Dynamics Solns Ch05

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Solutions Manual

Engineering Mechanics: Dynamics1st Edition

Gary L. GrayThe Pennsylvania State University

Francesco CostanzoThe Pennsylvania State University

Michael E. PleshaUniversity of WisconsinMadison

With the assistance of:

Chris Punshon

Andrew J. Miller

Justin High

Chris OBrien

Chandan Kumar

Joseph Wyne

Version: August 10, 2009

The McGraw-Hill Companies, Inc.


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Copyright 20022010

Gary L. Gray, Francesco Costanzo, and Michael E. Plesha

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It

may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon

request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without

the permission of McGraw-Hill, is prohibited.


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Dynamics 1e 3

Important Information about

this Solutions Manual

Even though this solutions manual is nearly complete, we encourage you to visit

http://www.mhhe.com/pgc

often to obtain the most up-to-date version. In particular, as of July 30, 2009, please note the following:

: The solutions for Chapters 1 and 2 have been accuracy checked and have been edited by us. They are

in their final form.

: The solutions for Chapters 4 and 7 have been accuracy checked and should be error free. We will be

adding some additional detail to these solutions in the coming weeks.

:The solutions for Chapters 3, 6, 8, and 9 are being accuracy checked and the accuracy checked versionsshould be available by the end of August 2009. We will be adding some additional detail to these

solutions in the coming weeks.

: The solutions for Chapter 10 should be available in their entirety by the end of August 2009.

All of the figures in Chapters 610 are in color. Color will be added to the figures in Chapters 15 over the

coming weeks.

Contact the Authors

If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the

authors and editors via email at:

[email protected]

We welcome your input.

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4 Solutions Manual

Accuracy of Numbers in Calculations

Throughout this solutions manual, we will generally assume that the data given for problems is accurate to

3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in

this solutions manual using the rounded intermediate numerical results that are reported, you should obtain

the final answers that are reported to 3 significant digits.

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568 Solutions Manual

Chapter 5 Solutions

Problem 5.1

Use the definition of impulse given in Eq. (5.5) to compute the impulse of the

forces shown during the interval 0 t 2 s.

Solution

Using the definition of impulse, we have

Impulse DZt2t1

EF . t / d t :

Hence, for the blue curve we have

Impulse DZ2 s0

0:3 t2 2t dt; ) Impulse D 0:3 t33

t22 s0

D 0:4 lbs.

For the green curve we have

Impulse D Z2 s

0 0:31 e2t dt ) Impulse D 0:3t C 0:3 e

2t

2 2 s

0 D 0:453 lbs.For the redcurve we have

Impulse DZ2 s0

.1 t / d t ) Impulse D t t2

2

2 s0

D 0 lbs.

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Dynamics 1e 569

Problem 5.2

The total mass of the Earth is me D 5:97361024 kg. Modeling the Earth (with everything in and on it)as an isolated system and assuming that the center of the Earth is also the center of mass of the Earth,

determine the displacement of the center of the Earth due to

(a) a 2 m jump off the surface by a 85 kg person;

(b) the Space Shuttle, with a mass of124;000 kg, reaching an orbit of 200 km;

(c) 170;000 km3 of water being elevated 50 m (these numbers are estimates based on publicly available

information about the Aswan Dam at the border between Egypt and Sudan). Use 1 g=cm3 for the

density of water.

Solution

We begin by choosing the center of mass of the system as the fixed origin of an inertial reference frame. Since

the system is isolated, the velocity of the systems center of mass must remain constant and therefore equal to

zero. This implies that the center of mass of the system must remain at the origin of our coordinate system.

Consequently, assuming that the motion of the Earth and of an object P occurs along a line, and designating

this line as the x axis of a coordinate system, we must have

0 D mexe C mPxPme C mP

) xe D mPxPme

; (1)

where xe and xP are the x coordinates of the center of the Earth and of the object P, respectively, and where

mP is the mass of the object P.

Now, let d denote the distance by which the Earth is displaced. Then the displacement of the Earth is

simply the absolute value of the result obtained in Eq. (1), i.e.,

d DmPxPme

: (2)

Using Eq. (2) in the three cases listed in the problem statement, and recalling that me D 5:97361024 kg,we have

(a) xe D 28:51024m, where we set mP D 85:0 kg and xP D 2:00 m;

(b) xe D 4:151012m, where we set mP D 124;000 kg and xP D 200 km D 200103m;

(c) xe D 1:42106m, where we set

mP D .170;000 km3/

1000m

km

31:00

g

cm3

0:001 kg=g

.0:01 m=cm/3D 170:01015 kg and xP D 50 m:

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Problem 5.3

Consider an elevator that moves with an operating speed of 2:5 m=s. Suppose

that a person who boards the elevator on the ground floor gets off on the fifth

floor. Assuming that the elevator has achieved operating speed by the time it

reaches the second floor and that it is still moving at its operating speed as itpasses the fourth floor, determine the momentum change of a person with a

mass of80 kg between the second and fourth floors if each floor is 4 m high. In

addition, determine the impulse of the persons weight during the same time

interval.

Solution

Let the subscripts 1 and 2 denote the time instants at which the elevator goes by floors 1 and 2, respectively.

Observe that the person riding in the elevator moves with the same velocity of the elevator. Hence, denoting

this velocity by Ev and observing that this velocity is constant, we have we have

mEv1 D mEv2;that is,

The change in momentum is equal to zero.

The impulse of the weight force isZt2t1

.mg O| / d t ) mg .t2 t1/ O| :

Letting d be the distance between floors 2 and 4, and letting v be the operating speed of the

elevator, then, since v is constant, we have that the time interval in traveling from the second to the fourthfloor is

t2 t1 D dv

:

Consequently, the impulse of the weight force acting on the person during the time interval .t1; t2/ isZt2t1

.mg O| / d t D mgdv

O| ;

so that we have

Impulse of the weight D .2510 Ns/ O| ;

where we have used the following numerical data: m D 80:0 kg, g D 9:81 m=s2, d D 8:00 m, andv D 2:5 m=s.

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Dynamics 1e 571

Problem 5.4

A 180 gr (7000 gr D 1 lb) bullet goes from rest to 3300 ft=s in 0:0011 s. Deter-mine the magnitude of the impulse imparted to the bullet during the given time

interval. In addition, determine the magnitude of the average force acting on

the bullet.

Solution

Let the subscripts 1 and 2 denote the time instants when the bullet is at rest and when the bullet has a speed

v2 D 3300 ft=s, respectively. Let EF represent the total force acting on the bullet. The magnitude of theimpulse of EF is equal to the magnitude of the change in momentum of the bullet during the time interval.t1; t2/. Hence, we have

Zt2t1

EF dtD mEv2 mEv1 D mEv2 ) Zt2

t1

EF dtD mv2 D 2:64 lbs,

where we have used the following numerical data:

m D .180 gr/ 1.7000 gr=lb/

1

.32:2 ft=s2/D 7:986104 lbs2=ft and v2 D 3300 ft=s: (1)

Recalling that the average force over the time interval .t1; t2/ can be computed by dividing the change in

momentum during the time interval by t2 t1, then the magnitude of the average force acting on the bullet is

EFavg D mv2t2 t1 D 2400 lb;where, in addition to the data in Eq. (1) we have used the fact that t2 t1 D 0:0011 s.

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Problem 5.5

A 3400 lb car is parked as shown. Determine the impulse of the normal reaction force acting on the car

during the span of an hour if D 15.

Solution

Using the FBD on the right and writing the equilibrium equation in the y direction, we

have XFy W N mg cos D 0 ) N D mg cos :

Then the impulse of the normal reaction for the time interval of duration t D 1 h is

Impulse ofN DZt0

.N O| / d t D mgt cos O| :

therefore, using the above equation we have

Impulse ofN D .11:8106 lbs/ O| @ 15;

where we have used the following numerical data: mg D 3400 lb, t D 3600 s, and D 15:0.

