dynamics solns ch06

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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

3significant digits. When calculations are performed, all intermediate numerical results are reported to4significant 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 to3 significant digits.

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Dynamics 1e 755

Chapter 6 Solutions

Problem 6.1

LettingRAD 7:2 in:andRBD 4:6 in:, and assuming that the belt does not slip relative to pulleysA andB , determine the angular velocity and angular acceleration of pulleyB when pulleyArotates at340 rad=s

while accelerating at120 rad=s2.

Solution

LetQ be a point on the periphery of pulleyAand letP be a point on the periphery of pulleyB . The no slip

condition between the pulleys and the belt requires

vQD vP ) RA!AD RB!B ) !BD RARB

!A: (1)

Substituting in given data, we obtain:

E!BD 532Ok rad=s :Differentiating Eq. (1) with respect to time, we obtain

BD RARB

A;

which, upon substituting in given data gives

EBD 188Ok rad=s2 :

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Dynamics 1e 757

Problem 6.3

Letting RAD 203 mm, RBD 107 mm, RCD 165 mm, and RDD 140 mm, determine the angularacceleration of gears B, C, and D when gear A has an angular acceleration with magnitudej Aj D47 rad=s2 in the direction shown. Note that gearsB andCare mounted on the same shaft and they rotate

together as a unit.

Solution

We let Oand Hbe the centers of gears A and Brespectively. Because

the gears can not slip relative to each other, we can enforce the no

slip conditionEvP=QDE0between the contact pointsP andQ. Usingthe rigid body velocity equation for gear A we have

EvPD EvOC !AOk ErP=O ) EvPD !ARAO| :Doing the same for gear B we have

EvQD EvHC !BOk ErQ=H ) EvQD !BRBO| :Enforcing the no slip condition we must have

EvPD EvQ C EvP=Q ) EvPD EvQ ) !ARAD !BRB ) !BD RARB

!A: (1)

Differentiating Eq. (1) with respect to time we get

BD RARB

A

Plugging in known values we get

EBD 89:2Ok rad=s2:Because gearsB andC are mounted on the same shaft, we have

ECD 89:2Ok rad=s2:Next, by considering gears C andD with contact pointsF andG and proceeding as we did for gearsAand

B, we get

DD RCRD

C: (2)


EDD 105Ok rad=s2:

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758 Solutions Manual

Problem 6.4

The bevel gearsA and B have nominal radii RAD 20 mmand RBD 5 mm,respectively, and their axes of rotation are mutually perpendicular. If the angular

speed of gearAis !AD 150 rad=s, determine the angular speed of gearB .

Solution

Since the gears do not slip over their nominal circles relative to one another, we can write

!ARAD !BRB ) !BD RARB

!AD 600 rad=s.

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Dynamics 1e 759

Problem 6.5

A rotor with a fixed spin axis identified by point O is accelerated from rest with

an angular acceleration OD 0:5 rad=s2. If the rotors diameter isdD 15 ft,determine the time it takes for the points at the outer edge of the rotor to reach

a speedv0D 300 ft=s. Finally, determine the magnitude of the acceleration ofthese points when the speedv0is achieved.

Solution

Points on the outer edge of the rotor have a speed vD d2

!, where! is the angular

speed of the rotor. Letting!0be the angular speed corresponding to v0, we must have

v0D 12d!0 ) !0D 2v0

d :

If the angular acceleration is constant, andt0is the time taken to go from rest to!0, we have

O t0D !0D 2v0d

) t0D 2v0Od

.

The angular acceleration of points a distance d2

fromO is

EadD E

O E

rd=O

!2

0 Erd=O ) E

adD

OO

k 1

2dOur

2v20

d Our ) E

adD

2v20

d OurC

1

2O

dOu

;

Ead Ds

4v40d2

C 2Od

2

4 D 12;000 ft=s2:

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Problem 6.6

Do pointsA andB on the surface of the bevel gear (bg), which rotates with angular velocity !bg, move

relative to each other? At what rate does the distance betweenAand B change?

Solution

PointsA and B move relative to one another in the sense that the velocity ofAis different from the velocity

ofB . The difference between the velocity ofAand the velocity ofB arises from the fact that pointsAand B

are at different distances from the axis of rotation. As far as the distance between AandB is concerned, if

we model the gear as a rigid body, then that distance must remain constant.

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Problem 6.8

The velocities of pointsA andB on a disk, which is undergoing planar motion,

are such that jEvAj D j EvB j. Is it possible for the disk to be rotating about a fixedaxis going through the center of the disk atO ? Explain.

Solution

The answer is no. The reason is that if the disk were spinning about a fixed axis perpendicular to the plane of

the figure, the velocity ofB would have been perpendicular to the segment OB. However, the figure shows

that the velocity vector ofB is not perpendicular to OB and therefore the motion of the disk cannot be a

rotation about a fixed axis perpendicular to the plane of the figure.

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Dynamics 1e 763

Problem 6.9

The velocity of a pointAand the acceleration of a pointB on a disk undergoing

planar motion are shown. Is it possible for the disk to be rotating about a fixed

axis going through the center of the disk atO ? Explain.

Solution

Yes, it is possible that the motion of the disk is a fixed-axis rotation about O . The velocity vector of point

Ais perpendicular to the segment OA, which does not guarantee that the motion of the disk is a fixed-axis

rotation, but it does not exclude it either.

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Problem 6.10

Assuming that the disk shown is rotating about a fixed axis going through

its center at O , determine whether the disks angular velocity is constant,

increasing, or decreasing.

Solution

The angular velocity of the disk is decreasing. In fact, if the angular velocity were constant, then the

acceleration of pointB would be directed toward the centerO . Since the figure indicates that this is not the

case, we know that the angular velocity of the disk must either be increasing or decreasing. With this in mind,

notice that the the vectorEaB has a component that is opposite to the vectorEvBD E!O ErB=O . This in turnindicates that the speed of point B is decreasing and therefore that the spin rate of the disk as a whole isslowing down.

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Dynamics 1e 765

Problem 6.11

The sprinkler shown consists of a pipe AB mounted on a hollow vertical shaft. The water comes in the

horizontal pipe atO and goes out the nozzles atA and B , causing the pipe to rotate. LettingdD 7 in:,determine the angular velocity of the sprinkler!s , and jEaB j, the magnitude of the acceleration ofB , ifBis moving with a constant speedvBD 20 ft=s. Assume that the sprinkler does not roll on the ground.

Solution

Because the arms of the sprinkler are modeled as rigid bodies, weknow that

EvBD !SOk ErB=O :Defining the position vector fromO toB as shown to the right, we

also have

EvBD !SOk .dOuR C hOu/ D d!SOu ) d!SD vB :

Plugging in known values and solving, we have

E!S

D

vB

d D34:3

Ourad=s:

SinceO is not in the plane of motion of B, we have

EaBD E!S . E!S ErB=O/ D !2SdOur ) jEaB j D v2B

d :

Plugging in known values we have

jEaB j D 686 ft=s2:

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Problem 6.12

In a carnival ride, two gondolas spin in opposite directions about a fixed axis. If` D 4 m, determine themaximum constant angular speed of the gondolas if the magnitude of the acceleration of pointAis not to

exceed2:5g.

