location is then defined by the position vector r = xi + yj + zk

P

(12–10)

Because of the particle motion and the shape of the path, the x, y, z components of r are generally all functions of time; i.e., x = x1t2, y = y1t2, z = z1t2, so that r = r1t2. In accordance with the discussion in Appendix C, the magnitude of r is always positive and defined from Eq. C–3 as

s k i

z r $ xi " yj " zk

y

j

x

y

x

Position (a)

r = 4x2 + y2 + z2

Fig. 12–17a

The direction of r is specified by the components of the unit vector ur = r>r.

Velocity. The first time derivative of r yields the velocity v of the particle. Hence,

z

dr d d d v = = 1xi2 + 1yj2 + 1zk2 dt dt dt dt

P

When taking this derivative, it is necessary to account for changes in both the magnitude and direction of each of the vector’s components. The derivative of the i component of v is therefore d dx di 1xi2 = i + x dt dt dt

dr = vxi + vy j + vzk dt

v $ vxi " vyj " vzk y x

The second term on the right side is zero, since 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 =

s

Velocity (b)

Fig. 12–17b

(12–11)

where # # # vx = x vy = y vz = z

(12–12)


34

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

# # # The “dot” notation x, y, z represents the first time derivatives of the parametric equations x = x1t2, y = y1t2, z = z1t2, respectively. The velocity has a magnitude defined as the positive value of

v = 4v2x + v2y + v2z

and a direction that is specified by the components of the unit vector uv = v>v. This direction is always tangent to the path, as shown in Fig. 12–17b.

z P s a $ axi " ayj " azk

y

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). Using dots to represent the derivatives of the components, we have

x Acceleration (c)

a =

Fig. 12–17 (cont.)

dv = axi + ay j + azk dt

(12–13)

where # $ ax = vx = x # $ ay = vy = y # $ az = vz = z

(12–14)

Here ax , ay , az represent, respectively, the first time derivatives of the functions vx = vx1t2, vy = vy1t2, vz = vz1t2, or the second time derivatives of the functions x = x1t2, y = y1t2, z = z1t2. The acceleration has a magnitude defined by the positive value of

a = 4a2x + a2y + a2z

and a direction specified by the components of the unit vector ua = a>a. Since a represents the time rate of change in velocity, in general a will not be tangent to the path, Fig. 12–17c. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

35


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 direction of motion of the particle is 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 of each component 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. • 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. C–3, and their directions from the components of their unit vectors, Eqs. C–4 and C–5. y

As the airplane takes off, its path of motion can be established by knowing its horizontal position x = x1t2, and its vertical position or altitude y = y1t2, both of which can be found from navigation equipment. By plotting the results from these equations the path can be shown, and by taking the time derivatives, the velocity and acceleration of the plane at any instant can be determined.

x


36

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

EXAMPLE 12.9 y

At any instant the horizontal position of the weather balloon in Fig. 12–18a is defined by x = 18t2 ft, where t is in seconds. If the equation of the path is y = x2>10, determine (a) the distance of the balloon from the station at A when t = 2 s, (b) the magnitude and direction of the velocity when t = 2 s, and (c) the magnitude and direction of the acceleration when t = 2 s.

B

y$

x2 10

SOLUTION Position. When t = 2 s, x = 8122 ft = 16 ft, and so y = 11622>10 = 25.6 ft

x

A 16 ft

The straight-line distance from A to B is therefore r = 411622 + 125.622 = 30.2 ft

(a)

Fig. 12–18a

v $ 26.8 ft/s uv $ 72.6%

B

Velocity. Using Eqs. 12–12 and application of the chain rule of calculus the components of velocity when t = 2 s are d # vx = x = 18t2 = 8 ft>s : dt d 2 # # vy = y = 1x >102 = 2xx>10 = 21162182>10 = 25.6 ft>s c dt When t = 2 s, the magnitude of velocity is therefore v = 41822 + 125.622 = 26.8 ft>s

(b)

Fig. 12–18b

ua $ 90% B (c)

Fig. 12–18c

Ans.

The direction is tangent to the path, Fig. 12–18b, where uv = tan-1

a $ 12.8 ft/s2

Ans.

vy vx

= tan-1

25.6 = 72.6° 8

Ans.

Acceleration. The components of acceleration are determined from Eqs. 12–14 and application of the chain rule, noting that $ x = d218t2>dt2 = 0. We have # ax = vx = 0 d # # # # $ ay = vy = 12xx>102 = 21x2x>10 + 2x1x2>10 dt = 21822>10 + 21162102>10 = 12.8 ft>s2 c Thus, a = 41022 + 112.822 = 12.8 ft>s2

Ans.

The direction of a, as shown in Fig. 12–18c, is 12.8 = 90° Ans. 0 NOTE: It is also possible to obtain vy and a y by first expressing y = f1t2 = 18t22>10 = 6.4t2 and then taking successive time derivatives. ua = tan-1


37


EXAMPLE 12.10 The motion of a box B moving along the spiral conveyor shown in Fig. 12–19 is defined by the position vector r = 50.5 sin12t2i + 0.5 cos12t2j - 0.2tk6 m, where t is in seconds and the arguments for sine and cosine are in radians 1p rad = 180°2. Determine the location of the box when t = 0.75 s and the magnitudes of its velocity and acceleration at this instant.

z

SOLUTION Position. Evaluating r when t = 0.75 s yields r ƒ t = 0.75 s = 50.5 sin11.5 rad2i + 0.5 cos11.5 rad2j - 0.210.752k6 m = 50.499i + 0.0354j - 0.150k6 m Ans. The distance of the box from the origin O is 2

2

2

r = 410.4992 + 10.03542 + 1-0.1502 = 0.522 m

g = cos-11-0.2872 = 107°

a v a

r

b B

y

x

Ans.

The direction of r is obtained from the components of the unit vector, 0.499 0.0354 0.150 r i + j k ur = = r 0.522 0.522 0.522 = 0.955i + 0.0678j - 0.287k Hence, the coordinate direction angles a, b, g, Fig. 12–19, are a = cos-110.9552 = 17.2° b = cos-110.06782 = 86.1°

g

O

Fig. 12–19

Ans. Ans. Ans.

Velocity. The velocity is defined by dr d v = = [0.5 sin12t2i + 0.5 cos12t2j - 0.2tk] dt dt = 51 cos12t2i - 1 sin12t2j - 0.2k6 m>s Hence, when t = 0.75 s the magnitude of velocity, or the speed, is v = 4v2x + v2y + v2z = 4[1 cos11.5 rad2]2 + [-1 sin11.5 rad2]2 + 1-0.222 = 1.02 m>s

Ans.

Acceleration.

The acceleration a of the box is dv = 5-2 sin12t2i - 2 cos12t2j6 m>s2 a = dt

At t = 0.75s, a = 2 m>s2

Ans.

NOTE: The velocity will be tangent to the path and its coordinate

angles can be determined from uv = v>v. The acceleration will not be tangent to the path, Fig. 12–19. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

38

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12.6 Motion of a Projectile The free-flight motion of a projectile is often studied in terms of its rectangular components, since the projectile’s acceleration always acts in the vertical direction. To illustrate the kinematic analysis, consider a projectile launched at point (x0 , y0), as shown in Fig. 12–20. The path is defined in the x–y plane such that the initial velocity is v0 , having components 1v02x and 1v02y . When air resistance is neglected, the only force acting on the projectile is its weight, which causes the projectile to have a constant downward acceleration of approximately ac = g = 9.81 m>s2 or g = 32.2 ft>s2.* y

a $ #gj vx v0

(v0)y

vy

(v0)x

v

r

y y0

x x0 x

Fig. 12–20

Horizontal Motion. Since ax = 0, application of the constant acceleration equations, 12–4 to 12–6, yields + 2 v = v + a t; 1: vx = 1v02x 0 c 1 2 + 2 x = x + vt + at; 1: x = x0 + 1v02xt 0 0 2 c + 2 v2 = v2 + 2a 1s - s 2; 1: v = 1v 2 0

c

0

x

0 x

The first and last equations indicate that the horizontal component of velocity always remains constant during the motion. 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 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.

Vertical Motion. Since the positive y axis is directed upward, then ay = -g. Applying Eqs. 12–4 to 12–6, we get

1+ c 2 1+ c 2 1+ c 2

v = v0 + act; y = y0 + v0t +

vy = 1v02y - gt 1 2 2 a ct ;

v2 = v20 + 2ac1y - y02;

y = y0 + 1v02yt - 12 gt2 v2y = 1v022y - 2g1y - y02

Recall that the last equation can be formulated on the basis of eliminating the time t between 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.


12.6 MOTION OF A PROJECTILE

39

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, is defined by the vector sum as shown in Fig. 12–20.

Procedure for Analysis Free-flight projectile motion problems can be solved using the following procedure. 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 the three unknowns. In all cases the acceleration of gravity acts downward. 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., 1vx2 = 1v02x , and

x = x0 + 1v02xt Vertical Motion.

• In the vertical or y direction only two of the following three equations can be used for solution.

vy = 1v02y + act y = y0 + 1v02yt + 21 act2 v2y = 1v022y + 2ac1y - y02

• For example, if the particle’s final velocity vy is not needed, then the first and third of these equations (for y) will not be useful.

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. Rectilinear coordinates are used for the analysis since the acceleration is only in the vertical direction.


40

C H A P T E R 12

EXAMPLE 12.11

K I N E M AT I C S

OF A

PA R T I C L E

A sack slides off the ramp, shown in Fig. 12–21, with a horizontal velocity of 12 m>s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where sacks begin to pile up. y A

12 m/s

x a $ #g

6m

B

C R

Fig. 12–21

SOLUTION Coordinate System. The origin of coordinates is established at the beginning of the path, point A, Fig. 12–21. The initial velocity of a sack has components 1vA2x = 12 m>s and 1vA2y = 0. Also, between points A and B the acceleration is ay = -9.81 m>s2. Since 1vB2x = 1vA2x = 12 m>s, the three unknowns are 1vB2y , R, and the time of flight tAB . Here we do not need to determine 1vB2y . Vertical Motion. The vertical distance from A to B is known, and therefore we can obtain a direct solution for tAB by using the equation 1+ c 2

y = y0 + 1v02ytAB + 12 act2AB -6 m = 0 + 0 + 211-9.81 m>s22t2AB tAB = 1.11 s

Ans.