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Dynamics 1e 573

Problem 5.6

An airplane performs a turn at constant speed and elevation so as to change its

course by 180. Let A and B designate the beginning and end points of the

turn. Assuming that the change in mass of the plane due to fuel consumption

is negligible, is the airplanes momentum at A different from the airplanesmomentum at B? In addition, again neglecting the change in mass between A

and B , is the total work done on the plane between A and B positive, negative,

or equal to zero?

Solution

Yes the momentum is different because the momentum of an object is a vector quantity. Hence, even if the

mass and the speed of the plane are constant, the fact that the direction of the plane at B is different from that

at A implies that the momentum of the plane has changed. By contrast the planes kinetic energy at A and B

is the same and therefore the total work done on the plane between A and B is equal to zero.

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Problems 5.7 and 5.8

The takeoff runway on carriers is much too short for a modern jetplane to take off on its own. For this

reason, the takeoff of carrier planes is assisted by hydraulic catapults (Fig. A). The catapult system is

housed below the deck except for a relatively small shuttle that slides along a rail in the middle of the

runway (Fig. B). The front landing gear of carrier planes is equipped with a tow bar that, at takeoff, isattached to the catapult shuttle (Fig. C). When the catapult is activated, the shuttle pulls the airplane along

the runway and helps the plane reach its takeoff speed. The takeoff runway is approximately 300 ft long,

and most modern carriers have three or four catapults.

Problem 5.7 In a catapult-assisted takeoff, assume that a 45;000 lb plane goes from 0 to 165 mph in 2 s

while traveling along a rectilinear and horizontal trajectory. Also assume that throughout the takeoff the

planes engines are providing 32;000 lb of thrust.

(a) Determine the average force exerted by the catapult on the plane.

(b) Now suppose that the takeoff order is changed so that a small trainer aircraft must take off first. If

the trainers weight and thrust are 13,000 and 5850 lb, respectively, and if the catapult is not reset to

match the takeoff specifications for the smaller aircraft, estimate the average acceleration to which the

trainers pilots would be subjected and express the answer in terms ofg. What do you think would

happen to the trainers pilot?

Problem 5.8 If the carrier takeoff of a 45;000 lb plane subject to the 32;000 lb thrust of its engines were

not assisted by a catapult, estimate how long it would take for a plane to safely take off, i.e., to reach a

speed of165 mph starting from rest. Also, how long a runway would be needed under these conditions?

Solution to 5.7

Part (a). Let t1 and t2 be the time instants at which the plane starts and reaches

the takeoff speed, respectively. Using the expression of average force in terms of

change in momentum, the average average force acting on the airplane during the

time interval of duration t2 t1 isEFavg D Ep.t2/ Ep.t1/

t2 t1 ;(1)

where Ep.t1/ and Ep.t2/ are the linear momenta of the airplane at time instants t1 and t2, respectively. Sincethe motion is in the horizontal direction and using the FBD on the right, the x component of Eq. (1) is

.Favg/c C Ft Dmf.vx2 vx1/

t2 t1 ) .Favg/c Dmf.vx2 vx1/

t2 t1 Ft D 137;100 lb; (2)

where .Favg/c is the average force due to the catapult, Ft D 32;000 lb is the engines thrust, vx1 D 0,vx2 D 165 mph D 242:0 ft=s, and mf D .45;000 lb/=.32:2 ft=s2/ D 1398 slug is the mass of the first

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Dynamics 1e 575

airplane. Expressing .Favg/c to three significant digits, we have

Favg

c

D 137;000 lb: (3)

Part (b). Repeating the calculations in Part (a) for the small trainer, we have

.Favg/c C Ft D mt .vx2 vx1/t2 t1 ;

(4)

where Ft and mt are the thrust and the mass of the trainer, respectively. Observe that the average acceleration

of the trainer is given by

.aavg/t D vx2 vx1t2 t1 : (5)

Therefore, using Eq. (4), we have

.aavg/t D.Favg/c C Ft

mt; (6)

from which we compute

.aavg/t D 354 ft=s2 D 11:0g;

where we have used the following numerical data: .Favg/c D 137;100 lb (see Eq (2)), Ft D 5850 lb, andmt D .13;000 lb/=.32:2 ft=s2/ D 403:7 slug.

As far as answering what would happen to the trainers pilot is concerned, we can speculate that the pilot

would likely black out.

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Solution to 5.8

Let t1 and t2 be the time instants at which the plane starts and reaches the takeoff

speed, respectively. Then, using the linear impulsemomentum principle, we have

that the average force acting on the plane is

EFavg D Ep.t2/ Ep.t1/t2 t1 ) FT D

m.vx2 vx1/t2 t1 ;

(7)

where, referring to the FBD shown, we have used the impulsemomentum principle only in the horizontal

direction.

Letting t1 D 0 and recalling that vx1 D 0, solving Eq. (7) for t2, we have

t2 D mvx2FT

D 10:6 s;

where we have used the following numerical data: m D .45;000 lb/=.32:2 ft=s2/ D 1398 slug, vx2 D165 mph D 242:0 ft=s, and FT D 32;000 lb.

Next, using the work-energy principle to find the total distance d needed to take off, we have

T1 C U1-2 D T2; (8)

where T1 D 0, U1-2 D FTd, and T2 D 12mv22 , so that we have

FTd D 12mv22 ) d Dmv222FT

) d D 1280 ft,

where we have used the following numerical data: m D .45;000 lb/=.32:2 ft=s2/ D 1398 slug, v2 D vx2 D165 mph D 242:0 ft=s, and FT D 32;000 lb.

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Dynamics 1e 577

Problem 5.9

A 60 ton railcar and its cargo, a 27 ton trailer, are moving to the right at 4 mph

when they come into contact with a bumper that is able to bring the system to

a stop in 0:78 s. Determine the magnitude of the average force exerted on the

railcar by the bumper.

Solution

Let the subscripts 1 and 2 denote the time instants when the train first comes into

contact with the spring and when the train first comes to a stop, respectively. Then,

using the relation between the change in linear momentum of a particle over a time

interval of duration t2 t1 and the average force over the same time interval, we havethat the magnitude of the average force is

EFavg D mEv2 mEv1

t2 t1 ) Favg Dmv1

t2 t1 D40;600 lb,

where we have used the fact that v2 D 0 and we have used the following numerical data: m D .60:0 ton C27:0 ton/.2000 lb=ton/.32:2 ft=s2/1 D 5404 slug, v1 D 4:00 mph D 5:867 ft=s, and t2 t1 D 0:780 s.

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Problem 5.10

In a simple force-controlled experiment, two curling stones A and B are made to slide

over a sheet of ice. Initially, A and B are at rest on the start line. Then they are acted

upon by identical and constant forces EF, which continually push A and B all the wayto the finish line. Let EpAFL and EpBFL denote the momentum ofA and B at the finishline, respectively, assuming that the forces EF are the only nonnegligible forces actingin the plane of motion. IfmA < mB , which of the following statements is true?

(a) EpAFL < EpBFL .

(b) EpAFL D EpBFL .

(c) EpAFL > EpBFL .

(d) There is not enough information given to make a comparison between EpAFL and

EpBFL

.

Solution

Since mB > mA, B will take more time to cross the finish line. From the definition of impulse for a constant

force Zt2t1

EF . t / d t D Ep .t2/ Ep .t1/ ) EF .t2 t1/ D Ep .t2/ Ep .t1/ ;

we see that a constant force acting over a longer time interval will cause B to have more momentum than A

at the finish line: (i) EpAFL < EpBFL is true.