Solution

Under the stated conditions, pointAwould be moving in uniform circular motion. Therefore, we would haveEaA D !2OA` 2:5g ) !OAp2:5g=`:Using the given numerical value of`, we therefore have

.!OA/maxD 2:48 rad=s:

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Dynamics 1e 767

Problem 6.13

The tractor shown is stuck with its right track off the ground, and therefore the track is able to move without

causing the tractor to move. Letting the radius of sprocketAbe dD 2:5 ftand the radius of sprocketB be` D 2 ft, determine the angular speed of wheelB if the sprocketA is rotating at1 rpm.

Solution

Because the track does not slip relative to the sprocket Awe must have

!Ad

2D !B `

2 ) !BD !Ad

`:


!BD 1:25 rpm:

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Problem 6.14

A battering ram is suspended in its frame via bars AD and B C, which are identical and pinned at their

endpoints. At the instant shown, point Eon the ram moves with a speed v0D 15 m=s. Letting HD 1:75 mandD 20, determine the magnitude of the angular velocity of the ram at this instant.

Solution

Since barsAD and B Care identical and parallel to one another, we must conclude that the velocities of

points A and B are equal to one another. Since pointA and B are coplanar and distinct, then we mustconclude that the ram is simply translating, which implies that

j E!ramj D 0:

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Dynamics 1e 769

Problem 6.15

A battering ram is suspended in its frame via bars OAand OB, which are

pinned atO . At the instant shown, pointG on the ram is moving to the right

with a speed v0D 15 m=s. Letting HD 1:75 m, determine the angular velocityof the ram at this instant.

Solution

The ram and its supportsOAand OB form a single rigid body,

therefore we must have

EvGD !ramOk EvG=OD !Ok .H /O|D !ramHO{:

We are told that jEvG j D v0;which means that

!ramHD v0:


E!ramD v0H

OkD 8:57Ok rad=s :

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Problems 6.16 and 6.17

The hammer of a Charpy impact toughness test machine has the geometry

shown, where G, is the mass center of the hammer head. Use Eqs. (6.8)

and (6.13) and write your answers in terms of the component system shown.

Problem 6.16 Determine the velocity and acceleration of G, assuming

` D 500 mm, h D 65 mm, d D 25 mm, P D 5:98 rad=s, and R D8:06 rad=s2.Problem 6.17 Determine the velocity and acceleration ofG as a function of

the geometric parameters shown,P, andR.

Solution to 6.16

Modeling the system as rigid and observing that pointO is fixed, we have

EvHD EvOC E!p ErH=OD E!p ErH=O ; (1)where the vectorsE!p andErH=O are given by

E!pDPOkD 5:98Ok rad=s and ErH=OD .` C h/Our dOuD .0:565 m/Our .0:025/Ou : (2)

Substituting the expressions in Eqs. (2) into Eq. (1) and carrying out the cross product, we have

EvHD .0:15Our 3:38Ou / m=s:

Again recalling thatO is a fixed point, for the acceleration, we have

EaHD Ep ErH=O !2p ErH=O ; (3)

whereEpDROkD 8:06Ok rad=sis the angular acceleration of the pendulum. Using this expression forEpand the expression forErH=O in the last of Eqs. (2), after simplifying, we have

EaHD .20:4Our 3:66Ou / m=s2:

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Dynamics 1e 771

Solution to 6.17

Modeling the system as rigid and observing that pointO is fixed, we have

EvHD EvOC E!p ErH=OD E!p ErH=O ; (4)

where the vectorsE!p andErH=O are given byE!pDPOk and ErH=OD .` C h/Our dOu : (5)

Substituting the expressions in Eqs. (5) into Eq. (4) and carrying out the cross product, we have

EvHDP dOurCP.` C h/Ou :

Again recalling thatO is a fixed point, for the acceleration, we have

EaHD Ep ErH=O !2p ErH=O ; (6)

whereEpDROkis the angular acceleration of the pendulum. Using this expression forEp and the expressionforErH=O in the second of Eqs. (5), after simplifying, we have

EaHDh R dP2.` C h/iOurC h R.` C h/ CP2diOu :

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Problem 6.18

At the instant shown the paper is being unrolled with a speed vpD 7:5 m=sand an acceleration apD1 m=s2. If at this instant the outer radius of the roll isrD 0:75 m, determine the angular velocity !s andaccelerations of the roll.

Solution

Referring to the figure shown to the right, the spool is in a fixed

axis rotation about its centerO . It is assumed that the paper is just

coming off the spool does not slip relative to the paper that is still

on the spool. LetPbe a point on the outmost layer of paper that is

just coming off the spool and letQ be the point on the paper layer

immediately below the outmost layer and on the same radial line as

P. Due to the no slip assumption, we haveEvPD EvQ, i.e.,

EvPD vPO{D EvQD E!s ErQ=O; (1)

whereE!sD !sOkis the angular velocity of the spool andErQ=OD rO| . Substituting these expressions inEq. (1) and solving for!s we have

E!sD .vP=r /Ok :

Enforcing the no slip condition for the acceleration gives thatEaQ O{D EaP O{, that is, the components of theaccelerations ofQ and P are the equal in the direction tangent to the contact. Now, observe that

EaPD aPO{ and EaQD Es .rO| / !2s .rO| / D sr O{ v2P

r O| : (2)

Using the expressions in Eq. (2) to enforce the no slip condition we have

sr

D aP, which can be solved

fors to obtain

EsD .ap=r /Ok :

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Dynamics 1e 773

Problem 6.19

At the instant shown, the propeller is rotating with an angular velocity !pD 400 rpmin the positivedirection and an angular accelerationpD 2 rad=s2 in the negative direction, where the axis is alsothe spin axis of the propeller. Consider the cylindrical coordinate system shown, with origin O on the

axis and unit vectorOuRpointing toward pointQ on the propeller which is 14 ftaway from the spin axis.Compute the velocity and acceleration ofQ. Express your answer using the component system shown.

Solution

The propeller is a rigid body, so we must have

EvQD E!P ErQ=OD !POu RQOuRD !PRQOuwhere!PD 400260D 41:89 rad=s andRQ = 14 ft. Plugging in known values we get

EvQD 586Ouft=s:Similarly forEaQ we get

EaQD !2PErQ=OC POk ErQ=OD !2PRQOuR C POk RQD !2PRQOuR C PRQ Ou :


EaQ

D

24:6

103

OuR

28:0

Ou ft=s

2:

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Problem 6.20

The wheelA, with diameterdD 5 cm, is mounted on the shaft of the motor shown and is rotating witha constant angular speed!AD 250 rpm. The wheelB , with center at the fixed point O , is connected toAwith a belt, which does not slip relative to A or B . The radius ofB isRD12:5 cm. At pointC thewheelB is connected to a saw. If pointC is at distance`D10 cmfromO , determine the velocity andacceleration ofC whenD 20. Express your answers using the component system shown.