Horizontal Motion. Since t has been calculated, R is determined as follows: + 2 1: x = x0 + 1v02xtAB R = 0 + 12 m>s 11.11 s2 R = 13.3 m

Ans.

NOTE: The calculation for tAB also indicates that if a sack were

released from rest at A, it would take the same amount of time to strike the floor at C, Fig. 12–21. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

12.6 MOTION OF A PROJECTILE

EXAMPLE 12.12 The chipping machine is designed to eject wood chips at vO = 25 ft>s as shown in Fig. 12–22. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if 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 represented as the height h, time of flight tOA , and vertical component of velocity 1vA2y . (Note that 1vA2x = 1vO2x .) With the origin of coordinates at O, Fig. 12–22, the initial velocity of a chip has components of 1vO2x = 125 cos 30°2 ft>s = 21.65 ft>s : 1vO2y = 125 sin 30°2 ft>s = 12.5 ft>s c Also, 1vA2x = 1vO2x = 21.65 ft>s and ay = -32.2 ft>s2. Since we do not need to determine 1vA2y , we have Horizontal Motion. + 2 1: x = x + 1v 2 t A

O

O x OA

20 ft = 0 + 121.65 ft>s2tOA tOA = 0.9238 s Vertical Motion. Relating tOA to the initial and final elevations of a chip, we have 1+ c 2

yA = yO + 1vO2ytOA + 21 act2OA

1h - 4 ft2 = 0 + 112.5 ft>s210.9238 s2 + 121-32.2 ft>s2210.9238 s22 h = 1.81 ft

Ans.

NOTE: We can determine 1vA2y by using 1vA2y = 1vO2y + a ctOA . Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

41

42

C H A P T E R 12

K I N E M AT I C S

EXAMPLE 12.13

OF A

PA R T I C L E

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 slope, the horizontal distance he travels before striking the ground, and the maximum height he attains. Neglect the size of the bike and rider.

(a)

y C 30% A

h 1m

x B

R (b)

Fig. 12–23

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 . Vertical Motion. Since the time of flight and the vertical distance between the ends of the path are known, we can determine vA . 1+ c 2 1sB2y = 1sA2y + 1vA2ytAB + 21 act2AB - 1 = 0 + vA sin 30°11.52 + 121-9.81211.522 vA = 13.38 m>s = 13.4 m>s

Ans.

Horizontal Motion. The range R can now be determined. + 2 1: 1sB2x = 1sA2x + 1vA2xtAB R = 0 + 13.38 cos 30°11.52 = 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 become the time of flight tAC , the horizontal distance from A to C, and the height h. At the maximum height 1vC2y = 0, and since vA is known, we can determine h directly without considering tAC using the following equation. 1vC22y = 1vA22y + 2ac[1sC2y - 1sA2y] 1022 = 113.38 sin 30°22 + 21-9.812[1h - 12 - 0] h = 3.28 m

Ans.

NOTE: Show that the bike will strike the ground at B with a velocity

having components of 1vB2x = 11.6 m>s : ,

1vB2y = 8.02 m>s T


43

PROBLEMS

PROBLEMS 12–66. A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = 56ti + 12t2k6 ft>s2. Determine the particle’s position (x, y, z) at t = 1 s.

12–71. 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

12–67. The velocity of a particle is given by v = 516t2i + 4t3j + 15t + 22k6 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? *"12–68. A particle is traveling with a velocity of 2 v = E 3 2te -0.2ti + 4e -0.8t j F m>s, where t is in seconds. Determine the magnitude of the particle’s displacement from t = 0 to t = 3 s. Use Simpson’s rule with n = 100 to evaluate the integrals. What is the magnitude of the particle’s acceleration when t = 2 s? 12–69. The position of a particle is defined by r = 551cos 2t2i + 41sin 2t2j6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particle when t = 1 s. Also, prove that the path of the particle is elliptical. 12–70. 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.

D 5m

15 m

B

C

10 m x

A

Prob. 12–71 *12–72. A car travels east 2 km for 5 minutes, then north 3 km for 8 minutes, and then west 4 km for 10 minutes. Determine the total distance traveled and the magnitude of displacement of the car. Also, what is the magnitude of the average velocity and the average speed? 12–73. 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

y x

B

vB $ 30 m/s B

C

45%

20 m

C

A

Prob. 12–70

vA $ 20 m/s

x A

Prob. 12–73


44

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–74. A particle moves along the curve y = e2x such that its velocity has a constant magnitude of v = 4 ft>s. Determine the x and y components of velocity when the particle is at y = 5 ft. 12–75. The path of a particle is defined by y2 = 4kx, and the component of velocity along the y axis is vy = ct, where both k and c are constants. Determine the x and y components of acceleration.

12–79. When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path 1y - 4022 = 160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy = 180 m>s, determine the magnitudes of the rocket’s velocity and acceleration when it reaches an altitude of 80 m. y (y # 40)2 $ 160x

*12–76. A particle is moving along the curve y = x - 1x2>4002, where x and y are in ft. If the velocity component in the x direction is vx = 2 ft>s and remains constant, determine the magnitudes of the velocity and acceleration when x = 20 ft. 12–77. The flight path of the helicopter as it takes off from A is defined by the parametric equations x = 12t22 m and y = 10.04t32 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.

40 m

y

x

Prob. 12–79

A

x

*12–80. Determine the minimum speed of the stunt rider, so that when he leaves the ramp at A he passes through the center of the hoop at B. Also, how far h should the landing ramp be from the hoop so that he lands on it safely at C? Neglect the size of the motorcycle and rider.

Prob. 12–77 12–78. At the instant shown particle A is traveling to the right at 10 ft>s and has an acceleration of 2 ft>s2. Determine the initial speed v0 of particle B so that when it is fired at the same instant from the angle shown it strikes A. Also, at what speed does it strike A?

B

y C B 3

4 ft

5 4

v0

100 ft

5 ft

3 ft h

24 ft

12 ft

Prob. 12–80

A

Prob. 12–78

A

12 ft

x 10 ft/s 2 ft/s2

12–81. Show that if a projectile is fired at an angle u from the horizontal with an initial velocity v0 , the maximum range the projectile can travel is given by Rmax = v20>g, where g is the acceleration of gravity. What is the angle u for this condition?


45

PROBLEMS 12–82. The balloon A is ascending at the rate vA = 12 km>h and is being carried horizontally by the wind at vw = 20 km>h. If a ballast bag is dropped from the balloon at the instant h = 50 m, determine the time needed for it to strike the ground. Assume that the bag was released from the balloon with the same velocity as the balloon. Also, with what speed does the bag strike the ground?

12–85. 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.

vA $ 12 km/h B A

vw $ 20 km/h

h

vA A

u

3.5 ft

18 ft

h

Prob. 12–85 Prob. 12–82 12–83. Determine the height on the wall to which the firefighter can project water from the hose, if u = 40° and the speed of the water at the nozzle is vC = 48 ft>s. *12–84. Determine the smallest angle u, measured above the horizontal, that the hose should be directed so that the water stream strikes the bottom of the wall at B. The speed of the water at the nozzle is vC = 48 ft>s.

12–86. The buckets on the conveyor travel with a speed of 15 ft>s. Each bucket contains a block which falls out of the bucket when u = 120°. Determine the distance s to where the block strikes the conveyor. Neglect the size of the block.

u = 120%

1 ft

3 ft

A s

vC $ 48 ft/s

h

u

C 3 ft

B 15 ft

Probs. 12–83/84

Prob. 12–86


46

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–87. Measurements of a shot recorded on a videotape during a basketball game are shown. The ball 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.

12–89. 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

30% A

C vA

B

v0

h

10 ft

7 ft 25 ft

h

u

x

R

5 ft

Prob. 12–87

Prob. 12–89

*12–88. 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.

12–90. The fireman standing on the ladder directs the flow of water from his hose to the fire at B. Determine the velocity of the water at A if it is observed that the hose is held at u = 20°.

A

A

40%

u vA 30 ft

3

B

5 4

B

R 60 ft

Prob. 12–88

Prob. 12–90


PROBLEMS 12–91. A ball bounces on the 30° inclined plane such that it rebounds perpendicular to the incline with a velocity of vA = 40 ft>s. Determine the distance R to where it strikes the plane at B.

47

12–93. The stones are thrown off the conveyor with a horizontal velocity of 10 ft>s as shown. Determine the distance d down the slope to where the stones hit the ground at B. 12–94. The stones are thrown off the conveyor with a horizontal velocity of 10 ft>s as shown. Determine the speed at which the stones hit the ground at B. A

vA $ 40 ft/s

10 ft/s

A 100 ft R B 30% 1

B

10

Prob. 12–91

d

Probs. 12–93/94 *12–92. The man stands 60 ft from the wall and throws a ball at it with a speed v0 = 50 ft>s. Determine the angle u at which he should release the ball so that it strikes the wall at the highest point possible. What is this height? The room has a ceiling height of 20 ft.

12–95. The drinking fountain is designed such that the 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.

vA 40% A 50 mm

v0 $ 50 ft/s

u

h

20 ft

B

100 mm

C 250 mm

5 ft 60 ft

Prob. 12–92

Prob. 12–95


48

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

*12–96. A boy at O throws a ball in the air with a speed vO at an angle u1 . If he then throws another ball at the same speed vO at an angle u2 6 u1 , determine the time between the throws so the balls collide in mid air at B.

12–98. 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 = 10.05x22 ft and neglect the size of the sprinkler.

y

B O

u1

u2

y $ (0.05x2) ft

y

15 ft/s B 60% x

x

Prob. 12–98

Prob. 12–96

12–97. 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 17uD2, then the second dart is thrown at uD .

12–99. The projectile is launched from a height h with a velocity v0 . Determine the range R.

y 5m uC A

v0

C

uD

u

D B

h x R

Prob. 12–97

Prob. 12–99


49

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS

12.7 Curvilinear Motion: Normal

O¿

and Tangential Components

n

O s

When the path along which a particle is moving is known, then it is often convenient to describe the motion using n and t coordinates which act normal and tangent to the path, respectively, and at the instant considered have their origin located at the particle.