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Dynamics 1e 579

Problem 5.11

A 30;000 lb airplane is flying on a horizontal trajectory with a speed v0 D650 mph when, at point A, it maneuvers so that at point B it is set on a steady

climb with D 40 and a speed of600 mph. Assuming that the change in massof the plane between A and B is negligible, determine the impulse that had tobe exerted on the plane in going from A to B .

NAVY

3

3

Solution

Referring to the figure on the right, we can describe the velocitities if the plane at A and B ,

respectively, as EvA D v0 O{ and EvB D vB.cos O{ C sin O| /, v0 is the speed at A and vB is thespeed at B . Then, applying the linear impulsemomentum principle, we have that the total

impulse exerted on the plane between A and B is

ZtB

tA

EFT dt D m.EvB EvA/;

where m is the mass of the plane and EFT is the total external force acting on the plane. Hence, using thecomponent system shown, we haveZtB

tA

EFT dt D Wg

.vB cos v0/ O{ C vB sin O|

;

so that our numerical answer is

ZtBtA

EFT dt D .260103 O{ C 527103 O| / lbs ;

where we have used the following numerical data:

W D 30;000 lb;g D 32:2 ft=s2;

vB D .600 mph/.5280 ft=mi/.3600 s=h/1 D 880:0 ft=s; D 40:0;

v0 D .650 mph/.5280 ft=mi/.3600 s=h/1 D 953:3 ft=s:

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Problem 5.12

A 1600 kg car, when on a rectilinear and horizontal stretch of road, can go from

rest to 100 km=h in 5:5 s.

(a) Assuming that the car travels on such a road, estimate the average value of

the force acting on the car for the car to match the expected performance.

(b) Recalling that the force propelling a car is caused by the friction between

the driving wheels and the road, and again assuming that the car travels on a

rectilinear and horizontal stretch of road, estimate the average value of the

friction force acting on the car for the car to match the expected performance.

Also estimate the coefficient of friction required to generate such a force.

Solution

Part (a). Let the subscripts 1 and 2 denote the time instants when the car starts from restand reaches the speed of100 km=h, respectively. Then, using the relation between average

force and changes in linear momentum over a given time interval, we have

EFavg D mEv2 mEv1t2 t1 :

(1)

Using the component system shown on the right and since the car moves along a straight line, we have

Ev1 D E0 and Ev2 D v2 O{; (2)where v2 D 100 km=h. Therefore, substituting Eqs. (2) into Eq. (1) and evaluating the magnitude of theresult, we have

EFavg D 8081 N (3)where we have used the following numerical data: m D 1600 kg, v2 D .100 km=h/.3600 s=hr/1 D27:78 m=s, and t2 t1 D 5:5 s. Expressing the above result using three significant figures, the answer toPart (a) of the problem is ad follows: EFavg D 8080 N:Part (b). The only force acting on the car is the force generated by friction between the wheels and the

road. Thus, the magnitude of the average force needed to accelerate the car from rest to 100 km=h is that in

Eq. (3). With this in mind, from the definition of static friction (since we are assuming no slip) we have that

jFfj sjNj; (4)where N is the normal reaction force between the car and the ground and Ff D EFavg. Since the car doesnot move in the vertical direction, the normal force N will have to equilibrate the cars weight, i.e., N D mg.Substituting this result into Eq. (4), we have

s EFavg

mgD 0:515;

where we have used the following numerical data: EFavg D 8081 N (see Eq. (3)), m D 1600 kg, and

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Dynamics 1e 581

Problem 5.13

A 1600 kg car, when on a rectilinear and horizontal stretch of road and when

the tires do not slip, can go from rest to 100 km=h in 5:5 s. Assuming that

the car travels on a straight stretch of road with a 40% slope and that no slip

occurs, determine how long it would take to attain a speed of 100 km=h if thecar were propelled by the same maximum average force that can be generated

on a horizontal road.

Solution

We start the solution of this problem by determining the average force that acts on the car when it travels on a

horizontal surface going from zero to 100 km=s in 5:5 s. Let the subscripts 1 and 2 denote the time instants

when the car starts from rest and reaches the speed of100 km=h, respectively. Then, letting vh1 and vh2 be

the horizontal components of velocity corresponding to the the time instants t1 and t2 respectively, using the

relation between the notion of average force and the change in momentum (over a given time interval), we

have

Favg D mvh2 mvh1t2 t1 ) Favg D 8081

N; (1)

where we have used the following numerical data: m D 1600 kg, vh2 D .100 km=h/.3600 s=hr/1 D27:78 m=s, vh1D 0, and t2 t1 D 5:50 s.

Now, we turn the analysis of the motion of the car along the incline. Referring to

the FBD shown on the right, as indicated by the problem statement, we assume that the

car is subject to a force equal to favg acting in the direction of motion. Hence, applying

Newtons second law in the x direction, we have

XFx W mg sin C Favg D max; ) ax D

Favg

m g sin :

Next, since ax is constant, we can apply the following constant acceleration equation: vx2 D vx1Cax.t2t1/.Let the initial time t1 D 0 so that vx1 D 0. Then, letting vx2 D 100 km=s, we can solve for t2, namely thetime needed to achieve vx2. Specifically, we have

vx2 D axt2 ) vx2ax

D t2 ) t2 D mvx2Favg mg sin D 19:7 s,

where we have used the following numerical data: m D 1600 kg, vx2 D .100 km=h/.3600 s=hr/1 D27:78 m=s, Favg D 8081 N (see Eq. (1)), g D 9:81 m=s2, D tan1.40=100/ D 21:80.

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Problem 5.14

A 518

oz baseball traveling at 80 mph rebounds off a bat with a speed of160 mph.

The ball is in contact with the bat for roughly 103 s. The incoming velocity of

the ball is horizontal, and the outgoing trajectory forms an angle D 31 anglewith respect to the incoming trajectory.

(a) Determine the impulse provided to the baseball by the bat.

(b) Determine the average force exerted by the bat on the ball.

(c) Determine how much the angle would change (with respect to 31) if we

were to neglect the effects of the force of gravity on the ball.

Solution

Part (a). let the subscripts 1 and 2 denote the time instants immediately before and afterthe bat his the ball, respectively. Hence, v1 and v2 are the pre and postimpact speeds

and the corresponding pre and, using the component system on the right, the postimpact

velocities can be written as

Ev1 D v1 O{ D .117:3 ft=s/ O{; (1)Ev2 D v2 .cos O{ C sin O| / D .201:2 O{ C 120:9 O| / ft=s; (2)

where we have used the following numerical data: v1 D 80 mph D 117:3 ft=s, v2 D 160 mph D 234:7 ft=s,and D 31. The impulse on the ball is given by the change in momentum. That is, using the FBD shown onthe right,

ImpulseD Z

tf

0 EFb mg O| dt D mEv2 mEv1 D 3:168 O{ C 1:203 O| lbs; (3)where in addition to the results in Eqs. (1) and (2), we have used the following numerical data m D.518

oz/.16:0 oz=lb/1.32:2 ft=s2/1 D 9:948103 slug. Expressing the answer to Part (a) using threesignificant digits, we have

Impulse D .3:17 O{ C 1:20 O| / lbs :

Part (b). Using the first of Eqs. (3), we can write

Impulse DZtf0

EFb mg O|

dt D

Ztf0

EFb dt Ztf0

mg O| dt D tf. EFb/avg Ztf0

mg O| dt (4)

) EFbavg D 1tf Impulse C Ztf

0 mg O| dt D 1tf Impulse C mgtf O| : (5)Using the value of the impulse in Eq. (3), and recalling that mg D 51

8oz D 0:3203 lb, and tf D 103 s, we

then have EFbavg D 3168 O{ C 1203 O| lb; (6)which, when expressed using three significant digits, becomes

EFbavg D .3170 O{ C 1200 O| / lb :August 10, 2009


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Dynamics 1e 583

Part (c). To answer the final question, we can again use Eq. (3), solve it for Ev2, and neglect mg to find

Ev2 D Ev1 C 1m

Ztf0

EFb dt: (7)

Now, taking advantage of the average force from Eq. (6), we haveZtf0

EFb dt D tf

EFb

avgD .3168 O{ C 1203 O| / lb.103 s/ D .3:168 O{ C 1:203 O| / lbs; (8)

so that,

Ev2 D .201:2 O{ C 120:9 O| / ft=s: (9)The above value for Ev2 is identical, to four significant figures, to that in Eq. (2). Therefore, within the accuracyof our calculation, the resulting value of would be unchanged with respect to the given value of31.