Solution

Since the belt does not slip relative to A or B , we have

!Ad

2D !BR ) !BD d

2R!A: (1)

Next, observing that the crank is in a fixed axis rotation aboutO , we have

EvCD E!B ErC=O ; (2)

where the

E!BD d!A2R

Ok and ErC=OD `.cos O{ C sin O| /: (3)Substituting Eqs. (3) into Eq. (2), after simplifying we obtain

EvCD d `!A2R

. sin O{ C cos O| / D .0:179 O{ C 0:492O| / m=s:

Recalling that the angular velocity of the wheelB is constant and thatO is a fixed point, we have

EaCD !2BErC=O : (4)

Hence we have

EaC

D d2`!2A

4R2 .cos

O{

Csin

O| /

D .2:58

O{

C0:938

O| / m=s2:

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Dynamics 1e 775

Problem 6.21

In a contraption built by a fraternity, a person is sitting at the center of a swinging

platform with lengthL D 12 ftthat is suspended via two identical arms each oflengthHD 10 ft. Determine the angle and the angular speed of the arms ifthe person is moving upward and to the left with a speed vpD 25 ft=sat theangleD 33.

Solution

The platform and its rider are undergoing a curvilinear translation. Therefore, the velocity vector of the

person,EvP, is equal to the velocity vectors of pointsC andD .

EvCD EvPD EvD ) D ) D 33:0:Also, we must have

vPD j!AB jHD j!CDjH:Plugging in known values we get

j!AB j D j!CDj D vPH

D 2:50 rad=s:

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Problem 6.22

Ageosynchronous equatorial orbitis a circular orbit above the Earths equator

that has a period of1 day (these are sometimes called geostationary orbits).

These geostationary orbits are of great importance for telecommunications

satellites because a satellite orbiting with the same angular rate as the rotationrate of the Earth will appear to hover in the same point in the sky as seen by a

person standing on the surface of the Earth. Using this information, modeling a

geosynchronous satellite as a rigid body, and noting that the satellite has been

stabilized so that the same side always faces the Earth, determine the angular

speed!s of the satellite.

Solution

The motion of the satellite is a fixed-axis rotation about the center of the Earth where the axis is perpendicular

to the plane of the satellites orbit. For the satellite to be geosynchronous, the satellites angular velocity must

be such that it moves through2 radians every day

!sD 2rad24 h

1 h

3600 sD 72:7106 rad=s:

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Dynamics 1e 777

Problem 6.23

The bucket of a backhoe is being operated while holding the arm OAfixed. At the instant shown, point

B has a horizontal component of velocityv0D 0:25 ft=sand is vertically aligned with point A. Letting` D 0:9 ft,wD 2:65 ft, andh D 1:95 ft, determine the velocity of pointC. In addition, assuming that, atthe instant shown, pointB is not accelerating in the horizontal direction, compute the acceleration of pointC. Express your answers using the component system shown.

Solution

Since the bucket is in a fixed axis rotation about pointA, we have

EvBD !ABOk ErB=AD v0 O{ ) !ABOk `O|D v0 O{ ) `!ABO{D v0 O{) !ABD v0

`D 0:2778 rad=s;

where!AB is the angular velocity of the bucket. SinceB is not accelerating in the horizontal direction,

ABD 0and we have

EvCD !ABOk ErC=AD !AB w O{ .h `/O| ) EvCD .h `/!ABO{ C !ABwO| :

Upon substituting in given quantities, we have

EvCD .0:292 O{ 0:736O| / ft=s:

SinceABD 0,EaCD !2ABErC=AD v20 w O{ .h `/O| , which means that

EaCD .0:204 O{ C 0:0810O| / ft=s2:

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Problem 6.24

WheelsAand C are mounted on the same shaft and rotate together. WheelsA

andB are connected via a belt and so are wheelsC andD. The axes of rotation

of all the wheels are fixed, and the belts do not slip relative to the wheels they

connect. If, at the instant shown, wheel Ahas an angular velocity!AD 2 rad=sand an angular acceleration AD 0:5 rad=s2, determine the angular velocityand acceleration of wheels B andD . The radii of the wheels areRAD 1 ft,RBD 0:25 ft,RCD 0:6 ft, andRDD 0:75 ft.

Solution

Since the belts do not slip relative to the wheels they connect, we must have that

!ARAD !BRB ; ARAD BRB ;!CRCD !DRD; CRCD DRD;

where!AD !C; and AD C:

Using the given quantities, we have

E!BD RARB

!AOkD 8:00Ok rad=s ;

EBD RARB

AOkD 2:00Ok rad=s2 ;

E!DD RCRD

!COkD 2:50Ok rad=s ;

EDD RCRD

COkD 0:0625Ok rad=s2 :

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Dynamics 1e 779

Problem 6.25

An acrobat lands at the end A of a board and, at the instant shown, point Ahas a downward vertical

component of velocityv0D 5:5 m=s. LettingD 15,` D 1 m, anddD 2:5 m, determine the verticalcomponent of velocity of pointB at this instant if the board is modeled as a rigid body.

Solution

Letting the pivot of the board point be O , and modeling the board

as a rigid body, we know that

EvAD E!AB ErA=OD !ABOk `.cos O{ C sin O| /D !AB`. sin O{ C cos O| /:

Looking at the y component ofEvAwe have

vAyD v0D !AB` cos ) !ABD v0` cos

:

Similarly, for point B we know that

EvBD !ABOk ErB=OD v0

` cos Ok .dcos O{ dsin O| / D v0d

` tan O{ C v0d

` O| :

Plugging known values into they component ofEvB we have

vByD v0d`

D 13:8 m=s :

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Problem 6.26

At the instant shown, Ais moving upward with a speed v0D 5 ft=s and accelerationa0D 0:65 ft=s2. Assuming that the rope that connects the pulleys does not slip relative tothe pulleys and letting ` D 6 in:and dD 4 in:, determine the angular velocity and angularacceleration of pulleyC.

Solution

We can define the length of the rope as

L D yDC 2yA:

Taking two time derivatives of this equation, and remembering that the overall length is

constant because the rope is inextensible, we have

0 D PyDC 2 PyA ) PyDD 2 PyAD 2v00 D RyDC 2 RyA ) RyDD 2 RyAD 2a0

Because the pointQ is contacting the rope, and the rope is not slipping, point Q has the

same velocity and vertical component of acceleration as D .

vQD vDD 2v0D !C`2

) E!CD 4 v0`D 40:0Ok rad=s ;

aQyD aDD 2a0D C`

2 ) ECD 4a0

` D 5:20Ok rad=s2 :

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Dynamics 1e 781

Problem 6.27

At the instant shown, the angle D 30, jEvAj D 292 ft=s, and the turbine is rotating clockwise. LettingOA D R,OBD R=2,RD 182 ft, and treating the blades as being equally spaced, determine the velocityof pointB at the given instant and express it using the component system shown.