Planar Motion. Consider the particle P shown in Fig. 12–24a, which is moving in a plane along a fixed curve, such that at a given instant it is at position s, measured from point O. We will now consider a coordinate system that has its origin at a fixed point on the curve, and at the instant considered this origin happens to coincide with the location of the particle. The t axis is tangent to the curve at P 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 and is directed from P 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 osculating plane, and in this case it is fixed in the plane of motion.*

un P

ut

(a)

Fig. 12–24a O¿

r

r

O¿

ds

r

r

ds

r

Radius of curvature (b)

Fig. 12–24b

O¿

indicated in Sec. 12.4, the particle’s velocity v has a direction that is always tangent to the path, Fig. 12–24c, and a magnitude that is determined by taking the time derivative of the path function s = s1t2, i.e., v = ds>dt (Eq. 12–8). Hence (12–15)

r

ds

Velocity. Since the particle is moving, s is a function of time. As

v = vut

t

Position

r

r

P v Velocity

where

(c)

# v = s

(12–16)

Fig. 12–24c

*The osculating plane may also be defined as that 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. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

O¿

50

C H A P T E R 12

K I N E M AT I C S

PA R T I C L E

Acceleration. The acceleration of the particle is the time rate of

O¿

du

OF A

change of the velocity. Thus,

r

r

un

# # # a = v = vut + vut

ds

u¿t

P

ut (d)

Fig. 12–24d

# In order to determine the time derivative ut , note that as the particle moves along the arc ds in time dt, ut preserves its magnitude of unity; however, its direction changes, and becomes utœ , Fig. 12–24d. As shown in Fig. 12–24e, we require utœ = ut + dut . Here dut stretches between the arrowheads of ut and utœ , which lie on an infinitesimal arc of radius ut = 1. Hence, dut has a magnitude of dut = 112 du, and its direction is defined by un . Consequently, # # dut = duun , and therefore the time derivative becomes ut = uun . Since # # ds = r du, Fig. 12–24d, then u = s>r, and therefore # # v s # ut = uun = un = un r r

un du u¿ t dut

ut

(12–17)

(e)

Fig. 12–24e

Substituting into Eq. 12–17, a can be written as the sum of its two components, a = atut + anun

(12–18)

where # at = v

P

a

at

Acceleration (f)

Fig. 12–24f

(12–19)

and

O¿

an

at ds = v dv

or

an =

v2 r

(12–20)

These two mutually perpendicular components are shown in Fig. 12–24f, in which case the magnitude of acceleration is the positive value of a = 4a2t + a2n

(12–21)


51


To summarize these concepts, consider the following two special cases of motion. 1. If the particle moves along a straight line, then r : q and from # Eq. 12–20, an = 0. Thus a = at = v, and we can conclude that the tangential component of acceleration represents the time rate of change in the magnitude of the velocity. 2. If the particle moves along a curve with a constant speed, then # at = v = 0 and a = an = v2>r. Therefore, the normal component of acceleration represents the time rate of change in the direction of the velocity. Since an always acts towards the center of curvature, this component is sometimes referred to as the centripetal acceleration. As a result of these interpretations, a particle moving along the curved path in Fig. 12–25 will have accelerations directed as shown.

at Increasing speed

a $ at

Change in direction of velocity an

an

a

a at

Change in magnitude of velocity

Fig. 12–25

Three-Dimensional Motion. If the particle is moving along a

space curve, Fig. 12–26, then at a given instant the t axis is uniquely specified; however, an infinite number of straight lines can be constructed normal to the tangent axis at P. As in the case of planar motion, we will choose the positive n axis directed from P toward the path’s center of curvature O¿. This axis is referred to as the principal normal to the curve at P. With the n and t axes so defined, Eqs. 12–15 to 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 a 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 so 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.

b

osculating plane

O

n s

O¿

ub un P

ut

t

Fig. 12–26


52

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

Procedure for Analysis

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.

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. • The n and t axes are particularly advantageous for studying the velocity and acceleration of the particle, because the t and n components of a are expressed by Eqs. 12–19 and 12–20, respectively. Velocity. • The particle’s velocity is always tangent to the path. • The magnitude of velocity is found from the time derivative of the path function. # v = s Tangential Acceleration. • The tangential component of acceleration is the result of the time rate of change in the magnitude of velocity. This component acts in the positive s direction if the particle’s speed is increasing or in the opposite direction if the speed is decreasing. • The relations between at , v, t and s are the same as for rectilinear motion, namely, # at = v at ds = v dv

• If at is constant, at = 1at2c , the above equations, when integrated, yield s = s0 + v0t + 211at2ct2 v = v0 + 1at2ct v2 = v20 + 21at2c1s - s02 Normal Acceleration. • The normal component of acceleration is the result of the time rate of change in the direction of the particle’s velocity. This component is always directed toward the center of curvature of the path, i.e., along the positive n axis. • The magnitude of this component is determined from v2 an = r

• If the path is expressed as y = f1x2, the radius of curvature r at any point on the path is determined from the equation [1 + 1dy>dx22]3>2 r = ƒ d2y>dx2 ƒ

derivation of this Prentice result isHall, given in any standard calculus text. River, Unpublished Work © 2007 by R. C. Hibbeler. ToThe be published by Pearson Pearson Education, Inc., Upper Saddle New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

53


EXAMPLE 12.14 When the skier reaches point A along the parabolic path in Fig. 12–27a, he has a speed of 6 m>s which is increasing at 2 m>s2. Determine the direction of his velocity and the direction and magnitude of his acceleration at this instant. Neglect the size of the skier in the calculation. SOLUTION Coordinate System. Although the path has been expressed in terms of its x and y coordinates, we can still establish the origin of the n, t axes at the fixed point A on the path and determine the components of v and a along these axes, Fig. 12–27a. Velocity. By definition, the velocity is always directed tangent to the 1 1 2 path. Since y = 20 x , dy>dx = 10 x, then dy>dx ƒ x = 10 = 1. Hence, at A,

v makes an angle of u = tan-1 1 = 45° with the x axis, Fig. 12–27. Therefore, vA = 6 m>s 45°dvA Ans. # Acceleration. The acceleration is determined from a = vut + 1v2>r2un . However, it is first necessary to determine the radius of curvature of the path at A (10 m, 5 m). Since d2y>dx2 = 2 3>2

r =

[1 + 1dy>dx2 ] 2

2

ƒ d y>dx ƒ

=

C1 +

1 10 ,

then

1 2 3>2 10 x 1 ƒ ƒ 10 x = 10 m

A

B D

`

n u

vA

A

t

5m x

= 28.28 m

10 m

The acceleration becomes

(a)

v2 # u a A = vut + r n = 2ut +

y $ 1 x2 20

y

Fig. 12–27a

16 m>s22 28.28 m

un

n 1.273 m/s2

2

= 52ut + 1.273un6 m>s

f

As shown in Fig. 12–27b,

a 2 m/s2

a = 41222 + 11.27322 = 2.37 m>s2

t

2 f = tan-1 = 57.5° 1.273

(b)

Fig. 12–27b

Thus, 57.5° - 45° = 12.5° so that, a = 2.37 m>s2

45%

12.5°da A

Ans.

NOTE: By using n, t coordinates, we were able to readily solve this

problem since the n and t components account for the separate changes in the magnitude and direction of v.


54

C H A P T E R 12

K I N E M AT I C S

OF A

EXAMPLE 12.15

PA R T I C L E

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 a

t

r $ 300 ft

Fig. 12–28

SOLUTION Coordinate System. The origin of the n and t axes is coincident with the car at the instant considered.The t axis is in the direction of motion, and the positive n axis is directed toward the center of the circle. This coordinate system is selected since the path is known. Acceleration.

The magnitude of acceleration can be related to its

components using a = 4a2t + a2n . Here at = 7 ft>s2. Since an = v2>r, the velocity as a function of time is v = v0 + 1at2ct v = 0 + 7t Thus an =

17t22 v2 = = 0.163t2 ft>s2 r 300

The time needed for the acceleration to reach 8 ft>s2 is therefore a = 4a2t + a2n 8 = 41722 + 10.163t222 Solving for the positive value of t yields 0.163t2 = 41822 - 1722 t = 4.87 s

Ans.

Velocity. The speed at time t = 4.87 s is v = 7t = 714.872 = 34.1 ft>s

Ans.

NOTE: Remember the velocity will be tangent to the path, whereas

the acceleration will be directed inwards, towards the +n axis.


55


EXAMPLE 12.16 The boxes in Fig. 12–29a travel along the industrial conveyor. If a box as in Fig. 12–29b starts from rest at A and increases its speed such that at = 10.2t2 m>s2, where t is in seconds, determine the magnitude of its acceleration when it arrives at point B. SOLUTION Coordinate System. The position of the box at any instant is defined from the fixed point A using the position or path coordinate s, Fig. 12–29b. The acceleration is to be determined at B, so the origin of the n, t axes is at this point. # Acceleration. To determine the acceleration components at = v and # 2 an = v >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

v

L0

(a)

Fig. 12–29a

(1)

t

A s

dv =

0.2t dt L0 v = 0.1t2

(2)

3m

The time needed for the box to reach point B can be determined by realizing that the position of B is sB = 3 + 2p122>4 = 6.142 m, Fig. 12–29b, and since sA = 0 when t = 0 we have v = 6.142

L0

ds =

2m

ds = 0.1t2 dt

n t

tB

L0

0.1t2 dt

B (b)

6.142 = 0.0333t3B

Fig. 12–29b

tB = 5.690 s Substituting into Eqs. 1 and 2 yields # 1aB2t = vB = 0.215.6902 = 1.138 m>s2 vB = 0.115.6922 = 3.238 m>s n

At B, rB = 2 m, so that 1aB2n =

5.242 m/s2

aB

13.238 m>s22 v2B = = 5.242 m>s2 rB 2m

t B

The magnitude of a B , Fig. 12–29c, is therefore aB = 411.13822 + 15.24222 = 5.36 m>s2

Ans.

1.138 m/s2

(c)

Fig. 12–29c


56

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

PROBLEMS *12–100. A car is traveling along a circular curve that has a radius of 50 m. If its speed is 16 m>s and is increasing uniformly at 8 m>s2, determine the magnitude of its acceleration at this instant. 12–101. A car moves along a circular track of radius 250 ft such that its speed for a short period of time, 0 … t … 4 s, is v = 31t + t22 ft>s, where t is in seconds. Determine the magnitude of its acceleration when t = 3 s. How far has it traveled in t = 3 s?

12–106. The jet plane travels along the vertical parabolic path. When it is at point A it has a speed of 200 m>s, which is increasing at the rate of 0.8 m>s2. Determine the magnitude of acceleration of the plane when it is at point A. y

y $ 0.4x2

A

12–102. At a given instant the jet plane has a speed of 400 ft>s and an acceleration of 70 ft>s2 acting in the direction shown. Determine the rate of increase in the plane’s speed and the radius of curvature r of the path.