Within the accuracy of our calculation (4 significant figures), remains equal to 31.

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Problem 5.15

In an unfortunate incident, a 2:75 kg laptop computer is dropped onto the floor from a height

of1 m. Assuming that the laptop starts from rest, that it rebounds off the floor up to a height

of5 cm, and that the contact with the floor lasts 103 s, determine the impulse provided by

the floor to the laptop and the average acceleration to which the laptop is subjected when incontact with the floor (express this result in terms of g, the acceleration of gravity).

Solution

Modeling the laptop as a particle of mass m, to determine the impulse provided by the floor

to the laptop we need to determine the change in velocity of the laptop due to its collision

with the ground. The velocity with which the laptop hits the ground can be related to the

height from which the laptop is dropped using the work-energy principle. Let and be the

positions of the laptop when it is first dropped and when it reached the ground, respectively.

Referring to the FBD shown, assuming that the laptop is only subject to gravity, and setting the datum from

gravity at the ground, we have

T1 C V1 D T2 C V2; (1)where

T1 D 0; V1 D mgh1; T2 D 12v22; V2 D 0: (2)Substituting Eqs. (2) into Eq. (1) and solving for v2, we have

v2 Dp

2gh1 D 4:429 m=s; (3)where we have used the following numerical data: g D 9:81 m=s2 and h1 D 1:00 m. Similarly, letting and be the position of the laptop right after impact and when reaching the maximum rebound height,

respectively, then applying the work-energy principle again, we have that have

v3D p2gh4 D 0:9905 m; (4)

where we have used the following numerical data: g D 9:81 m=s2 and h4 D 5:00 cm D 0:0500 m. Hence,using the component system shown on the right, the velocity of the laptop right before impact and right after

impact are

Ev2 D .4:429 m=s/ O| and Ev3 D .0:9905 m/ O| : (5)Now, applying the impulsemomentum principle, we can say that the impulse provided by the floor is

Impulse D Ep3 Ep2 D mEv3 mEv2:Recalling that m D 2:75 kg, and using the numerical results in Eqs. (5), we have

Impulse

D.14:9

O| / N

s :

Next, applying the formula for the average acceleration, we have

Eaavg D Ev3 Ev2t2 t1 ;

(6)

which, recalling that t2 t1 D 1103 s and using again the numerical results in Eqs. (5), gives

Eaavg D .5420 O| / m=s D 552g O| :

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Dynamics 1e 585

Problem 5.16

A train is moving at a constant speed vt relative to the ground, when a person who initially at rest (relative

to the train) starts running and gains a speed v0 (relative to the train) after a time interval t . Had the

person started from rest on the ground (as opposed to the moving train), would the magnitude of the total

impulse exerted on the person during t be smaller than, equal to, or larger than the impulse needed tocause the same change in relative velocity in the same amount of time on the moving train? Assume that

the person always moves in the direction of motion of the train.

Solution

The two impulses in question are the same. This is because the train is moving at a constant velocity with

respect to the ground. Hence, if the ground can be chosen as an inertial reference frame then the train can

also be chosen as an inertial reference frame. Therefore a given change in velocity in a given amount of time,

either on the train or on the ground, will require the same impulse.

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Problem 5.17

A train is decelerating at a constant rate, when a person who initially at rest (relative to the train) starts

running and gains a speed v0 (again relative to the train) after a time interval t . Had the person started

from rest on the ground (as opposed to the moving train), would the magnitude of the total impulse exerted

on the person during t be smaller than, equal to, or larger than the impulse needed to cause the samechange in velocity in the same amount of time on the moving train? Assume that the person always moves

in the direction of motion of the train and that the train does not reverse its motion during the time interval

t .

Solution

The magnitude of the total impulse in question would be greater. Suppose the ground is chosen as an inertial

frame. Then the train cannot be taken as an inertial frame. The use of the impulse-momentum principle

requires that velocity changes be measured by an inertial observer. One such observer would perceive thechange in speed of the person on the train to be smaller than it would have been on the ground due to the fact

that the train is decelerating. Therefore, for a given change in speed over a given time interval, a person on a

decelerating train requires a smaller impulse than a person on the ground.

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Dynamics 1e 587

Problem 5.18

A car of mass m collides head-on with a truck of mass 50m. What is the ratio between the magnitude of

the impulse provided by the car to the truck and the magnitude of the impulse provided by the truck to the

car during the collision?

Solution

The ratio of the impulse imparted onto the truck by the car is necessarily equal and opposite to the impulse

provided by the truck. This result is a direct consequence of Newtons third law.

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Problems 5.19 through 5.21

These problems are an introduction to perfectly plastic impact (which we will

cover in Section 5.2). In each problem, model the vehicles A and C as particles

and treat the swarm of bugs B hitting the vehicles as a single particle. Also

assume that the swarm of bugs sticks perfectly to each vehicle (this is what ismeant by a perfectly plastic impact).

Problem 5.19 An 80;000 lb semitruckA (the maximum weight allowed in

many states) is traveling at 70 mph when it encounters a swarm of mosquitoes

B . The swarm is traveling at 1 mph in the opposite direction of the truck.

Assuming that the entire swarm sticks to the truck, the mass of each mosquito

is 2 mg, and that all of these mosquitoes do not significantly damage the truck,

how many mosquitoes must have hit the truck if it slows down by 2 mph on

impact? If the same number of mosquitoes hit a small SUV C weighing 3000 lb,

by how much will the SUV slow down?

Problem 5.20 An 80;000 lb semitruckA (the maximum weight allowed inmany states) is traveling at 70 mph when it encounters a swarm of worker bees


Assuming that the entire swarm sticks to the truck, the mass of each bee is

0:1 g, and that all of these bees do not significantly damage the truck, how many

bees must have hit the truck if it slows down by 2 mph on impact? If the same

number of bees hit a small SUV C weighing 3000 lb, by how much will the

SUV slow down?

Problem 5.21 An 80;000 lb semitruckA (the maximum weight allowed in

many states) is traveling at 70 mph when it encounters a swarm of dragonflies


Assuming that the entire swarm sticks to the truck, the mass of each dragonfly

is 0:25 g, and that all of these dragonflies do not significantly damage the truck,

how many dragonflies must have hit the truck if it slows down by 2 mph on

impact? If the same number of dragonflies hit a small SUV C weighing 3000 lb,

by how much will the SUV slow down?