Solution

At the instant shownD 30, so the positions ofAand B in the given cartesian coordinate system are

ErA

D ErA=O

DR.sin

O{C

cos O| /

ErBD ErB=OD R

2O{:

Because the windmill is being treated as a rigid body, we also know

EvAD E!T ErA=OD !TOk ErA=OD !TR. cos O{ C sin O| / ) vAD j!TjRD 292 ft=s

whereE!T is the angular velocity of the turbine andvAis the speed ofA. Solving for!Twe have

!TD vAR

because the turbine is rotating clockwise. The velocity of pointB is then

EvBD E!T ErB=OD vA

ROk R

2O{

EvBD vA2

O|D 146O|ft=s:

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vcyclistD RCRS

RW!CD 17:6 m=s:

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Dynamics 1e 785

Solution to 6.30

We let!C; !W; and!S be the angular velocities andRC; RWandRS be the radii of the crank, wheel, and

sprocket respectively. The chain does not slip with respect to the crank or sprocket, so we must have

!CRC

D!SRS :

Given!CD 68 rpm, we are asked to pick a sprocket and crank combination such that!S is close to128 rpm.In other words, RC

RSmust be close to

!S

!CD 127

68D 1:868:

By trial and error we find that pairing C1 with S3 gives

RC

RSD 52:6 mm

28:3 mmD 1:86;

which is the closest pairing to the desired ratio. For this choice of gears and the given cadence,

!WD !SD RCRS

!CD 52:6mm

28:3 mm68 rpm D 126 rpm:

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Dynamics 1e 787

Problem 6.32

A carrier is maneuvering so that, at the instant shown, jEvAj D 25 knots(1 kn isexactlyequal to 1:852 km=h)andD 33. Letting the distance betweenA and B be220 m andD 22, determineEvB at the giveninstant if the ships turning rate at this instant isPD 2=s clockwise.

Solution

SinceA and B are on the same rigid body, we must have

EvBD EvA C E!carrier ErB=A (1)where

EvAD vA.sin O{ C cos O| /; (2)E!carrierD !OkD .2 =s/OkD .0:03491 rad=s/Ok; (3)

and

ErB=AD AB.cos O{ C sin O| /: (4)

Substituting Eqs. (2)(4) into Eq. (1) we have

EvBD vA.sin O{ C cos O| / C !Ok AB.cos O{ C sin O| /D .vAsin !ABsin / O{ C .vAcos C !ABcos /O| :

Substituting in known values we get

EvBD .9:88 O{ C 3:67O| / m=s:

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Problem 6.33

At the instant shown, the pinion is rotating between two racks with an angular

velocity!PD 55 rad=s. If the nominal radius of the pinion is RD 4 cmand ifthe lower rack is moving to the right with a speed vLD 1:2 m=s, determine thevelocity of the upper rack.

Solution

We defineP andQ to be the contact points between the pinion and the

top and bottom racks, respectively. Then we have

EvQD EvLD .1:2 m=s/ O{:

BecauseQ and P are points on the same rigid body, we must also have

EvPD EvQ C E!P ErP=QD vL O{ C !POk .2R/O|D .vL 2R!P/ O{:

Substituting known values we get

EvPD 3:2 O{m=s:

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Dynamics 1e 789

Problem 6.34

At the instant shown the lower rack is moving to the right with a speed of

vLD 4 ft=s, while the upper rack is fixed. If the nominal radius of the pinion isRD2 :5 in:, determine!P, the angular velocity of the pinion, as well as thevelocity of pointO , i.e., the center of the pinion.

Solution

We defineP andQ to be the contact points between the pinion and the

top and bottom racks, respectively. Then we have

EvQD EvLD .1:2 m=s/ O{ and EvPDE0:

BecauseQ and P are points on the same rigid body, we must also have

EvQD E!P ErQ=PD EvL ) !POk .2R/O|D EvL ) 2R!PO{D EvL ) !PD vL2R

: (1)


E!PD 9:60Ok rad=s:PointsO andP are also on the same rigid body, so we also must have

EvOD E!P ErO=PD !POk .R/O|D !PR O{: (2)

Substituting Eq. (1) into Eq. (2) we have

EvOD 2:00 O{ft=s:

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Problem 6.35

A bar of lengthL D 2:5 m is pin-connected to a roller at A. The roller is moving along a horizontal rail asshown withvAD5 m=s. If at a certain instant D 33 andPD 0:4 rad=s, compute the velocity of thebars midpointC.

Solution

BecauseC andA are points on the same rigid body, we must have

EvCD EvA C E!AC ErC=A; (1)

where

EvAD vA O{; E!ACDPOk; ErC=AD L

2.sin O{ cos O| /: (2)

Plugging Eqs. (2) into Eq. (1) we have

EvC

DvA

O{

C!AC

Ok

L

2.sin

O{

cos

O| /

D .vA C L2

!ACcos / O{ C !ACL2

sin O| :


EvCD .5:42 O{ C 0:272O| / m=s:

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Dynamics 1e 791

Problem 6.36

If the motion of the bar is planar, what would the speed ofAneed to be forEvCto be perpendicular to thebarAB ? Why?

Solution

The kinematic relation between points C andA is

EvCD EvA C E!AC ErA=C:

ForEvC to be perpendicular toErA=C in planar motion, we must have

EvADE0 ) vAD 0:

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Problem 6.37

PointsAand B are both on the trailer part of the truck. If the relative velocity of pointB with respect toA

is as shown, is the body undergoing a planar rigid body motion?

Solution

For the motion to be a planar rigid body motion,EvB=Amust be perpendicular to the vectorErB=A. The figureindicates thatEvB=AandErB=Aare not perpendicular. Therefore, the motion of the trailer is not a planar rigidbody motion.

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Dynamics 1e 793

Problem 6.38

A truck is moving to the right with a speed v0D 12 km=hwhile the pipe section with radiusRD 1:25 mand center atC rolls without slipping over the trucks bed. The center of the pipe sectionCis moving to

the right at2 m=srelative to the truck. Determine the angular velocity of the pipe section and the absolute

velocity ofC.

Solution

We letE!P be the angular velocity of the pipe. BecauseC andQ can be considered pointson the same rigid body, we must have

EvCD EvQ C E!P ErC=Q (1)where

E!PD !POk; EvQD EvTD v0 O{; and ErC=QD RO| : (2)

Substituting Eqs. (2) into Eq. (1) we have

EvCD .v0 !PR/ O{: (3)

We are also told that

EvC=QD E!P ErC=QD vpipe/truckO{ ) !PR O{D 2:000 O{m=s:Plugging in known values we haveE!PD 1:600;which to three significant figures is

E!PD 1:60Ok rad=s:

Plugging the unrounded value forE!Pand all known values into Eq. (3), and rounding to three significantfigures we have

EvCD 5:33 O{m=s:

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Problem 6.39

A wheelWof radiusRWD 7 mmis connected to point O via the rotating armOC, and it rolls withoutslip over the stationary cylinder S of radius RS D 15 mm. If, at the instant shown, D 47 and!OCD 3:5 rad=s, determine the angular velocity of the wheel and the velocity of pointQ, where pointQlies on the edge ofW and along the extension of the lineOC.