10 km 400 ft/s

60% a $ 70 ft/s2

r

x

5 km

Prob. 12–106

Prob. 12–102 12–103. A particle is moving along a curved path at a constant speed of 60 ft>s. The radii of curvature of the path at points P and P¿ are 20 and 50 ft, respectively. If it takes the particle 20 s to go from P to P¿, determine the acceleration of the particle at P and P¿. *12–104. 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. "12–105. Starting from rest, a bicyclist travels around a horizontal circular path, r = 10 m, at a speed of v = 10.09t2 + 0.1t2 m>s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m.

12–107. The car travels along the curve having a radius of 300 m. If its speed is uniformly increased from 15 m>s to 27 m>s in 3 s, determine the magnitude of its acceleration at the instant its speed is 20 m>s. 20 m/s

300 m

Prob. 12–107


57

PROBLEMS *12–108. The satellite S travels around the earth in a 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.

12–111. 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 direction shown. Determine the rate of increase in the train’s speed and the radius of curvature r of the path.

S v $ 20 m/s h 75% E

2

a $ 14 m/s

r

Prob. 12–108 Prob. 12–111 12–109. A particle P moves along the curve y = 1x2 - 42 m with a constant speed of 5 m>s. Determine the point on the curve where the maximum magnitude of acceleration occurs and compute its value. 12–110. The Ferris wheel turns such that the speed of the # passengers is increased by v = 14t2 ft>s2, where t is in seconds. If the wheel starts from rest when u = 0°, determine the magnitudes of the velocity and acceleration of the passengers when the wheel turns u = 30°.

*12–112. 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.

A

vA

40 ft u

1500 ft

B

Prob. 12–110

Prob. 12–112


58

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–113. The automobile is originally at rest at s = 0. If its # speed is increased by v = 10.05t22 ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration when t = 18 s. 300 ft

*12–116. 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.

s y (mm)

240 ft

y$

20(103) x

Prob. 12–113 12–114. The automobile is originally at rest s = 0. If it # then starts to increase its speed at v = 10.05t22 ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration at s = 550 ft.

v P x (mm)

300 ft s

Prob. 12–116 240 ft

12–117. 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.

Prob. 12–114 12–115. The truck travels in a circular path havng a radius of 50 m at a speed of 4 = m>s. For a short distance from # s = 0, its speed is increased by v = 10.05s2 m>s2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m. . v $ (0.05s) m/s2 v $ 4 m/s

y 2 x2 " y $ 1 (60)2 (40)2

40 m x 60 m

50 m

Prob. 12–115

Prob. 12–117


59

PROBLEMS 12–118. 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.

*12–120. The car B turns such that its speed is increased # by vB = 10.5et2 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. Also, through what angle u has it traveled?

y x2

(60)2

"

y2

(40)2

v

$1 B 40 m x 60 m

5m u

A

Prob. 12–120

Prob. 12–118

"12–119. The car B turns such that its speed is increased # by vB = 10.5et2 m>s2, where t is in seconds. If the car starts from rest when u = 0°, determine the magnitudes of its velocity and acceleration when the arm AB rotates u = 30°. Neglect the size of the car.

"12–121. 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.

v y

B

y $ 0.5x2

s 5m A

u

Prob. 12–119

x

A

Prob. 12–121


60

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–122. The ball is ejected horizontally from the tube with a speed of 8 m>s. Find the equation of the path, y = f1x2, and then find the ball’s velocity and the normal and tangential components of acceleration when t = 0.25 s. y

12–125. The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively. Determine in t = 2 s, (a) the displacement along the path of each particle, (b) the position vector to each particle, and (c) the shortest distance between the particles. y

vA $ 8 m/s

x

A

Prob. 12–122 5m

"12–123. 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 = 10.06t2 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. *12–124. 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 = 10.002s2 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.

B

A

vB $ 1.5 m/s

x

O

vA $ 0.7 m/s

Prob. 12–125 12–126. The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens. y

y

r

A

5m

x B vB $ 1.5 m/s

Prob. 12–124

Prob. 12–126

A O

x vA $ 0.7 m/s


61

PROBLEMS 12–127. 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 at = 10.4s2 m>s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m.

"12–131. Particles A and B are traveling counterclockwise around a circular track at a constant speed of 8 m>s. If at the instant shown the speed of A is increased by # vA = 14sA2 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?

A r

sA u $ 120%

s

A

sB B

r$5m

Prob. 12–127

Prob. 12–131

*12–128. A boy sits on a merry-go-round so that he is always located at r = 8 ft from the center of rotation. The merry-go-round is originally at rest, and then due to rotation the boy’s speed is increased at 2 ft>s2. Determine the time needed for his acceleration to become 4 ft>s2.

*12–132. 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 increased by vB = 4 m>s2, and at the same instant A # has an increase in speed vA = 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?

12–129. A particle moves along the curve y = sin x with a constant speed v = 2 m>s. Determine the normal and tangential components of its velocity and acceleration at any instant. 12–130. The motion of a particle along a fixed path is defined by the parametric equations r = 8 ft, u = 14t2 rad, and z = 16t22 ft, where t is in seconds. Determine the unit vector that specifies the direction of the binormal axis to the osculating plane with respect to a set of fixed x, y, z coordinate axes when t = 2 s. Hint: Formulate the particle’s velocity vP and acceleration aP in terms of their i, j, k components. Note that x = r cos u and y = r sin u. The binormal is parallel to vP * aP . Why?

A sA u $ 120%

sB B

r$5m

Prob. 12–132


62

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–133. The truck travels at a speed of 4 m>s along a circular road that has a radius of 50 m. For a short distance # from s = 0, its speed is then increased by v = 10.05s2 m>s2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

12–135. A particle P travels along an elliptical spiral path such that its position vector r is defined by r = 52 cos10.1t2i + 1.5 sin10.1t2j + 12t2k6 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? z

50 m

Prob. 12–133

P

"12–134. A go-cart moves along a circular track of radius 100 ft such that its speed for a short period of time, 2 0 … t … 4 s, is v = 6011 - e -t 2 ft>s. Determine the magnitude of its acceleration when t = 2 s. How far has it traveled in t = 2 s? Use Simpson’s rule with n = 50 to evaluate the integral.

r y

x

Prob. 12–135

12.8 Curvilinear Motion:

Cylindrical Components

Some engineering problems involve angular position and a radial distance. For these cases, it is often convenient to express the path of motion in terms of cylindrical coordinates, r, u, z. If motion is restricted to the plane, the polar coordinates r and u are used.

u uu

P

r u O

r

Position (a)

Fig. 12–30a

ur

Polar Coordinates. We can specify the location of particle P

shown in Fig. 12–30a using both the 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 or the radial direction +r extends from P along increasing r, when u is held fixed, and uu or +u extends from P in a direction that occurs when r is held fixed and u is increased. Note that these directions are perpendicular to one another.


63

12.8 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS

Position. At any instant the position of the particle, Fig. 12–30a, is defined by the position vector r = rur

u uu

(12–22)

Velocity. The instantaneous velocity v is obtained by taking the time

u O

# To evaluate ur , notice that ur changes only its direction with respect to time, since by definition the magnitude of this vector is always one unit. Hence, during the time ¢t, a change ¢r will not cause a change in the direction of ur ; however, a change ¢u will cause ur to become urœ , where urœ = ur + ¢ur , Fig. 12–30b. The time change in ur is then ¢ur . For small angles ¢u this vector has a magnitude ¢ur L 11¢u2 and acts in the uu direction. Therefore, ¢ur = ¢u uu , and so ¢ur ¢u # ur = lim = a lim b uu ¢t : 0 ¢t ¢t : 0 ¢t # # ur = uuu

ur

P

r

derivative of r. Using a dot to represent time differentiation, we have # # # v = r = rur + rur

r

Position (a)

Fig. 12–30a uu

u¿r 'u

'ur ur

(b)

Fig. 12–30b

(12–23)

Substituting into the above equation for v, the velocity can be written in component form as v = vrur + vuuu

(12–24)

# vr = r # vu = ru

(12–25)

where

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 = 41r22 + 1ru22 (12–26) and the direction of v is, of course, tangent to the path at P, Fig. 12–30c.

v vu vr P

r u O Velocity (c)

Fig. 12–30c


64

C H A P T E R 12 'uu

K I N E M AT I C S

OF A

PA R T I C L E

Acceleration. Taking the time derivatives of Eq. 12–24, using

uu

Eqs. 12–25, we obtain the particle’s instantaneous acceleration,

u¿u

$ # # # $ ## ## a = v = rur + rur + ruuu + ruuu + ruuu

ur

'u (d)

Fig. 12–30d

# To evaluate the term involving uu , it is necessary only to find the change made 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 , although a change ¢u will cause uu to become uuœ , where uuœ = uu + ¢uu , Fig. 12–30d. The time change in uu is thus ¢uu . For small angles this vector has a magnitude ¢uu L 11¢u2 and acts in the -ur , direction; i.e., ¢uu = - ¢uur . Thus, ¢uu ¢u # uu = lim = - a lim b ur ¢t : 0 ¢t ¢t : 0 ¢t # # uu = -uur

(12–27)

Substituting this result and Eq. 12–23 into the above equation for a, we can write the acceleration in component form as a = arur + auuu

(12–28)

# $ ar = r - ru2 # # $ au = ru + 2ru

(12–29)

where

a au

P

r u O Acceleration (e)

Fig. 12–30 (cont.)e

ar

$ The term u = d2u>dt2 = d>dt1du>dt2 is called the angular acceleration since it measures the change made in the angular velocity during an instant of time. Units for this measurement are rad>s2. Since a r , and a u are always perpendicular, the magnitude of acceleration is simply the positive value of # $ $ ## a = 41r - ru222 + 1ru + 2ru22

(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.