Solution to 5.19

Referring to the FBD shown, the systems linear momentum is conserved. Using the subscripts

1 and 2 to denote the pre- and postimpact conditions, respectively, in the x direction we have

mA.vAx/1 C nmm.vBx /1 D mA.vAx/2 C nmm.vBx /2 D .mA C nmm/vx2; (1)where mm is the mass of a single mosquito, n is the number of mosquitoes, and where we have set

.vAx/2 D .vBx /2 D vx2 since the mosquitoes will move with the truck after impact. Solving Eq. (1) for nand recalling that vx2 D .vAx/1 2 mph, we have

n D mA.vAx/1 vx2mmvx2 .vBx /1

D 5:317108mosquitos; (2)

where we have used the following numerical data: mA D .80;000 lb/=.32:2 ft=s2/ D 2484 slug, .vAx/1 D70:0 mph D 102:7 ft=s, vx2 D 68:0 mph D 99:73 ft=s, mm D .2:00 106 kg/.14:59 kg=slug/1 D

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Dynamics 1e 589

1:371107 slug, and .vBx /1 D 1:00 mph D 1:467 ft=s. Expressing the above result to three significantfigures, we have

n D 5:32108mosquitos: (3)

To determine the effect of the mosquitoes on the SUV, we solve Eq. (1) for vx2 after having replaced mA

with mC. This gives,

vx2 D mC.vCx /1 C nmm.vBx /1mCC nmm

D 56:98 ft=s D 38:85 mph; (4)

where .vCx /1 D .vAx/1 and where we have used the following numerical data: mC D 3000 lb32:2 ft=s2 D93:17 slug, .vCx /1 D 70:0 mph D 102:7 ft=s, n D 5:317108mosquitos (see Eq. (2)), mm D 2:00 mg D.106 kg/.14:59 kg=slug/1 D 1:371107 slug, and .vBx /1 D 1:00 mph D 1:467 ft=s. Thus the SUVwill slow down by .vCx/1 vx2, i.e.,

Slowdown D 45:7 ft=s D 31:2 mph:

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Solution to 5.20



mA.vAx/1

Cnmb.vBx /1

DmA.vAx/2

Cnmb.vBx /2

D.mA

Cnmb/vx2; (5)

where mb is the mass of a single bee, n is the number of bees, and where we have set .vAx/2 D .vBx /2 D vx2since the bees will move with the truck after impact. Solving Eq. (5) for n and recalling that vx2 D.vAx/1 2 mph, we have

n D mA.vAx/1 vx2mbvx2 .vBx /1

D 9:174106 bees; (6)

where we have used the following numerical data: mA D .80;000 lb/=.32:2 ft=s2/ D 2484 slug, .vAx/1 D70:0 mph D 102:7 ft=s, vx2 D 68:0 mph D 99:73 ft=s, mb D 0:100 g D .10:04 kg/.14:59 kg=slug/1 D6:854106 slug, and .vBx /1 D 12:0 mph D 17:60 ft=s. Expressing the above result to three significantfigures, we have

n D 9:170106

bees: (7)

To determine the effect of the bees on the SUV, we solve Eq. (5) for vx2 after having replaced mA with

mC. This gives,

vx2 D mC.vCx/1 C nmb.vBx /1mCC nmb

D 54:23 ft=s D 36:98 mph;

where .vCx /1 D .vAx/1 and where we have used the following numerical data: mC D 3000 lb32:2 ft=s2 D93:17 slug, .vCx/1 D 70:0 mph D 102:7 ft=s, n D 9:174 106 bees (see Eq. (6)), mb D 0:100 g D.103 kg/.14:59 kg=slug/1 D 6:854106 slug, and .vBx /1 D 12:0 mph D 17:60 ft=s. Thus the SUVwill slow down by .vCx/1 vx2, i.e.,


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Dynamics 1e 591

Solution to 5.21



mA.vAx/1

Cnmd.vBx /1

DmA.vAx/2

Cnmd.vBx /2

D.mA

Cnmd/vx2; (8)

where md is the mass of a single dragonfly, n is the number of dragonflies, and where we have set .vAx/2 D.vBx /2 D vx2 since the dragonflies will move with the truck after impact. Solving Eq. (5) for n and recallingthat vx2 D .vAx/1 2 mph, we have

n D mA.vAx/1 vx2mdvx2 .vBx /1

D 2:906106 dragonflies; (9)

where we have used the following numerical data: mA D .80;000 lb/=.32:2 ft=s2/ D 2484 slug, .vAx/1 D70:0 mph D 102:7 ft=s, vx2 D 68:0 mph D 99:73 ft=s, md D 0:250 g D .103 kg/.14:59 kg=slug/1 D17:14106 slug, and .vBx /1 D 33:0 mph D 48:40 ft=s. Expressing the above result to three significantfigures, we have

n D 2:91106 dragonflies: (10)To determine the effect of the dragonflies on the SUV, we solve Eq. (5) for vx2 after having replaced mA

with mC. This gives,

vx2 D mC.vCx /1 C nmd.vBx /1mCC nmd

D 50:06 ft=s D 34:13 mph;

where .vCx /1 D .vAx/1 and where we have used the following numerical data: mC D 3000 lb32:2 ft=s2 D93:17 slug, .vCx /1 D 70:0 mph D 102:7 ft=s, n D 2:906106 dragonflies (see Eq. (9)), md D 0:250 g D.103 kg/.14:59 kg=slug/1

D17:14

106 slug, and .vBx /1

D 33:0 mph

D 48:40 ft=s. Thus the SUV

will slow down by .vCx/1 vx2, i.e.,


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Problem 5.22

Solve Example 5.4 by directly applying Eq. (5.14), using the same assumptions

made in that solution. Note that, unlike Example 5.4, the velocity of the person

relative to the platform does not appear in the solution.

Solution

Equation (5.14) is a direct consequence of the impulsemomentum principle for

a system, i.e.,

EF D ddt

mEvG

; (1)

where EvG is the velocity of the systems center of mass G. Observe that EFx D 0.Hence, using Eq. (1), we have

0 D mvGx ) C D mvGx ; (2)

where C is a constant. Since the system is initially stationary, we have C D 0. In turn, this implies that

0 D mvGx D m xGdt

) xG D constant: (3)

If the position of the G does not change with time, then letting the subscripts 1 and 2 denote the initial time

instant and a generic subsequent time instant, respectively, we must have

xG1

DxG2: (4)

Applying the definition of center of mass to the platform-person

system, we have that, at any instant,

xG D xAmA C xBmBmA C mB :

(5)

Substituting Eq. (5) into Eq. (4), and using the coordinate system

shown to the right, we have

Lfpmp C 0mfpmp C mfp D

xpmp C xpmfpmp C mfp :

(6)

Letting the time t2 be the instant when the person reaches the right end of the platform, i.e., letting xp D xfpin Eq. (6) and then solving for .xfp/2, we have

.xfp/2 D Lfpmpmp C mfp ;

which is exactly the same result expressed by Eq. (10) in Example 5.4.

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Dynamics 1e 593

Problems 5.23 through 5.25

Two persons A and B weighing 140 and 180 lb, respectively, jump off a floating

platform (in the same direction) with a velocity relative to the platform that

is completely horizontal and with magnitude v0 D 6 ft=s for both A and B .The floating platform weighs 800 lb. Assume that A, B , and the platform areinitially at rest.

Problem 5.23 Neglecting the water resistance to the horizontal motion of the

platform, determine the speed of the platform after A and B jump at the same

time.


platform, and knowing that B jumps first, determine the speed of the platform

after both A and B have jumped.


platform, and knowing that A jumps first, determine the speed of the platformafter both A and B have jumped.

Solution to 5.23

Referring to the FBD on the right, there are no external forces acting on the system.

Therefore, the momentum is conserved in the x direction. Let the subscripts 1 and

2 represent the time instants immediately before and after the jump, respectively.

Then we must have

mA.vAx/1 C mB.vBx /1 C mP.vPx/1D mA.vAx/2 C mB.vBx /2 C mP.vPx/2: (1)

All masses are initially at rest and .vAx/2 D .vBx /2. Because the velocity v0 is a relative velocity we musthave

v0 D .vAx/2 .vPx/2 ) .vAx/2 D .vPx/2 v0:Consequently, Eq. (1) becomes

0 D .vPx/2 v0.mA C mB/ C mP.vPx/2;which can be solved for .vPx/2 to obtain

.vPx/2 D.mA

CmB/v0

mA C mB C mP D.WA

CWB/v0

WA C WB C WP ; (2)where we have multiplied both the numerator and denominator by the acceleration due to gravity g to obtain

the last expression. Observing that the platform moves only in the x direction, then vP2 D j.vPx/2j so thatwe have

vP2 D 1:71 ft=s;where we have used the following numerical data: WA D 140 lb, WB D 180 lb, WP D 800 lb, andv0 D 6:00 ft=s.