Solution

PointsC andO are two points on the arm, which is a rigid body, so we must

have

EvCD EvOC E!OC ErC=O; (1)where

E!OCD !OCOk and ErC=OD .RSC RW/Our : (2)Substituting Eqs. (2) into Eq. (1) we have

EvCD .RSC RW/!OCOu : (3)We letE be the contact point between cylinderSand wheelW, soEvEDE0, and

EvCD EvEC E!W ErC=ED !WOk RWOurD !WRWOu : (4)Equating Eqs. (3) and (4) we have

!WRWOuD !OC.RSC RW/ Ou ) !WD .RSC RW/

RW!OC:


E!W

D11:0

Ok rad=s:

BecauseQ is a point on the same rigid body asE , we also must have

EvQD EvEC !W Ok 2RWOurD 2RW!WOuD 2.RSC RW/!OCOu :


EvQD 0:154Oum=s:

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Dynamics 1e 795

Problem 6.40

For the slider-crank mechanism shown, let RD 20 mm, LD 80 mm, andHD 38 mm. Use the concept of instantaneous center of rotation to determinethe values of, with0 360, for whichvBD 0. Also, determine theangular velocity of the connecting rod at these values of.

Solution

WhenvBD 0,B is the instantaneous center of rotation ofAB , andEvAmustbe perpendicular to the lineBA. However,EvAmust also be perpendicularto the lineOA, sinceO is the center of rotation of the crank. Therefore,O ,

A, andB must be collinear. The two cases where this happens are shown

to the left. In the first, case we have that

.L C R/ cos 1D H ) 1D cos1 HL C R ) 1D 67:7

:

In the second case we have that

.L R/ cos 2D H ) 2D 180 C cos1 HL R

) 2D 231:

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Problem 6.41

At the instant shown barsAB andB Care perpendicular to each other, and bar B Cis rotating counter-

clockwise at20 rad=s. LettingL D 2:5 ftand D 45, determine the angular velocity of barAB as wellas the velocity of the sliderC.

Solution

BecauseD 45, the instantaneous center of rotation of bar B Cis pointA.Therefore,

vCDp

2L!BCD 70:7 ft=sBecauseCis constrained to move along the guide with vector .O{ C O| /, andis moving up and to the right, we have

EvCD .50:0 O{ C 50:0 O{/ ft=s:

A is also the center of rotation of barAB . Therefore, in the present configura-

tion

vB D !ABL D !BCL ) !AB D !BC ) E!ABD 20:0Ok rad=s:

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Dynamics 1e 797


A ball of radius RAD 3 in: is rolling without slip in a stationary spherical bowl of radius RBD 8 in:Assume that the balls motion is planar.

Problem 6.42 If the speed of the center of the ball isvAD

1:75 ft=sand if the ball is moving down and

to the right, determine the angular velocity of the ball.

Problem 6.43 If the angular speed of the ball j!Aj D 4 rad=sis counterclockwise, determine the velocityof the center of the ball.

Solution to 6.42

ConvertingRAto feet we have RAD 0:2500 ft:Since the ball rolls without slip over a stationary surface, thepoint of contact between the ball and the bowl is the instantaneous center of rotation of the ball. Therefore,

vAD j E!AjRA ) !AD vARA

) !AD 7:000 rad=s

Observing that the ball will be rotating counterclockwise and given thatOkD Our Ou , to three significantfigures we have

E!AD 7:00Ok rad=s:

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Solution to 6.43

Since the ball rolls without slip over a stationary surface, the point of contact between the ball and the bowl is

the instantaneous center of rotation of the ball. Therefore,

vAD j E

!AjRA

D4RA

) EvA

D 12

Ouin=s:

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Dynamics 1e 799

Problem 6.44

One way to convert rotational motion into linear motion and vice versa is via the use of a mechanism

called a Scotch yoke, which consists of a crankCthat is connected to a sliderB via a pin A. The pin

rotates with the crank while sliding within the yoke, which, in turn, rigidly translates with the slider. This

mechanism has been used, for example, to control the opening and closing valves in pipelines. Letting theradius of the crank beRD 1:5 ft, determine the angular velocity !Cof the crank so that the maximumspeed of the slider isvBD 90 ft=s.

Solution

Calling the center of the wheel the origin, from the geometry of the problem we

have

ErAD R.cos O{ C sin O| /: (1)By construction, the velocity of the slider is equal to the horizontal component of

the velocity of pointA. We can getEvAby taking the derivative of Eq. 1 with respect to time. Doing so wehave

EvAD R P . sin O{ C cos O| / ) EvBD R Psin O{:BecauseR

Psin varies as a function of, with its maximum value equal toR

j Pj, for.vB/max we have

.vB/maxD Rj Pj D R!C ) !CD .vB/maxR

) E!CD 60:0Ok rad=s:

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Problems 6.45 through 6.47

The system shown consists of a wheel of radius RD 14 in: rolling on ahorizontal surface. A barAB of length LD 40 in: is pin-connected to thecenter of the wheel and to a sliderAthat is constrained to move along a vertical

guide. PointCis the bars midpoint.

Problem 6.45 If, whenD 72, the wheel is moving to the right so thatvBD 7 ft=s, determine the angular velocity of the bar as well as the velocity ofthe sliderA.

Problem 6.46 If, whenD 53, the slider is moving downward with a speedvAD 8 ft=s, determine the velocity of pointsB andC.Problem 6.47 If the wheel rolls without slip with a constant counterclockwise

angular velocity of10 rad=s, determine the velocity of the sliderA whenD45.

Solution to 6.45

Converting units on known values, we have

RD 1:167 ft L D 3:333 ft:We are told the wheel is moving to the right. Therefore, using the component system shown

EvBD 7 O{ft=s: (1)In addition, the slider must be moving down, so we haveEvAD vAO| ;and the angular velocity of the bar isE!ABD !ABOk: Also, the position vector ofB with respect toAisErB=AD L.cos O{ sin O| /:From rigidbody kinematics, we must have

EvBD EvA C E!AB ErB=AD vAO|C !ABOk L.cos O{ sin O| /D !ABL cos O{ C .!ABL sin vA/O| : (2)

Equating Eq. (1) and Eq. (2) we have

vBD !ABL sin ) !ABD vBL sin

; (3)

vAD !ABL cos : (4)Plugging known values into Eq. (3) we get

!ABD 2:208 rad=s:To three significant figures this is

E!ABD 2:21Ok rad=s:Plugging the unrounded value for!AB into Eq. (4) we get

vAD 2:27 ft=s ) EvAD 2:27O|ft=s:

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Dynamics 1e 801

Solution to 6.46

Converting units on known values, we have

RD 1:167 ft L D 3:333 ft:

We are told the slider is moving down. Therefore, using the componentsystem shown,

EvAD 8O|ft=s:In addition, because the slider is moving down, the the wheel must be

moving to the right, so we haveEvBD vB O{; E!ABD !ABOk, andE!WD!W Ok: Also, the position vector ofB relative toA isErB=AD L.cos O{sin O| /;and ofB relative toQis ErB=QD RO| :From rigid body kinematicswe must have

EvBD EvA C E!AB ErB=A

D vAO

|C

!ABO

k

L.cos O{

sin O| /

D !ABL cos O{ C .!ABL sin vA/O| : (5)

We must also have

EvBD EvQ C E!W ErB=QD !WOk RO|D !WR O{: (6)

Equating components of Eq. (5) with those of Eq. (6) we have

!WR

D!ABL sin ; (7)

vAD !ABL cos : (8)

Solving Eq. (8) gives!ABD 3:988 rad=s:Plugging!AB into Eq. (7) gives !WD 9:100 rad=s:Plugging!Winto Eq. (6) gives

EvBD 10:6 O{ft=s:

Realizing that the position vector fromA to C isErC=ADErB=A2 , from rigid body kinematics we have

EvCD EvA C E!AB ErC=A

D vA

O|

C!ABOk

L

2

.cos

O{

sin

O| /

D !ABL cos O{ C .!ABL sin vA/O| : (9)

Plugging all known values into Eq. (9) gives

EvCD .5:31 O{ 4O| / ft=s:

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Dynamics 1e 803

Problem 6.48

At the instant shown, the lower rack is moving to the right with a speed of

2:7 m=s while the upper rack is moving to the left with a speed of1:7 m=s.

If the nominal radius of the pinion O isRD 0:25 m, determine the angularvelocity of the pinion, as well as the position of the pinions instantaneouscenter of rotation relative to point O .

Solution

From the no slip condition at pointsU andQ we know that

EvUD 1:7 O{m=s (1)and

EvQD 2:7 O{m=s: (2)

From rigid body kinematics we have

EvUD EvQ C E!P ErU=QD vL O{ C !POk 2RO|D vL O{ 2R!PO{; (3)

where!P is the angular speed of the pinion. Comparing Eq. (3) with Eq. (1) component by component, we

have

1:7 D vL 2R!P ) !PD vL C 1:7

2R ) !PD 8:800 rad=s:

In vector form, and to three significant figures this is

E!PD 8:80Ok rad=s:

We also know that

EvICD EvQ C E!P ErIC=QE0 D vL O{ C !POk .dIC=Q/x O{ C .dIC=Q/yO| E0 D vL .dIC=Q/y!P O{ C .dIC=Q/x!PO| : (4)

Breaking Eq. (4) into two scalar equations by components we have

0 D vL .dIC=Q/y!P ) .dIC=Q/yD vL

!P;

0

D.dIC=Q/x!P

) .dIC=Q/x

D0:

Substituting known values we have

ErIC=QD 0:307O|rad=s:

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Problem 6.49

A carrier is maneuvering so that, at the instant shown, jEvAj D 22 kn,D 35, jEvB j D 24 kn(1 kn is equalto 1 nautical mile (nml) per hour or 6076 ft=h). LettingD 19 and the distance between A and B be720 ft, determine the ships turning rate at the given instant if the ship is rotating clockwise.

Solution

We let

EvA

D vA.sin

O{

Ccos

O| /,

E!A

D !A

Ok, and

ErB=A

D dAB.cos

O{

Csin

O| /. From rigid body

kinematics we know

EvBD EvA C E!AB ErB=AD vA.sin O{ C cos O| / C !ABOk dAB.cos O{ C sin O| /D .vAsin !ABdABsin / O{ C .vAcos C !ABdABcos /O| : (1)

Taking the magnitude of both sides of Eq. (1) we have

v2BD .vAsin !ABdABsin /2 C .vAcos C !ABdABcos /2: (2)

All values in Eq. (2) are known except for !AB . Solving, we have!ABD 0:06805 rad=s:Since we aretold that the ship is turning clockwise, using the component system shown we must have

E!ABD 0:0681Ok rad=s:

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Breaking Eq. (5) into two scalar equations by components we have

0 D !ABR sin !BCcos

sin1

R

Lcos

(6)

vCD 2R cos .!ABC !BC/: (7)

Eqs. (6) and (7) are two equations invC,!BC, and known values. Solving these equations forvCwe get

EvCD 61:1O|ft=s:

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Dynamics 1e 807

Solution to 6.51

We let ErB=AD R.cos O{ C sin O| /andE!ABD !ABOk. Converting all given values to ft=sor rad=swe haveRD 0:1583 ft; L D 0:5083 ft; HD 0:1000 ft; and !ABD 507:9 rad=s:


EvBD E!AB ErB=AD !ABOk R.cos O{ C sin O| / D !ABR sin O{ C !ABR cos O| : (8)

If we letE!BCD !BCOk, ErC=BD L.sin O{ C cos O| /, andEvCD vCO| , then from rigid body kinematics wehave

EvCD vCO|D EvBC E!BC ErC=B :D EvBC !BCOk L.sin O{ C cos O| /D EvB !BCL cos O{ C !BCL sin O| : (9)

Substituting Eq. (8) into Eq. (9) we get

vCO|D .!ABR sin C !BCL cos / O{ C .!ABR cos C !BCL sin /O| : (10)From the geometry of the problem we also have

L sin D R cos ) D sin1

R

Lcos

: (11)


vCO|D

!ABR sin C !BCL cos

sin1

R

Lcos

O{ C 2R cos .!ABC !BC/O| : (12)

Equating components of Eq. (12) we have

0 D !ABR sin !BCcos

sin1

R

Lcos

(13)

vCD 2R cos .!ABC !BC/: (14)Eqs. (13) and (14) are two equations in vC,!BC, and known values. Solving these equations for !BC, we

get

!BCD 74:76 rad=s:Because measures the angle ofB C with respect to the horizontal axis,PD !BC. Therefore,

P

D 74:8 rad=s:

We letErD=BD H.sin O{ C cos O| /. Therefore, from rigid body kinematics we haveEvDD EvBC E!BC ErD=B

D EvBC !BCOk H.sin O{ C cos O| /D EvB !BCHcos O{ C !BCHsin O| : (15)

Substituting Eq. (8) into Eq. (15) we get

EvDD .!ABR sin C !BCHcos / O{ C .!ABR cos C !BCHsin /O| : (16)August 10, 2009


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Combining Eq. (11) with Eq. (16) to eliminate gives

EvDD

!ABR sin C !BCHcos

sin1

R

Lcos

O{ C

R cos

!ABC !BCH

L

O| : (17)

Substituting all known values into Eq. (17) we get

EvDD .29:3 O{ C 69:6O| / ft=s:

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Dynamics 1e 809

Problem 6.52

A wheelW of radius RWD 7 mmis connected to point O via the rotatingarm OC, and it rolls without slip over the stationary cylinder S of radius

RSD 15 mm. If, at the instant shown,D 63 and !WD 9 rad=s, determinethe angular velocity of the armO Cand the velocity of pointP, where pointPlies on the edge ofW and is vertically aligned with pointC.