65


Cylindrical Coordinates. If the particle P moves along a space

uz

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:

uu

P

ur rP z

rp = rur + zuz

O

# # # v = rur + ruuu + zuz

u

r

# $ $ ## $ a = 1r - ru22ur + 1ru + 2ru2uu + zuz Fig. 12–31

Time Derivatives. The equations $ of kinematics 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 coordinates are specified as time parametric equations, r = r1t2 and u = u1t2, then the time derivatives can be found directly. For example, consider r = 4t2 # r = 8t $ r = 8

u = 18t3 + 62 # u = 24t2 $ u = 48t z

2. If the time-parametric equations are not given, then we have to specify the path r = f1u2 and find the relationship between the time derivatives using the chain rule of calculus. Consider the following examples. r

r = 5u2

# # r = 10uu # # $ $ r = 10[1u2u + u1u2] # $ = 10u2 + 10uu

The spiral motion of this boy can be followed by using cylindrical components. Here the radial coordinate r is constant, the transverse coordinate u will increase with time as the boy rotates about the vertical, and his altitude z will decrease with time.


66

C H A P T E R 12

K I N E M AT I C S

OF A

or

PA R T I C L E

r2 # 2rr # # $ 2[1r2r + r1r2] $ # r2 + rr

= 6u3

# = 18u2u

# # $ = 18[12uu2u + u21u2] $ # = 912uu2 + u2u2 $ # $ # If two of the four time derivatives r, r, u, and u are known, then the other two can be obtained from the equations for first and second time derivatives of r = f1u2. See Example 12.19. In some problems, however, two of these time derivatives may not be known; instead the magnitude of the particle’s velocity or acceleration may be specified. If this is the case, Eqs. 12–26 and # $ # ## $ ## 12–30 [v2 = r2 + 1ru22 and a2 = 1r - ru222 + 1ru + 2ru22] may be $ # $ # used to obtain the necessary relationships involving r, r, u, and u. See Example 12.20.

Procedure for Analysis Coordinate System.

• Polar coordinates are a suitable choice for solving problems when

• •

data regarding the angular motion of the radial coordinate r is given to describe the particle’s motion. Also, some paths of motion can conveniently be described in terms of these coordinates. To use polar coordinates, the origin is established at a fixed point, and the radial line r is directed to the particle. The transverse coordinate u is measured from a fixed reference 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 = f1u2, it is very important to use the chain rule of calculus. Motion in three dimensions requires a simple extension of the # $ above procedure to include z and z.

Besides the examples which follow, further examples involving the calculation of ar and au can be found in the “kinematics” sections of Examples 13.10 through 13.12. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.


EXAMPLE 12.17 The amusement park ride shown in Fig. 12–32a consists of a chair that is rotating in a horizontal circular path of radius r such $that the arm # OB has an angular velocity u and angular acceleration u. Determine the radial and transverse components of velocity and acceleration of the passenger. Neglect his size in the calculation.

n

O

· ·· u, u

r

· v $ ru

· ar $ #ru2

u B

·· au $ ru r

r

u,t (b)

(a)

Fig. 12–32a

Fig. 12–32b

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. Equations 12–25 and 12–29 will be used for the solution, and so 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: Also shown are the n, t axes, which in this special case of circular motion happen to be collinear with the# r and u axes, respectively. In particular note that v = vu = vt = ru. Also,

# # 1ru22 v2 -ar = an = = = ru2 r r # $ du d # dr # dv = 1ru2 = u + r = 0 + ru au = at = dt dt dt dt


67

68

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

EXAMPLE 12.18 The rod OA in Fig. 12–33a is rotating in the horizontal plane such that u = 1t32 rad. At the same time, the collar B is sliding outward along OA so that r = 1100t22 mm. If in both cases t is in seconds, determine the velocity and acceleration of the collar when t = 1 s. SOLUTION Coordinate System. Since time-parametric equations of the path are given, it is not necessary to relate r to u. O u

Velocity and Acceleration. Determining the time derivatives and evaluating when t = 1 s, we have

r

B

r = 100t2 ` A

# r = 200t `

(a)

Fig. 12–33a

u $ 57.3%

$ r = 200 `

vu $ 300 mm/s

u

d

t=1 s

# = 200 mm>s u = 3t2 ` = 200 mm>s2

$ u = 6t `

= 1 rad = 57.3°

t=1 s

t=1 s

= 3 rad>s = 6 rad>s2.

The magnitude of v is

(b)

f

t=1 s

t=1 s

= 200ur + 100132uu = 5200ur + 300uu6 mm>s

r

Fig. 12–33b

u $ 57.3%

t=1 s

As shown in Fig. 12–33b, # # v = rur + ruuu

v vr $ 200 mm/s

= 100 mm u = t3 `

v = 4120022 + 130022 = 361 mm>s 300 b = 56.3° d + 57.3° = 114° d = tan-1 a 200 a

Ans.

As shown in Fig. 12–33c, # $ $ ## a = 1r - ru22ur + 1ru + 2ru2uu au $ 1800 mm/s2 u

= [200 - 1001322]ur + [100162 + 2120023]uu

ar $ 700 mm/s2

= 5-700ur + 1800uu6 mm>s2

r (c)

Ans.

The magnitude of a is

Fig. 12–33c

a = 4170022 + 1180022 = 1930 mm>s2 f = tan-1 a

1800 b = 68.7° 700

1180° - f2 + 57.3° = 169°

Ans. Ans.

NOTE: The velocity is tangent to the path; however, the acceleration is directed towards the center of curvature, as expected. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

69


EXAMPLE 12.19 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 is rotating at a constant rate of u = 4 rad>s.

u

r

SOLUTION Coordinate System. Polar coordinates will be used to solve this problem since the angular rate of the searchlight is given. To find the necessary time derivatives it is first necessary to relate r to u. From Fig. 12–34a, this relation is

· u $ 4 rad/s

100 m (a)

Fig. 12–34a

r = 100>cos u = 100 sec u Velocity and Acceleration. Using the chain rule of calculus, noting that d1sec u2 = sec u tan u du, and d1tan u2 = sec2 u du, we have # # r = 1001sec u tan u2u # # # # $ r = 1001sec u tan u2u1tan u2u + 100 sec u1sec2 u2u1u2 $ + 100 sec u tan u1u2 # # $ = 100 sec u tan2 u1u22 + 100 sec3 u1u22 + 1001sec u tan u2u # $ Since u = 4 rad>s = constant, then u = 0, and the above equations, when u = 45°, become r = 100 sec 45° = 141.4 # r = 400 sec 45° tan 45° = 565.7 $ r = 16001sec 45° tan2 45° + sec3 45°2 = 6788.2

r vr

v

u

u (b)

Fig. 12–34b r ar

= 565.7ur + 141.4142uu

a

= 5565.7ur + 565.7uu6 m>s v =

+

v2u

100 m

vu

As shown in Fig. 12–34b, # # v = rur + ruuu

2 4vr

u

r

2

u

2

= 41565.72 + 1565.72

= 800 m>s As shown in Fig. 12–34c, # $ $ ## a = 1r - ru22ur + 1ru + 2ru2uu

r

Ans. u

(c)

Fig. 12–34c a

= 54525.5ur + 4525.5uu6 m>s2 Ans. $ NOTE: It is also possible to find a without having to calculate r (or a r). As shown in Fig. 12–34d, since au = 4525.5 m>s2, then by vector resolution, a = 4525.5>cos 45° = 6400 m>s2.

100 m

au

= [6788.2 - 141.41422]ur + [141.4102 + 21565.724]uu a = 4a2r + a2u = 414525.522 + 14525.522 = 6400 m>s2

u

u $ 45% ar

au $ 4525.5 m/s2

(d)

Fig. 12–34d


70

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

EXAMPLE 12.20 r $ 0.5 (1 # cosu) ft

u

A r

· ·· u, u

(a)

Fig. 12–35a

Due to the rotation of the forked rod, the ball A in Fig. 12–35a travels around the slotted path, a portion of which is in the shape of a cardioid, r = 0.511 - cos u2 ft, where u is in radians. If the ball’s velocity is v = 4 ft>s and its acceleration is a = 30 ft>s2 at the instant u = 180°, # $ determine the angular velocity u and angular acceleration u of the fork. SOLUTION Coordinate System. This path is most unusual, and mathematically it is best expressed using polar coordinates, as done here, rather than # $ rectangular coordinates. Also, u and u must be determined so r, u coordinates are an obvious choice. Velocity and Acceleration. Determining the time derivatives of r using the chain rule of calculus yields r = 0.511 - cos u2 # # r = 0.51sin u2u # # $ $ r = 0.51cos u2u1u2 + 0.51sin u2u Evaluating these results at u = 180°, we have r = 1 ft

# r = 0

# $ r = -0.5u2

# Since v = 4 ft>s, using Eq. 12–26 to determine u yields # # v = 41r22 + 1ru22 # 4 = 41022 + 11u22 # u = 4 rad>s

Ans.

$ In a similar manner, u can be found using Eq. 12–30. r v $ 4 ft/s a $ 30 ft/s2 u (b)

Fig. 12–35b

# $ $ ## a = 41r - ru222 + 1ru + 2ru22 $ 30 = 4[-0.51422 - 11422]2 + [1u + 2102142]2 $ 13022 = 1-2422 + u2 $ u = 18 rad>s2

Ans.

Vectors a and v are shown in Fig. 12–35b. NOTE: At this location A, the u and t (tangential) axes will coincide. The +n (normal) axis is directed to the right, opposite to +r.


PROBLEMS

71

PROBLEMS *12–136. 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. 12–137. If a particle’s position is described by the polar coordinates r = 411 + sin t2 m and u = 12e -t2 rad, where t is in seconds and the argument for the sine is in radians, determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s. 12–138. # The slotted fork is rotating about O at a constant rate of u = 3 rad>s. Determine the radial and transverse components of the velocity and acceleration of the pin A at the instant u = 360°. The path is defined by the spiral groove r = 15 + u>p2 in., where u is in radians. 12–139. The slotted fork is rotating about O at $ # u = 3 rad>s, which is increasing at u = 2 rad>s2 when u = 360°. Determine the radial and transverse components of the velocity and acceleration of the pin A at this instant. The path is defined by the spiral groove r = 15 + u>p2 in., where u is in radians.

•

u $ 3 rad/s O

A

12–141. If a particle’s position is described by the polar coordinates r = 12 sin 2u2 m and u = 14t2 rad, where t is in seconds, determine the radial and transverse components of its velocity and acceleration when t = 1 s.

12–142. A particle is moving along a circular path having a 400-mm radius. Its position as a function of time is given by u = 12t22 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°.

12–143. A particle moves in the x–y plane such that its position is defined by r = 52ti + 4t2j6 ft, where t is in seconds. Determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s.