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Solution to 5.24

There are no external forces acting on the system. Therefore, the momentum is

conserved in the x direction. Let the subscripts 1, 2, 3, and 4 represent the time

instants immediately before B jumps, right after B jumps, right before A jumps,

and right after A jumps, respectively. Then we must have


All masses are at rest at time t1 and .vAx/2 D .vPx/2. Because v0 is a relative velocity, we have

.vBx /2 .vPx/2 D v0 ) .vBx /2 D .vPx/2 v0:

Consequently, Eq. (3) becomes

0 D mB.vPx/2 v0C .mA C mP/.vPx/2;which can be solved for .vPx/2 to obtain

.vPx/2 DmBv0

mA C mB C mPD .vPx/3 D .vAx/3:

Since momentum is conserved in the x direction between time instants t3 and t4, we have

mA.vAx/3 C mP.vPx/3 D mA.vAx/4 C mP.vPx/4: (4)

Again, because v0 is a relative velocity, we must write

.vAx/4 .vPx/4 D v0 ) .vAx/4 D .vPx/4 v0:

Then, Eq. (4) can be rewritten as

.mA C mP/mBv0mA C mB C mP

D mA

.vPx/4 v0C mP.vPx/4;

which can be solved for .vPx/4 to obtain

.vPx/4 D mAv0mA C mP

C mBv0mA C mB C mP

D WAv0WA C WP

C WBv0WA C WB C WP

; (5)

where we have multiplied both the numerator and denominator of each fraction by the acceleration due to

gravity g to obtain the last expression. Observing that the platform moves only in the x direction, then

vP4 D j.vPx/4j so that we havevP4 D 1:86 ft=s;

where we have used the following numerical data: WA D 140 lb, WB D 180 lb, WP D 800 lb, andv0 D 6:00 ft=s.

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Dynamics 1e 595

Solution to 5.25

There are no external forces acting on the system. Therefore, the momentum is

conserved in the x direction. Let the subscripts 1, 2, 3, and 4 represent the time

instants immediately before A jumps, right after A jumps, right before B jumps,

and right after B jumps, respectively. Then we must have


All masses are at rest at time t1 and .vBx /2 D .vPx/2. Because v0 is a relative velocity, we have

.vAx/2 .vPx/2 D v0 ) .vAx/2 D .vPx/2 v0:

Consequently, Eq. (6) becomes

0 D mA.vPx/2 v0C .mB C mP/.vPx/2;which can be solved for .vPx/2 to obtain

.vPx/2 DmAv0

mA C mB C mPD .vPx/3 D .vBx /3:

Since momentum is conserved in the x direction between time instants t3 and t4, we have

mB.vBx /3 C mP.vPx/3 D mB.vBx /4 C mP.vPx/4: (7)

Again, because v0 is a relative velocity, we must write

.vBx /4 .vPx/4 D v0 ) .vBx /4 D .vPx/4 v0:

Then, Eq. (7) becomes

.mB C mP/mAv0mA C mB C mP

D mB

.vPx/4 v0C mP.vPx/4;

which can be solved for .vPx/4 to obtain

.vPx/4 D mBv0mB C mP

C mAv0mA C mB C mP

D WBv0WB C WP

C WAv0WA C WB C WP

; (8)

where we have multiplied both the numerator and denominator by the acceleration due to gravity g to obtain

the last expression. Observing that the platform moves only in the x direction, then vP4 D j.vPx/4j so thatwe have

vP4 D 1:85 ft=s:

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Problem 5.26

At the instant shown a group of three space-junk fragments with masses m1 D7:45 kg, m2 D 3:22 kg, and m3 D 8:45 kg are traveling as shown with v1 D7701 m=s, v2 D 6996 m=s, and v3 D 6450 m=s. Assume that the velocity vectorsof the fragments are coplanar, that Ev1 and Ev2 are parallel, and that is measuredwith respect to a line perpendicular to the direction of both Ev1 and Ev2. Furthermore,assume that the system is isolated and that, because of gravity, the fragments will

eventually form a single body. Determine the common velocity of the fragments

after they come together if D 25.

Solution

Since the system is isolated, the velocity of the center of mass of the system is

conserved. When the objects form a single body they will travel with a velocity

equal to the initial velocity of the mass center. We write the initial velocities as

Ev1 D v1 O| ; Ev2 D v2 O| ; and v3 D v3 cos O{ v3 sin O| : (1)

The velocity of the center of mass of the system is therefore

EvG D m1Ev1 C m2Ev2 C m3Ev3m1 C m2 C m3 D

m3v3 cos O{ C .m2v2 m1v1 m3v3 sin / O|m1 C m2 C m3 :

Using the problems given we then have

EvG D .2580 O{ 3030 O| / m=s;where we have used the following numerical data , m1 D 7:45 kg, m2 D 3:22 kg, m3 D 8:45 kg, v1 D7701 m=s, v2 D 6996 m=s, v3 D 6450 m=s, and D 25.

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Dynamics 1e 597

Problem 5.27

A 180 lb man A and a 40 lb child C are at the opposite ends of a 250 lb floating

platform P with a length Lfp D 15 ft. The man, child, and platform are initiallyat rest at a distance D 1 ft from a mooring dock. The child and the man movetoward each other with the same speed v0 relative to the platform. Determinethe distance d from the mooring dock where the child and man will meet.

Assume that the resistance due to the water to the horizontal motion of the

platform is negligible.

Solution

There are no external forces in the x direction so momentum is conserved in this

direction. Since the system is initially at rest, the position of the center of mass must

remain constant. The man and child will meet in the center of the platform because

they move with equal speed relative to the platform. Let the subscripts 1 and 2 denote

the initial time and the time at which A and C meet, respectively. Then we must have,

mAxA1 C mCxC1 C mPxP1 D mAxA2C mCxC2 C mPxP2; (1)

where, referring to the coordinate system shown to the below,

xC1 D D 1 ft, mP1 D xC1C 12Lfp, xA1 D xC1CLfp, and xA2 D xC2 D xP2 D dC 12Lfp. Consequently,Eq. (1) becomes

mA.xC1 C Lfp/ C mCxC1 C mP

xC1 C 12Lfp D .mA C mCC mP/d: (2)

Multiplying Eq. (2) through by the acceleration due to gravity g and solving for d we obtain

d DWA

.xC1 C

Lfp

/C

WC

xC1 C

WPxC1 C 12LfpWA C WCC WP D 10:7 ft;

where we have used the following numerical data: WA D 180 lb, WC D 40:0 lb, WP D 250 lb, Lfp D 15:0 ft,xC1 D 1:00 ft.

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Problem 5.28

A man A, with a mass mA D 85 kg, and a child C, with a mass mC D 18 kg,are at the opposite ends of a floating platform P, with a mass mP D 150 kgand a length Lfp D 6 m. Assume that the man, child, and platform are initiallyat rest and that the resistance due to the water to the horizontal motion of theplatform is negligible. Suppose that the man and child start moving toward

each other in such a way that the platform does not move relative to the water.

Determine the distance covered by the child until meeting the man.