Solution

Let E be the contact point betweenW and S. From the no slip condition,

EvE DE0. We let E!W D !W Ok and from the geometry of the problemErC=ED RW.cos O{ C cos O| /. From rigid body kinematics we have

EvCD EvEC E!W ErC=ED !W Ok RW.cos O{ C sin O| /D !WRW. sin O{ C cos O| /: (1)

PointsC andO are also points on the arm where

ErC=O D .RS C RW/.cos O{C sin O| /. Therefore, we must have

EvCD EvOC E!OC ErC=OD !OCOk .RSC RW/.cos O{ C sin O| /

D!OC.RS

CRW/.

sin

O|

Ccos

O| /: (2)

Equating Eqs. (1) and (2) we have

!WRWD !OC.RSC RW/ ) E!OCD RWRSC RW

!W OkD 2:86Ok rad=s:

PointsP andC are both points on the wheel withErP=CD RW O| . Therefore, we must have

EvPD EvCC E!W ErP=C:

Substituting in Eq. (1) forEvCwe have

EvPD !WRW. sin O{ C cos O| / C !W Ok RW O|D !WRW. sin 1/ O{ C cos O| : (3)

Substituting known values into Eq. (3) we get

EvPD .0:119 O{ C 0:0286O| / m=s:

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Problem 6.53

At the instant shown, the center O of a spool with inner and outer radiirD 1 mandRD 2:2 m, respectively, is moving up the incline at speed vOD 3 m=s.If the spool does not slip relative to the ground or relative to the cable C,

determine the rate at which the cable is wound or unwound, that is, the lengthof rope being wound or unwound per unit time.

Solution

From the no slip condition,EvQDE0. We also knowEvO D vOO{ ,ErO=QD RO| , andE!SD !SOk. From rigid body kinematics we have

EvOD EvQ C E!S ErO=Q ) vOO{D !SOk RO|) !SD vO

R :

We also know thatErD=OD rO| , andEvCD EvD. From rigid bodykinematics we have

EvDD EvOC E!S ErD=OEvCD vOO{ C !SOk rO|EvCD .vO !Sr/ O{ ) EvCD vO

1 C r

R

O{:

Since

EvC and

EvO are in the same direction, andvC > vO , the cord is being unwound at a rate of

jEvC EvO j D rR

vOD 1:36 m=s:

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Problem 6.55

The bucket of a backhoe is the element AB of the four-bar linkage system ABCD. As-

sume that the points A and D are fixed and that, at the instant shown, point B is ver-

tically aligned with point A, point C is horizontally aligned with point B, and point B

is moving to the right with a speed vB D 1:2 ft=s. Determine the velocity of point Cat the instant shown, along with the angular velocities of elements BC and CD. Let hD 0:66 ft,eD 0:46 ft,lD 0:9 ft, andwD 1:0 ft.

Solution

PointCis a point on both bar B Cand barCD , and therefore can be related to both points B andD through

rigid body kinematics. We have that

EvCD EvDC E!CD ErC=DD !CDOk .w h/ O{ C .e C `/O| D !CD.e C `/ O{ !CD.w h/O| (1)

SinceEvBD vBO{, we ahve

EvCD EvBC E!BC ErC=BD vBO{ C !BC .w/ O{D vBO{ !BCwO| : (2)

Equating Eqs. (1) and (2) component by component we have

!CD.e C `/ O{D vBO{ and !CD.w h/O|D !BCwO| : (3)

Solving Eqs. (3) we have

!CDD vBe C ` and !BCD

!CD.w h/w

) !BCD vBe C `

w hw

: (4)

Substituting known values into Eq. (4), we have

E!CDD 0:882Ok rad=s; E!BCD 0:300Ok rad=s:

Substituting Eqs. (4) into Eq. (2) we have

EvCD vBO{ C vBe C ` .w h/O| : (5)

Substituting known values into Eq. (5) we have

EvCD .1:20 O{ C 0:300O| / ft=s:

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Dynamics 1e 813

Problem 6.56

Bar AB is rotating counterclockwise with an angular velocity of 15 rad=s.

LettingL D 1:25 m, determine the angular velocity of bar CD whenD 45.

Solution

Because barAB is rotating about the fixed point A, at the given instant point B is only moving in theO{direction, and we have

EvBD !ABL O{:PointsB and Care both on the barB Cso we have

EvCD EvBC E!BC ErC=BD !ABL O{ C !BCOk L O{D !ABL O{ C !BCLO| : (1)

PointsC andD are also both on the barCD so we have

EvCD EvDC E!CD ErC=DD !CDOk L

2.cos O{ C sin O| / D L

2!CD. sin O{ C cos O| /: (2)

Equating the O{ components of Eqs. (1) and (2) we have

!ABL D L

2!CDsin ) !CDD 2!AB

sin :

Substituting in known values we have

E!CDD 42:4Ok rad=s:

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65/187

Dynamics 1e 815

Problem 6.58

Collars A and B are constrained to slide along the guides shown and are

connected by a bar with length LD 0:75 m. Letting D 45, determinethe angular velocity of the bar AB at the instant shown if, at this instant,

vBD 2:7 m=s.

Solution

From rigid body kinematics, we must have

EvAD EvBC E!AB ErA=B;whereEvAD vAO| ,EvBD vBO{, andErA=BD Lp2O{ C

Lp2O| . Plugging these values in we have

vAO|D vBO{ C !ABOk Lp

2O{ C Lp

2O|

vAO|D vBO{ !AB Lp2

O{ !AB Lp2

O| : (1)

From the O{ component of Eq. (1) we have

vBD !AB Lp2

) !ABD vBL

p2:

Substituting known values we have

E!ABD 5:09Ok rad=s:

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67/187

Dynamics 1e 817

Problem 6.60

A spool with inner radiusRD 1:5 m rolls without slip over a horizontal rail as shown. If the cable on thespool is unwound at a rate vAD 5 m=sin such a way that the unwound cable remains perpendicular to therail, determine the angular velocity of the spool and the velocity of the spools centerO .

Solution

We letP be the contact point between the rope and the spool. By

the no slip condition, theO| component ofEvPis equal tovA. Also,vQD 0because it is the contact point between the spool and therail. From rigid body kinematics we then have

EvPD EvQCE!spool ErP=QD !spoolOk .R O{ C RO| /D R!spool. O{ C O| /:

EquatingvPy tovAwe have

vAD R!spool ) !spoolD vAR

D 3:33 rad=s:

In vector form this is

E!spoolD 3:33Ok rad=s:

We must also have

EvOD EvQ C E!spool ErO=QD !spoolOk RO|D !spoolR O{ ) EvOD 5:00 O{rad=s:

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Problem 6.61

In the four-bar linkage system shown, the lengths of the barsAB andCD are

LABD 46 mm andLCDD 25 mm, respectively. In addition, the distancebetween pointsAand D is dADD 43 mm. The dimensions of the mechanismare such that when the angleD 132, the angleD 69. ForD 132andPD 27 rad=s, determine the angular velocity of bars B C andCDas well asthe velocity of the point E , the midpoint of bar B C. Note that the figure is

drawn to scale and that barsB C andCD are not collinear.