*12–144. A truck is traveling along the horizontal circular curve of radius r = 60 m with a constant# speed v = 20 m>s. Determine the angular rate of rotation u of the radial line r and the magnitude of the truck’s acceleration.

12–145. 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.

r $ (5 " u/p)in.

· u r $ 60 m

Probs. 12–138/139

*12–140. If a particle moves along a path such that r = 12 cos t2 ft and u = 1t>22 rad, where t is in seconds, plot the path r = f1u2 and determine the particle’s radial and transverse components of velocity and acceleration.

u

Probs. 12–144/145


72

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–146. 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 is in radians, the argument for the sine is in degrees, and t is in seconds. Determine the acceleration of the particle at u = 30°. The particle starts from rest at u = 0°.

12–151. A particle travels along a portion of the “four-leaf rose” defined by the equation r = 15 cos 2u2 m. If the angular velocity of the radial coordinate line is # u = 13t22 rad>s, where t is in seconds, determine the radial and transverse components of the particle’s velocity and acceleration at the instant u = 30°. When t = 0, u = 0°.

12–147. 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 = 10.4u2 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.

r $ 5 cos 2u u

*12–148. Solve $ Prob. 12–147 if the # slotted link has an angular acceleration u = 8 rad>s2 when u = 3 rad>s at u = p>3 rad.

r

12–149. 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 = 10.4u2 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

Prob. 12–151

r · u $ 3 rad/s

r $ 0.4 u u

O

Probs. 12–147/148/149 12–150. A train is traveling along the circular curve of radius r# = 600 ft. At the instant shown, its angular $ rate of rotation is u = 0.02 rad>s, which is decreasing at u = -0.001 rad>s2. Determine the magnitudes of the train’s velocity and acceleration at this instant.

*12–152. At the instant # shown, the water sprinkler is rotating with an angular speed u = 2 rad>s and an angular acceleration $ u = 3 rad>s2. If the nozzle lies in the vertical 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 = 0.2 m.

r $ 0.2 m u

r $ 600 ft

Prob. 12–150

· u $ 0.02 rad/s

u

Prob. 12–152


PROBLEMS 12–153. The boy slides down the slide at a constant speed of 2 m>s. If the slide is in the form of a helix, defined by the equations r = 1.5 m and z = -u>p, determine the boy’s # angular velocity about the z axis, u, and the magnitude of his acceleration. z

73

12–155. For a short distance the train travels along a track having the shape of a spiral, r = 11000>u2 m, where u is in radians. If it maintains a constant speed v = 20 m>s, determine the radial and transverse components of its velocity when u = 19p>42 rad.

1.5 m

r 2m

u 1000 r$ u

u

x

Prob. 12–155

r

Prob. 12–153 12–154. A cameraman standing at A is following the movement of a race car, B, which is traveling along a straight track at a constant speed of 80 ft>s. Determine the angular rate at which he must turn in order to keep the camera directed on the car at the instant u = 60°. vB $ 80 ft/s

*12–156. For a short distance the train travels along a track having the shape of a spiral, r = 11000>u2 m, where u is in radians. If the angular rate is constant, # u = 0.2 rad>s, determine the radial and transverse components of its velocity and acceleration when u = 19p>42 rad.

B

r u r

100 ft

1000 r$ u

u A

Prob. 12–154

Prob. 12–156


74

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–157. The arm of the robot has a fixed length so that r = 3 ft and its grip A moves along the path z = 13 sin 4u2 ft, where u is in radians. If u = 10.5t2 rad, where t is in seconds, determine the magnitudes of the grip’s velocity and acceleration when t = 3 s.

12–161. The searchlight on the boat anchored 2000 ft from shore is turned on the automobile, which is traveling along the straight road at a constant speed of 80 ft>s. Determine the angular rate of rotation of the light when the automobile is r = 3000 ft from the boat.

12–158. For a short time the arm of the robot is extending # at a constant rate such that r = 1.5 ft>s when r = 3 ft, 2 z = 14t 2 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.

12–162. If the car in Prob. 12–161 is accelerating at 15 ft>s2 at the instant $r = 3000 ft, determine the required angular acceleration u of the light at this instant.

A 80 ft/s r

z r

u

u u

Probs. 12–157/158 12–159. The rod OA rotates counterclockwise with a # constant angular velocity of u = 5 rad>s. Two pinconnected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 10012 - cos u2 mm. Determine the speed of the slider blocks at the instant u = 120°. *12–160. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–159 when u = 120°. . u $ 5 rad/s

y

2000 ft

Probs. 12–161/162 12–163. For a short time the bucket of the backhoe traces the path of the cardioid r = 2511 - cos u2 ft. Determine the magnitudes of the velocity and acceleration of the bucket when velocity # u = 120° if the boom is rotating with an angular $ of u = 2 rad>s and an angular acceleration of u = 0.2 rad>s2 at the instant shown.

A B r

u O

x r

u $ 120%

r $ 100 (2 # cos u)

Probs. 12–159/160

Prob. 12–163


75

PROBLEMS *12–164. A car is traveling 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.

12–166. The roller coaster is traveling down along the spiral ramp with a constant speed v = 6 m>s. If the track descends a distance of 10 m for every full revolution, u = 2p rad, determine the magnitude of the roller coaster’s acceleration as it moves along the track, r = 5 m. Hint: For part of the solution, note that the tangent to the ramp at any point is at an angle f = tan-1[10>2p152] = 17.66° from the horizontal. Use this to determine the velocity #components # vu and vz , which in turn are used to determine u and z.

r $ 400 ft . u

r$5m

Prob. 12–164 10 m

12–165. The mechanism of a machine is constructed so that for a short time the roller at A follows the surface of the cam described by the equation r = 10.3 + 0.2 cos u2 m. If # $ u = 0.5 rad>s and u = 0, determine the magnitudes of the roller’s velocity and acceleration at the instant, u = 30°. Neglect the size of the roller. Also determine the velocity components 1vA2x and 1vA2y of the roller at this instant. The rod to which the roller is attached remains vertical and can slide up or down along the guides while the guides move horizontally to the left.

Prob. 12–166 12–167. A cameraman standing at A is following the movement of a race car, B, which is traveling around a curved track #at a constant speed of 30 m>s. Determine the angular rate u at which the man must turn in order to keep the camera directed on the car at the instant u = 30°.

y vB $ 30 m/s B

A

u

r & u $ 0.5 rad/s

r

20 m

G (vA)y (vA)x

Prob. 12–165

A x

20 m

u 20 m

20 m

r $ 0.3 " 0.2 cos u

Prob. 12–167


76

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

*12–168. The pin follows the path described by the equation r = 10.2 + 0.15 cos u2 m. At the instant u = 30°, $ # u = 0.7 rad>s and u = 0.5 rad>s2. Determine the magnitudes of the pin’s velocity and acceleration at this instant. Neglect the size of the pin.

12–170. The small washer is sliding down the cord OA.When it is at the midpoint, its speed is 200 mm>s and its acceleration is 10 mm>s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical components .

z A y

v, a r $ 0.2 " 0.15 cos u z

r

700 mm

O

u & u $ 0.7 rad/s

r

u

x x

300 mm

400 mm

y

Prob. 12–168 Prob. 12–170 12–169. For a short time the position of the roller-coaster car along its path is defined by the equations r = 25 m, u = 10.3t2 rad, and z = 1-8 cos u2 m, where t is in seconds. Determine the magnitude of the car’s velocity and acceleration when t = 4 s.

u

A

12–171. 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 = 14 cos 2u2 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.

r

z

r2 $ 4 cos 2u & u $ 6 rad/s O

Prob. 12–169

r

A C

Prob. 12–171


77

12.9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES

12.9 Absolute Dependent Motion 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 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) referenced from a fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of block A and block B, and (3) has a positive sense from C to A and D to B. If the total cord length is lT , the position coordinates are related by the equation 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 lengths of the changing segments of the cord, we have dsB dsA + = 0 dt dt

or

Datum sA

C D

Datum sB

O A

B

Fig. 12–36

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 involving dependent motion of two blocks 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 . Here we have chosen coordinate axes which are (1) referenced from fixed points or datums, (2) measured in the direction of motion of each block, and (3) positive to the right 1sA2 and positive downward 1sB2. 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 Since l and h are constant during the motion, the two time derivatives yield 2vB = -vA 2aB = -aA Hence, when B moves downward 1+sB2, A moves to the left 1-sA2 with two times the motion.

Datum sB

B

h

A

Datum

sA

(a)

Fig. 12–37a


78

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

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 21h - sB2 + h + sA = l

Datum

Time differentiation yields 2vB = vA 2aB = aA sB

B

Here the signs are the same. Why?

h

Procedure for Analysis Datum

A

Datum

sA

(b)

Fig. 12–37 (cont.)b

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. The following procedure is required. Position-Coordinate Equation.

• Establish position coordinates which have their origin located at • • •

•

a fixed point or datum. The coordinates are directed along the path of motion and extend to a point having the same motion as each of the particles. 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 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

The motion of the traveling block on this equations yield the required velocity and acceleration equations oil rig depends upon the motion of the which relate the motions of the particles. cable connected to the winch which operates it. It is important to be able to • The signs of the terms in these equations will be consistent with relate these motions in order to determine those that specify the positive and negative sense of the position the power requirements of the winch and coordinates. the force in the cable caused by Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, accelerated motion.

New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.


EXAMPLE 12.21 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 are changing 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

Ans.


79

80

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

EXAMPLE 12.22 Determine the speed of block A in Fig. 12–39 if block B has an upward speed of 6 ft>s. Datum sA sC A

sB C

D

B

6 ft/s

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 which change length, it will be necessary to use a third coordinate, sC , in order to relate sA to sB . In other words, the length of one of the cords can be expressed in terms of sA and sC , and the length of the other cord can be expressed in terms of sB and sC . The red colored segments of the cords in Fig. 12–39 do not have to be considered in the analysis. Why? For the remaining cord lengths, say l1 and l2 , we have sA + 2sC = l1 sB + 1sB - sC2 = l2 Eliminating sC yields an equation defining the positions of both blocks, i.e., sA + 4sB = 2l2 + l1 Time Derivative. The time derivative gives vA + 4vB = 0 so that when vB = -6 ft>s (upward), vA = +24 ft>s = 24 ft>s T

Ans.