Solution

There are no external forces in the x direction so momentum is conserved in this

direction. Since the system is initially at rest, the center of mass does not move. Let

the subscripts 1 and 2 denote the time instants corresponding to the initial and final

positions, respectively. Since we want the platform to remain stationary we can use

it as an inertial frame of reference. Choosing a coordinate system with origin at xC1 ,as shown, we have

mAxA1 C mPxP1 C mCxC1 D mAxA2C mPxP2 C mCxC2; (1)) mAxA1 D mAxA2C mCxC2; (2)

given that xC1 D 0. When the man and child meet xA2 D xC2 and Eq. (2) becomes

mAxA1 D .mA C mC/xC2;

which, recalling that xA1 D Lfp, can be solved for xC2 to obtain

xC2 DmA

Lfp

mA C mC : (3)

Observing that the distance covered by the child is d D jxC2j, we have

d D mALfpmA C mC

D 4:95 m;

where we have used the following numerical data: mA D 85:0 kg, mC D 18:0 kg, mP D 150:0 kg, andLfp D 6:00 m.

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Recursion via a For Loop

vInitial 375.0 5280.

3600.;

Fori 1, i 281, i, Ifi 1, vFinal vInitial;;

vFinal 1

28 000.01i 13.216.

28 000.01i13.2

16.vFinal

13.2

16.3250. ;

v vFinalvInitial

26.9233

Recursion via Function Definition

vAirplane

i_

:

1

28 000.01i 13.216.

28 000.01i13.2

16.vAirplanei1

13.2

16.3250. ;

vAirplane0 375.0 5280.

3600.;

v Block$RecursionLimit Infinity, vAirplane280vAirplane026.9233

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Dynamics 1e 601

Problem 5.30

A person P on a cart on rails is receiving packages from people standing on a

stationary platform. Assume that person P and the cart have a combined weight

of 350 lb and start from rest. In addition suppose that a person PA throws a

package A weighing 60 lb, which is received by person P with a horizontalspeed vA D 4:5 ft=s. After person P has received the package from person PA,a second person PB throws a package B weighing 80 lb, which is received by

person P with a horizontal speed relative to P and in the same direction as the

velocity ofP of5:25 ft=s. Determine the final velocity of the person P and the

cart. Neglect any friction or air resistance acting on P and the cart.

Solution

The FBD of the system when P receives the package from PA shows that there are no

forces in the horizontal direction. Then, recalling that the cart starts from rest, we must

have

mA.vAx/1 D .mPC mA/.vPx/2 ) .vPx/2 D mAmPC mA

.vAx/1; (1)

where the subscripts 1 and 2 identify the states of the system before and after P receives the package from

PA, respectively, .vAx/1 D 4:5 ft=s, and where we have accounted for the fact that, after P receives thepackage, P and the package have a common velocity.

Let 3 denote the system after P receives the package from PB . The FBD of the

system in going from 2 to 3 also indicates that the linear momentum of the system in

the x direction is conserved, so that we have

.mPC mA/.vPx/2 C mB.vBx /2 D .mPC mA C mB/.vPx/3; (2)

where, based on the problem statement

.vBx /2 .vPx/2 D .vB=P/2 D 5:25 ft=s ) .vBx /2 D .vB=P/2 C .vPx/2: (3)Substituting Eq. (3) into Eq. (2) and then using the second of Eqs. (1), we have

mA.vAx/1 C mB.vB=P/2 CmAmB

mPC mA .vAx/1D .mPC mA C mB/.vPx/3; (4)

which can be rewritten as

mA.vAx/1mPC mA C mB

mPC mAC mB.vB=P/2 D .mPC mA C mB/.vPx/3: (5)

Multiplying each mass terms in Eq. (5) by the acceleration due to gravity g, i.e., replacing each mass term by

the corresponding weight term, and solving for .vPx

/3

, we have

.vPx/3 D .vPx/final D WAWPC WA

.vAx/1 C WBWPC WA C WB

.vB=P/2; (6)

which, given that the motion fo P is only in the x direction gives

.EvP/final D .1:52 ft=s/ O{ ;

where we have used the following numerical data: WA D 60:0 lb, WP D 350 lb, WB D 80:0 lb, .vAx/1 D4:50 ft=s, and .vB=P/2 D 5:25 ft=s.

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Problem 5.31

The spacecraft shown is out in space and is far enough from any other mass (e.g., planets, etc.) so as not to

be affected by any gravitational influence (i.e., the net external force on the rocket is approximately zero).

The system (i.e., the spacecraft and all its fuel) is at rest when it starts at A, and it thrusts all the way to B

along the straight line shown using internal chemical rockets (which work by ejecting the fuel mass at veryhigh speeds out the tail of the rocket). We are given that the mass of the system at A is m and that it has

ejected half of its mass in thrusting from A to B . What will be the location of the systems mass center

when the spacecraft reaches B?

Solution

The systems center of mass will not move from its initial position. the reason for this is that the rocket and

its fuel form an isolated system, that is, a system with no external forces acting on it. The linear momentumof an isolated system is conserved. Since the momentum is conserved we know that the center of mass cannot

change its velocity. Since the system was initially at rest, then the center of mass of the system will have to

remain in its initial position.

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Dynamics 1e 603

Problem 5.32

Energy storage devices that use spinning flywheels to store energy are starting

to become available. To store as much energy as possible, it is important that

the flywheel spin as fast as possible. Unfortunately, if it spins too fast, internal

stresses in the flywheel cause it to come apart catastrophically. Therefore, it isimportant to keep the speed at the edge of the flywheel below about 1000 m=s.

In addition, it is critical that the flywheel be almost perfectly balanced to avoid

the tremendous vibrations that would otherwise result. With this in mind, let the

flywheel D, whose diameter is 0:3 m, rotate at ! D 60;000 rpm. In addition,assume that the cart B is constrained to move rectilinearly along the guide

tracks. Given that the flywheel is not perfectly balanced, that the unbalanced

weight A has mass mA, and that the total mass of the flywheel D, cart B , and

electronics package E is mB , determine the following as a function , the

masses, the diameter, and the angular speed of the flywheel:

(a) the amplitude of the motion of the cart,

(b) the maximum speed achieved by the cart.

Neglect the mass of the wheels, assume that initially everything is at rest, and

assume that the unbalanced mass is at the edge of the flywheel. Finally, evaluate

your answers to Parts (a) and (b) for mA D 1 g (about the mass of a paper clip)and mB D 70 kg (the mass of the flywheel might be about 40 kg).

Solution

Part (a). The position of the unbalanced mass A attached to the flywheel is

described in polar coordinates. Expressing the unit vectors Our and Ou in termsofO{ and O| , we have

Our D cos O{ C sin O| ; (1)Ou D sin O{ C cos O| : (2)

The velocity A relative to B can now be expressed as

EvA=B D r P Ou D r Psin O{ C r Pcos O| ; (3)

where r D d=2 D 0:1500 m is the radius of the flywheel. We can determine the absolute velocities of A andB as

EvB D vB O{ and EvA D EvB C EvA=B D .vB r! sin /O{ C .r! cos / O| ;where the angular velocity P is a constant: !. Therefore, the total momentum of the system is

Ep D mAEvA C mB EvB D mA.vB r! sin / C mBvB O{ C mAr! cos O| (4)

Since there are no external forces acting the in the x direction, the component of momentum in this direction

is conserved. Thus,

.mA C mB/vB mAr! sin D C; (5)

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where C is a constant. We can express the angle as !t since the angular velocity ! is a constant. Thus, we

can rewrite Eq. (5) as

.mA C mB/ vB mAr! sin !t D C; (6)Since we know the cart starts from rest, we have

EvB.0/ D 0; ) C D 0: (7)Knowing the value of the constant in Eq. (6) from Eq. (7), we can solve for the velocity of the cart

vB D mAr!sin !t

mA C mB: (8)

Integrating the velocity of the cart with respect to time, we find that

xB D mArmA C mB

cos !t C K; (9)

where K is a constant. Recalling that the amplitude of the oscillation is the term in front of the term cos !t ,

we have

Amplitude D mArmA C mB

D 2:14106m;

where we have used the following numerical data: mA D 1 g D 1:00 103 kg, r D 0:1500 m, andmB D 70:0 kg.Part (b). The maximum speed achieved by the cart occurs when sin !t D 1 in Eq. (8). This gives

.vB/max D mAr!mA C mB

D 0:0135 m=s;

where we have used the following numerical data: mA D 1 g D 1:00 103 kg, r D 0:1500 m, ! D.60;000 rpm/.2 rad=rev/=.60 s=min/ D 6283 rad=s, and mB D 70:0 kg.