EXAMPLE 12.23 Determine the speed with which block B rises 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 2 m/s

B

Fig. 12–40

SOLUTION Position-Coordinate Equation. The position of point A is defined by sA , and the position of block B is specified by sB since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fixed pin at pulley D. Since the system consists of two cords, the coordinates sA and sB cannot be related directly. Instead, by establishing a third position coordinate, sC , we can now express the length of one of the cords in terms of sB and sC , and the length of the other cord in terms of sA , sB , and sC . Excluding the red colored segments of the cords in Fig. 12–40, the remaining constant cord lengths l1 and l2 (along with the hook and link dimensions) can be expressed as sC + sB = l1 1sA - sC2 + 1sB - sC2 + sB = l2 Eliminating sC yields sA + 4sB = l2 + 2l1 As required, this equation relates the position sB of block B to the position sA of point A. Time Derivative. The time derivative gives vA + 4vB = 0 so that when vA = 2 m>s (downward), vB = -0.5 m>s = 0.5 m>s c

Ans.


81

82

C H A P T E R 12

K I N E M AT I C S

EXAMPLE 12.24

SOLUTION Position-Coordinate Equation. This problem is unlike the previous examples since rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the positions of S and A, are specified by means of the x and y coordinates 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

E

have lDA = 411522 + x2 ; also, lCD = 15 - y. Hence,

C

10 m

PA R T I C L E

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 at E. The rope is 30 m long and passes over a small pulley at D.

D

15 m

OF A

l = lDA + lCD

S y

30 = 411522 + x2 + 115 - y2

A vA $ 0.5 m/s x

y = 4225 + x2 - 15 (1) Time Derivatives. Taking the time derivative, using the chain rule, where vS = dy>dt and vA = dx>dt, yields dy 1 2x dx = c d 2 dt dt 2 4225 + x x = vA 2 225 + x 4

Fig. 12–41

vS =

(2)

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 2 4225 + 1202

10.52 = 0.4 m>s = 400 mm>s c

Ans.

The acceleration is determined by taking the time derivative of Eq. 2. Since vA is constant, then aA = dvA>dt = 0, and we have aS =

d2y dt2

= c

-x1dx>dt2

dxvA + c 1225 + x223>2

1 2 4225 + x

da

dx bvA + c dt

1 2 4225 + x

dx

dvA 225v2A = dt 1225 + x223>2

At x = 20 m, with vA = 0.5 m>s, the acceleration becomes 22510.5 m>s22 aS = = 0.00360 m>s2 = 3.60 mm>s2 c [225 + 120 m22]3>2

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 as well as its length. Unpublished Work © 2007 by R. C. Hibbeler.direction To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

83

12.10 RELATIVE-MOTION ANALYSIS OF TWO PARTICLES USING TRANSLATING AXES

12.10 Relative-Motion Analysis of Two Particles Using Translating Axes

Throughout this chapter the absolute motion of a particle has been determined using a single fixed reference frame for measurement. There are many cases, however, where the path of motion for a particle is complicated, so that it may be feasible 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. Any type of coordinates—rectangular, cylindrical, etc.—can be chosen to describe these two different motions. In this section translating frames of reference will be considered for the analysis. Relative-motion analysis of particles using rotating frames of reference will be treated in Secs. 16.8 and 20.4, since such an analysis depends on prior knowledge of the kinematics of line segments.

z¿

paths aa and bb, respectively, as shown in Fig. 12–42a.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 relative position of “B with respect to A” is designated by a relative-position vector rB>A . Using vector addition, the three vectors shown in Fig. 12–42a can be related by the equation* rB = rA + rB>A

(12–33)

Velocity. An equation that relates the velocities of the particles can be determined by taking the time derivative of Eq. 12–33, i.e., vB = vA + vB>A

a

z

Position. Consider particles A and B, which move along the arbitrary

a rA

Fixed observer

A

Translating observer rB/A y

O

rB

y¿

b B

x¿

b

x (a)

Fig. 12–42a

(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 this vector’s components will only have to account for the change in the vector’s magnitude. Equation 12–34 therefore states that the velocity of B is *An easy way to remember the setup of this equation, and others like it, is to note the Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, “cancellation” ofAll therights subscript A between the two terms, i.e., rB = by rA Copyright + rB>A . and written permission should be obtained from the New Jersey. reserved. This publication is protected publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

84

C H A P T E R 12

vB/A vB

K I N E M AT I C S

OF A

PA R T I C L E

equal to the velocity of A plus (vectorially) the relative velocity of “B with respect to A,” as measured by the translating observer fixed in the x¿, y¿, z¿ reference, Fig. 12–42b.

Acceleration. The time derivative of Eq. 12–34 yields a similar vector

vA

relationship between the absolute and relative accelerations of particles A and B.

(b)

a B = a A + a B>A

Fig. 12–42b

Here a B>A is the acceleration of B as seen by the observer located at A and translating with the x¿, y¿, z¿ reference frame. The vector addition is shown in Fig. 12–42c.

aB/A aA

(12–35)

aB

(c)

Fig. 12–42 (cont.)c

Procedure For Analysis • When applying the relative-position equation, rB = rA + rB>A ,

it is first necessary to specify the locations of the fixed x, y, z, and translating x¿, y¿, z¿ axes.

• Usually, the origin A of the translating axes is located at a point having a known position, rA , Fig. 12–42a.

• A graphical representation of the vector addition rB = rA + rB>A can be shown, and both the known and unknown quantities labeled on this sketch.

• 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 rB , rA , and rB>A into rectangular or Cartesian components, thereby generating a set of scalar equations.

• The

relative-motion equations vB = vA + vB>A and a B = a A + a B>A are applied in the same manner as explained above, except in this case the origin O of the fixed x, y, z axes does not have to be specified, Figs. 12–42b and 12–42c.

The pilots of these jet planes flying close to one another must be aware of their relative positions and velocities at all Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, times in New orderJersey. to avoid collision. All arights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

85


EXAMPLE 12.25 A train, traveling at a constant speed of 60 mi>h, crosses over a road as shown in Fig. 12–43a. If automobile A is traveling at 45 mi>h along the road, determine the magnitude and direction of the relative velocity of the train with respect 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 and a Cartesian vector analysis, we have vT = vA + vT>A

45%

T vT $ 60 mi/h y¿

y x

A

x¿ vA $ 45 mi/h (a)

Fig. 12–43a

60i = 145 cos 45°i + 45 sin 45°j2 + vT>A vT>A = 528.2i - 31.8j6 mi>h The magnitude of vT>A is thus

Ans.

vT>A = 4128.222 + 1-31.822 = 42.5 mi>h

Ans.

From the direction of each component, Fig. 12–43b, the direction of vT>A defined from the x axis is 1vT>A2y 31.8 tan u = = 1vT>A2x 28.2 u = 48.5° cu Ans. 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.

28.2 mi/h

u

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

d = c

45 mi>h 45°

d + c

1vT>A2x

d + c

vT/A

31.8 mi/h (b)

Fig. 12–43b

1vT>A2y d c

: : a Resolving each vector into its x and y components yields + 2 1: 60 = 45 cos 45° + 1v 2 + 0

vA $ 45 mi/h

T>A x

1+ c 2 0 = 45 sin 45° + 0 + 1vT>A2y Solving, we obtain the previous results,

u

45% vT $ 60 mi/h

1vT>A2x = 28.2 mi>h = 28.2 mi>h : 1vT>A2y = -31.8 mi>h = 31.8 mi>h T

vT/A

(c)

Ans.

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education,Fig. Inc.,12–43c Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

86

C H A P T E R 12

K I N E M AT I C S

EXAMPLE 12.26 y¿ 600 km/h A

50 km/h2

x¿

B

PA R T I C L E

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.

y

700 km/h

OF A

x

100 km/h2 400 km

4 km (a)

SOLUTION Velocity. The x, 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 relativevelocity equation in scalar form since the velocity vectors of both planes are parallel at the instant shown, we have 1+ c 2

Fig. 12–44a

vB = vA + vB>A 600 = 700 + vB>A

vB/A

vB>A = -100 km>h = 100 km>h T

vA $ 700 km/h v $ 600 km/h B

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)

Fig. 12–44b

1aB2n = 900 km/h2

1600 km>h22 v2B = = 900 km>h2 r 400 km

Applying the relative-acceleration equation, we have aB = a A + a B>A

u

900i - 100j = 50j + a B>A

aB/A

150 km/h2 (c)

Fig. 12–44c

Ans.

Thus, a B>A = 5900i - 150j6 km>h2 From Fig. 12–44c, the magnitude and direction of a B>A are therefore aB>A = 912 km>h2

u = tan-1

150 = 9.46° c 900

Ans.

NOTE: The solution to this problem was possible using a translating frame of reference, since the pilot in plane A is “translating.” Observation of 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.


87


EXAMPLE 12.27 At the instant shown in Fig. 12–45 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 a 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¿

2 m/s2 60% x¿

A 3 m/s2

18 m/s

r $ 100 m 12 m/s B y

vB = vA + vB>A -12j = 1-18 cos 60°i - 18 sin 60°j2 + vB>A

60%

x

vB>A = 59i + 3.588j6 m>s Thus, vB>A = 41922 + 13.58822 = 9.69 m>s

Noting that vB>A has +i and +j components, Fig. 12–45b, its direction is 1vB>A2y 3.588 tan u = = 1vB>A2x 9 u = 21.7° a

(a)

Ans.

Ans.

Fig. 12–45a

3.588 m/s

Acceleration. Car B has both tangential and normal components of acceleration. Why? The magnitude of the normal component is

vB/A

u 9 m/s

2

112 m>s2 v2B 1aB2n = = = 1.440 m>s2 r 100 m Applying the equation for relative acceleration yields a B = a A + a B>A

(b)

Fig. 12–45b

1-1.440i - 3j2 = 12 cos 60°i + 2 sin 60°j2 + a B>A

2.440 m/s2

2

a B>A = 5-2.440i - 4.732j6 m>s Here aB>A has -i and -j components. Thus, from Fig. 12–45c, aB>A = 412.44022 + 14.73222 = 5.32 m>s2 1aB>A2y 4.732 tan f = = 1aB>A2x 2.440 f = 62.7° d

f

Ans. aB/A

Ans.

NOTE: Is it possible to obtain the relative acceleration of a A>B using this

4.732 m/s2 (c)

Fig. 12–45c

method? Refer to the comment made at the end of Example 12.26. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

88

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

PROBLEMS *12–172. 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.

12–175. 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. A

C

vA A 2 m/s

B

Prob. 12–172

B

12–173. 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.

vB

Prob. 12–175 *12–176. If the hydraulic cylinder at H draws rod BC in by 8 in., determine how far the slider at A moves.

D

A

C

2 m/s

A

H

B

C

Prob. 12–176 B

Prob. 12–173 12–174. Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load at B 15 ft in 5 s.

12–177. The crate is being lifted up the inclined plane using the motor M and the rope and pulley arrangement shown. Determine the speed at which the cable must be taken up by the motor in order to move the crate up the plane with a constant speed of 4 ft>s.

A

B

vA

M

A C

Prob. 12–174

B

vB

Prob. 12–177


89

PROBLEMS 12–178. Determine the displacement of the block at B if A is pulled down 4 ft.

12–181. If block A is moving downward with a speed of 4 ft>s while C is moving up at 2 ft>s, determine the speed of block B. 12–182. If block A 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.

C A

B

Prob. 12–178 B

12–179. The hoist is used to lift the load at D. If the end A of the chain is traveling downward at vA = 5 ft>s and the end B is traveling upward at vB = 2 ft>s, determine the velocity of the load at D.

A

C

Prob. 12–182

A

B

D

12–183. 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.

Prob. 12–179 C

*12–180. 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. D

B

B C P

4 ft/s A

A

Prob. 12–180

Prob. 12–183


90

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

*12–184. If block A of the pulley system is moving downward with a speed of 4 ft>s while block C is moving up at 2 ft>s, determine the speed of block B.

12–186. The cylinder C is being lifted using the cable and pulley system shown. If point A on the cable is being drawn toward the drum with a speed of 2 m>s, determine the speed of the cylinder.

A vA

A C C

B

s

Prob. 12–184

Prob. 12–186 12–185. If the end A of the cable is moving upwards at vA = 14 m>s, determine the speed of block B.

12–187. The 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. Determine the velocity and acceleration of the end of the cord at B if at the instant sA = 4 ft the collar is moving upwards at 5 ft>s, which is decreasing at 2 ft>s2.

3 ft

F

3 ft

C

vA $ 14 m/s

D sB

A

E

sA

B

D C

A

B

Prob. 12–185

Prob. 12–187


91

PROBLEMS *12–188. 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 downwards 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

3 ft

12–190. 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

C

B

D xB

sB sA

Prob. 12–190

B

12–191. The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when xA = 0 and moves backward with a constant acceleration aA = 0.2 m>s2, determine the speed of the boy at the instant yB = 4 m. Neglect the size of the limb. When xA = 0, yB = 8 m, so that A and B are coincident, i.e., the rope is 16 m long.

A

C

Prob. 12–188 yB

12–189. 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 B xA xC 4m

8m B A

xA

Prob. 12–191

*12–192. Collars A and B are connected to the cord that passes over the small pulley at C. When A is located at D, B is 24 ft to the left of D. If A moves at a constant speed of 2 ft>s to the right, determine the speed of B when A is 4 ft to the right of D. C

A

10 ft

C s

B

A D

Prob. 12–189

2 ft/s

Prob. 12–192


92

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–193. 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

*12–196. The roller at A is moving upward with a velocity of vA = 3 ft>s and has an acceleration of aA = 4 ft>s2 when sA = 4 ft. Determine the velocity and acceleration of block B at this instant.

h vB, aB A

B

Prob. 12–193 12–194. Vertical motion of the load is produced by movement of the piston at A on the boom. Determine the distance the piston or pulley at C must move to the left in order to lift the load 2 ft. The cable is attached at B, passes over the pulley at C, then D, E, F, and again around E, and is attached at G. A

6 ft/s

E

sA $ 4 ft

B

Prob. 12–196

B

C

G

3 ft

D

F

Prob. 12–194 12–195. The motion of the collar at A is controlled by a motor at B such that when the collar is at sA = 3 ft it is moving upwards at 2 ft>s and slowing down at 1 ft>s2. Determine the velocity and acceleration of the cable as it is drawn into the motor B at this instant.

12–197. 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.

4 ft A vA B

u

sA

vB A B

Prob. 12–195

Prob. 12–197


93

PROBLEMS 12–198. At the instant shown, cars A and B are traveling at speeds of 30 mi>h and 20 mi>h, respectively. If B is increasing its speed by 1200 mi>h2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A. 12–199. At the instant shown, cars A and B are traveling at speeds of 30 m>h and 20 mi>h, respectively. If A is increasing its speed at 400 mi>h2 whereas the speed of B is decreasing at 800 mi>h2, determine the velocity and acceleration of B with respect to A.

12–201. 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. vB $ 18.5 m/s B

A 100 m

B

vA $ 10 m/s

100 m

30%

45

0.3 mi

vB $ 20 mi/h A

vA $ 30 mi/h

Probs. 12–198/199

Prob. 12–201

*12–200. Two boats leave the shore at the same time and travel in the directions shown. If vA = 20 ft>s and vB = 15 ft>s, determine the speed of boat A with respect to boat B. How long after leaving the shore will the boats be 800 ft apart?

12–202. An aircraft carrier is traveling forward with a velocity of 50 km>h. At the instant shown, the plane at A has just taken off and has attained a forward horizontal air speed of 200 km>h, measured from still water. If the plane at B is traveling along the runway of the carrier at 175 km>h in the direction shown, determine the velocity of A with respect to B.

vA A B

15%

30% O

vB

B

A

45% 50 km/h

Prob. 12–200

Prob. 12–202


94

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

12–203. 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.

12–205. At the instant shown car A is traveling with a velocity of 30 m>s and has an acceleration of 2 m>s2 along the highway. At the same instant B is traveling on the trumpet interchange curve with a speed of 15 m>s, which is decreasing at 0.8 m>s2. Determine the relative velocity and relative acceleration of B with respect to A at this instant.

vA A B rA $ 300 ft

vB

60% rB $ 250 ft

A B

r $ 250 m

60%

Prob. 12–203

Prob. 12–205

*12–204. The airplane has a speed relative to the wind of 100 mi>h. If the wind relative to the ground is 10 mi>h, determine the angle u at which the plane must be directed in order to travel in the direction of the runway. Also, what is its speed relative to the runway.

12–206. The boy A is moving in a straight line away from the building at a constant speed of 4 ft>s. The boy C throws the ball B horizontally when A is at d = 10 ft. At what speed must C throw the ball so that A can catch it? Also determine the relative speed of the ball with respect to boy A at the instant the catch is made.

C u

vC

10 mi/h

B

20% 20 ft

A 4 ft/s

d

Prob. 12–204

Prob. 12–206


95

DESIGN PROJECT 12–207. The boy A is moving in a straight line away from the building at a constant speed of 4 ft>s. At what horizontal distance d must he be from C in order to make the catch if the ball is thrown with a horizontal velocity of vC = 10 ft>s? Also determine the relative speed of the ball with respect to the boy A at the instant the catch is made.

*12–208. At a given instant, two particles A and B are moving with a speed of 8 m>s along the paths shown. If B is decelerating at 6 m>s2 and the speed of A is increasing at 5 m>s2, determine the acceleration of A with respect to B at this instant. y

1m C

vC

y2 $ x3 A

B

1m

20 ft

x A

1m 4 ft/s

B

d

y $ #x

Prob. 12–208

Prob. 12–207

Design Project 12–1D. Design of a Marble-Sorting Device Marbles roll off the production chute at 0.5 ft>s. Determine the range for the angle 0 … u … 30° for a selected position s for the placement of the hopper relative to the end of the chute. Submit a drawing of the device that shows the path the marbles take. u

0.5 ft/s

3 ft

2 ft s

Fig. 12–1D


96

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

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. (See page 5.)

P

s

O s

The average velocity is a vector quantity, defined as the displacement divided by the time interval.

vavg =

- ¢s ¢t

's

#'s P¿

P

O

The average speed is a scalar, and is the total distance traveled divided by the time of travel. (See page 6.) 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. (See page 8.)

1vsp2avg =

v =

ds , dt

a =

sT ¢t

dv , dt

s

sT

a ds = v dv

s = s0 + v0t + 12 act2 v2 = v20 + 2ac1s - s02

v = v0 + act

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. (See page 18.)

v = ds>dt, a = dv>dt, a ds = v dv.


97

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. (See pages 33 and 34.)

# vx = x # vy = y # vz = z

z

# ax = vx # ay = vy # az = vz

P 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 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. (See pages 38 and 39.)

1+ c 2

vy = 1v02y + act

1+ c 2

y = y0 + 1v02y t + 12 act2

1+ c 2

v2y = 1v02y2 + 2ac1y - y02

+ 2 1:

x = x0 + 1v02x t + 12 act2

y

a $ #gj vx v0

(v0)y

vy

(v0)x

y

v

r

y0 x x0 x


98

C H A P T E R 12

K I N E M AT I C S

OF A

PA R T I C L E

Curvilinear Motion n, t If normal and tangential axes are used for the analysis, then v is always in the positive t direction. The acceleration has two components. The tangential component, a t, 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 a n accounts for the change in the direction of the velocity.This component is always in the positive n direction. (See pages 49–51.)

Curvilinear Motion, r, U, z

O¿ n

O an

s

# at = v

or an =

P

at ds = v dv

a at

v

t

2

v r

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.

# vr = r # vu = ru

To apply the time-derivative equations, it # $ # $ is necessary to determine r, r, r, u, u at the instant considered.

# $ ar = r - ru2 $ ## au = ru + 2ru

v vu vr P

r u

If the path r = f1u2 is given, then the chain rule of calculus must be used to obtain the time derivatives. 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. (See pages 62–64.)

O Velocity

a au

r

P

ar

u O Acceleration


99

CHAPTER REVIEW

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 so that they are directed along the line of motion of the blocks. Using geometry and/or trigonometry, the coordinates are then related to the cable length in order to formulate a position coordinate equation.

Datum sB

B

h

2sB + h + sA = l

A

The first time derivative of this equation gives a relationship between the velocities of the blocks, and a second time derivative gives the relationship between their accelerations. (See page 77.)

2vB = -vA

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 equations 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. (See pages 83 and 84.)

sA

Datum

2aB = -aA

z¿ a

z

rB = rA + rB>A vB = vA + vB>A aB = aA + aB>A

a rA

Fixed observer

A

Translating observer rB/A y

O x¿

rB

y¿

b B

b

x


Kinematics of a Particle

Kinematics of a Particle

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