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Dynamics 1e 605

Problem 5.33

The 135 lb woman A sits atop the 90 lb cart B , both of which are initially at rest. If

the woman slides down the frictionless incline of length L D 11 ft, determine thevelocity of both the woman and the cart when she reaches the bottom of the incline.

Ignore the mass of the wheels on which the cart rolls and any friction in their bearings.The angle D 26.

Solution

Let be when the person is at the top of the incline and be when

the person is at the bottom of the incline. All forces doing work are

conservative. Choosing the datume for the potential energy of gravity at

the , then at and the kinetic and potential energies of the person

and the cart are:

T1 D 0; T2 D1

2mAv2

A2 C1

2mBv2

B2 ; V1 D mAgL sin ; V2 D 0:Hence, using the work-energy principle, we have

T1 C V1 D T2 C V2 ) 2gmAL sin D mAv2A2 C mBv2B2 : (1)

The above equation has two unknowns, namely the speeds vA2 and vB2. Now observe that the problem

requires that we find the velocities of the cart and the woman. Hence, we need to derive additional equations

to accomplish this task. Clearly, we need to take into account the slope of the incline. Second, referring to

the FBD on the right, we notice that there are no external forces acting in the x direction so that the linear

momentum of the system is conserved in the x direction, i.e.,

mAvAx1 C mBvBx1 D mAvAx2 C mBvBx2 ) 0 D mAvAx2 C mBvBx2; (2)where we have used the fact that A and B are initially at rest.

Next we need to deal with the kinematics of the problem. Using the component system shown, observing

that the cart will be moving in the positive x direction, and that the relative velocity of A with respect to be

must B must be in the direction of the unit vector Out , we have

EvB2 D vB2 O{ andEvA=B2 D vA=B2 Out D vA=B2 cos O{ vA=B2 sin O| ; (3)

where

vA=B2

is the component of the relative velocity ofA with respect to B in the direction of Out . Then,using relative kinematics, we have

EvA2 D EvA=B2 C EvB2 ) EvA2 D vB2 vA=B2 cos O{ vA=B2 sin O| : (4)The last of Eqs. (4) implies that

v2A2 DEvA22 D vB2 vA=B2 cos 2C vA=B2 sin 2 D v2B2 2vB2vA=B2 cos C vA=B22: (5)

Furthermore, the first of Eqs. (3) along with the last of Eqs. (4) imply thatvAx

2

D vB2

vA=B2

cos and

vBx2

D vB2 : (6)

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Finally, substituting the result in Eq. (5) into the last of Eqs. (1) and substituting Eqs. (6) into the last of

Eqs. (2) we have

2mAgL sin D mA

v2B2 2vB2

vA=B2

cos C vA=B22C mBv2B2 ; (7)0 D mAvB2 vA=B2 cos C mBvB2 ; (8)

which is a system of two equations in the two unknowns

vA=B2

and vB2 . We can solve this system by first

solving Eq. (8) with respect to

vA=B2

. This gives

vA=B

2

D mA C mBmA cos

vB2 : (9)

Substituting the result in Eq. (9) into Eq. (7) and simplifying, we have

2mAgL sin D .mA C mB/.mA C mB mAcos2 /

mA cos2 v2B2

) vB2 D s 2g m2AL sin cos2 .mA C mB/.mA C mB mA cos2 / D 13:24 ft=s; (10)where the quantity vB2 is necessarily positive since it is a speed, and where we have used the following

numerical data: mA D .135 lb/=.32:2 ft=s2/ D 4:193 slug, mB D .90:0 lb/=.32:2 ft=s2/ D 2:795 slug, D 26:0, L D 11:0 ft, g D 32:2 ft=s2. Substituting this result into Eq. (9), we then obtain

vA=B2

D 24:55 ft=s; (11)

where we have used the appropriate values listed right below Eq. (10). Finally, substituting the results of

Eqs. (10) and (11) into the first of Eqs. (3) and the last of Eqs. (4), we have

EvA D .8:83 O{ 10:8 O| / ft=s and EvB D .13:2 ft=s/ O{ ;

where, again, we have used the appropriate values listed right below Eq. (10).

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Dynamics 1e 607

Problem 5.34

An Apollo Lunar Module A and Command and Service Module B are moving

through space far from any other bodies (so that their gravitational effects can

be ignored). When D 30, the two craft are separated using an internal linearelastic spring whose constant is k D 200;000 N=m and is precompressed 0:5 m.Noting that the mass of the Command and Service Module is about 29;000 kgand that the mass of the Lunar Module is about 15;100 kg, determine their post-

separation velocities if their common preseparation velocity is 11;000 m=s.

Solution

Let be when the spring is compressed and be post-separation. Referring

to the FBD on the right, observe that the system is isolated. Furthermore,

considering the individual FBDs of A and B , we observe that no force is

acting on either A or B in the x direction. These observations, imply that the

x components of the moment ofA and B individually are conserved and thatthe y component of the linear momentum of the system is conserved, i.e.,

mA.vAx/1 D mA.vAx/2; mA.vBx /1 D mA.vBx /2; (1)mA.vAy/1 C mB.vBy/1 D mA.vAy/2 C mB.vBy/2: (2)

Next, we observe that the system is conservative so that the work-energy

principle gives:

T1 C V1 D T2 C V2; (3)where

V1 D 12k21 ; T1 D 12mAv2A1 C 12mBv2B1 D 12.mA C mB/v20 ; (4)V2 D 0; T2 D 12mAv2A2 C 12mBv2B2 D 12mA.vAx/22 C .vAy/22C 12mB.vBx /22 C .vBy/22; (5)

and where the x and y components of the velocities for A and B at are:

.vAx/1 D v0 sin D .vBx /1 and .vAy/1 D v0 cos D .vBy/1: (6)Substituting the first of Eqs. (6) into Eqs. (1) and simplifying, we have

.vAx/2 D v0 sin D .vBx /2: (7)Next, substituting Eqs. (4) and (5) into Eq. (3), we have

12

.mA C mB/v20 C 12k21 D 12mA.vAx/22 C .vAy/22C

12

mB.vBx /22C .vBy/22 (8)

The, substituting the last of Eqs. (6) into Eq. (2), and substituting Eqs. (7) into Eq. (8), we obtain the following

system of two equations in the two unknowns .vAy/2 and .vBy/2:

.mA C mB/v0 cos D mA.vAy/2 C mB.vBy/2; (9)12

.mA C mB/v20 C 12k21 D 12mA

v20 sin2 C .vAy/22

C 12

mB

v20 sin2 C .vBy/22

: (10)

To solve this system, we being with solving Eq. (9) for .vBy/2 to obtain

.vBy/2 D mA C mBmB

v0 cos mAmB

.vAy/2: (11)

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Substituting the result in Eq. (11) into Eq. (10), simplifying, and rearranging, we have

12

mA

1 C mA

mB

.vAy/

22

mA

mB.mA C mB/v0 cos

.vAy/2C mA

2mBcos2 .mA C mB/v20 12k21

D 0; (12)

which, upon recognizing that the terms , , and are constants, can simply be given the form of the familiar

the second order algebraic equation

.vAy/22 .vAy/2 C D 0 ) .vAy/2 D

8

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Dynamics Solns Ch05