Solution

We let pointD be the origin. From the geometry of the problem we have

ErAD dAD O{D 0:04300 O{m;ErB=AD LAB.cos O{ C sin O| / D .0:03078 O{ C 0:03418O| / m;

ErBD ErA C ErB=AD .0:01222 O{ C 0:03418O| / m;ErCD LCD.cos O{ C sin O| / D .0:008959 O{ C 0:02334O| / m;and

ErC=BD ErC ErBD .LCDcos dAD LABcos / O{C .LCDsin LABsin /O|D .0:003261 O{ 0:01085O| / m:


EvBD EvA C E!AB ErB=A:D !ABOk LAB.cos O{ C sin O| /

D!AB

LAB

.

sin O{C

cos O| /

D.

0:9230O{

0:8311O| / m=s:

We must also have

EvCD EvBC E!BC ErC=BD !ABLAB. sin O{ C cos O| / C !BCOk .LCDcos dAD LABcos / O{

C .LCDsin LABsin /O|D !BC.LABsin LCDsin / !ABLABsin O{

C !BC.LCDcos dAD LABcos / C !ABLABcos O| : (1)

PointCis also on barCD where ErC=DD LCD.cos O{ C sin O| /. From rigid body kinematics we then have

EvCD EvDC E!CD ErC=DD !CDLCD. sin O{ C cos O| /: (2)

Equating Eqs. (1) and (2) component by component we have

!BC.LABsin LCDsin / !ABLABsin D !CDLCDsin (3)and

!BC.LCDcos dAD LABcos / C !ABLABcos D !CDLCDcos : (4)

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Dynamics 1e 819

Eqs. (3) and (4) are two equations in!BC and!CD . Dividing the left hand side of Eq. (3) by the left hand

side of Eq. (4) and the right hand side of Eq. (3) by the right hand side of Eq. (4) and switching the sides of

the result we have

tan D !BC.LABsin LCDsin / !ABLABsin !BC.LCDcos

dAD

LABcos /

C!ABLABcos

)!ABLAB.sin tan cos / D !BC.LABsin LCDsin /

C tan .LCDcos dAD LABcos /) !BCD !ABLAB.sin tan cos /

LABsin LCDsin C tan .LCDcos dAD LABcos / : (5)

Substituting known values into Eq. (5) we get

E!BCD 1310Ok rad=s:

Solving Eq. (3) for!CD we have

!CDD !ABLABsin !BC.LABsin LCDsin /

LCDsin : (6)

Substituting Eq. (5) for!BC in Eq. (6) we have

!CDD !ABLABsin LCDsin

!ABLAB.sin tan cos /.LABsin LCDsin /LABsin LCDsin C tan .LCDcos dAD LABcos /LCDsin

(7)

Substituting known values into Eq. (7) we have

E!CDD 571Ok rad=s:

We can calculateEvE from the rigid body kinematics of barB C, whereErE=BD 12 ErC=D. We have

EvED EvBC !BCOk ErE=BD vBx O{ C vByO|C !BCOk 1

2ErC=D

D vBx O{ C vByO|C !BC2

Ok .rC=B /x O{ C .rC=B/yO|

DvBx

!ABLAB.sin tan cos /

2LABsin LCDsin C tan .LCDcos dAD LABcos /.rC=B/y

O{

C vByC !ABLAB.sin tan cos /

2LABsin LCDsin C tan .LCDcos dAD LABcos /.rC=B /xO| : (8)

Substituting all known values into Eq. (8) we have

EvED .6:20 O{ 2:97O| / m=s:

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Problem 6.62

A person is closing a heavy gate with rusty hinges by pushing the gate with car A. IfwD 24 m andvAD 1:2 m=s, determine the angular velocity of the gate whenD 15.

Solution

From the geometry of the problem, we have

yHD w tan : (1)

Taking the first derivative of Eq. (1) we have

PyHD wcos2

P :

Realizing thatPyHD vAwe havevAD w

cos2 P :

Solving forPwe are left with

PD vAcos2

w : (2)

Finally, substituting known values into Eq. (2) we have

PD 0:0467 rad=s:

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Dynamics 1e 821


In the four-bar linkage system shown, let the circular guide with center at O be fixed and such that, when

D 0, the barsAB andB Care vertical and horizontal, respectively. In addition, let RD 2 ft,L D 3 ft,andHD 3:5 ft.Problem 6.63 WhenD 0, the collar at C is sliding downward with a speed of23 ft=s. Determine theangular velocities of the barsAB andBCat this instant.

Problem 6.64 WhenD 37,D 25:07,D 78:71, and the collar is sliding clockwise with a speedvCD 23 ft=s. Determine the angular velocities of the barsAB andB C.Problem 6.65 Determine the general expression for the angular velocities of barsAB andB C as a

function of,,,R,L,H, andP.

Solution to 6.63

Because pointB is the instantaneous center of rotation of barCB,

it has zero velocity. Point Ais pinned and cannot move, so barABcannot be moving. Therefore we have

E!ABDE0:

From the instantaneous center of rotation method, we then have

vCD L!BC ) !BCD vCL

D 7:667 rad=s:

Following the given component system, in vector form, and to three significant figures we have

E!BCD 7:67Ok rad=s:

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Solution to 6.64

From the geometry of the problem, we have the following position

vectors:

ErC=O

DR.cos

O{

Csin

O| /;

ErB=CD L.cos O{ sin O| ;ErB=AD H. cos O{ C sin O| /;

and the velocity vector

EvCD vC.sin O{ cos O| /:

We also defineE!BCD !BCOkandE!ABD !ABOk. PointsB andCare both on barB Cso from rigid bodykinematics we have

EvBD EvCC !BCOk ErB=CD vC.sin O{ cos O| / C !BCL.sin O{ C cos O| /

D .vCsin C !BCL sin / O{ C .!BCL cos vCcos /O| : (1)Also, because pointsAand B are both on bar AB , we have

EvBD EvA C E!AB ErB=AD !ABH.sin O{ C cos O| /: (2)Equating Eq. (1) with Eq. (2) component by component we have

vCsin C !BCL sin D !ABHsin (3)!BCL cos vCcos D !ABHcos : (4)

Eqs. (3) and (4) are two equations in !AB and!BC. To solve them we divide the left hand side of Eq. (3) by

the left hand side of Eq. (4) and the right hand side of Eq. (3) by the right hand side of Eq. (4). We have

vCsin C !BCL sin !BCL cos vCcos

D tan ) vCsin C !BCL sin D tan .!BCL cos vCcos /

) !BCD vCsin C tan cos

L.tan cos sin / :

Substituting in known values, we have

!BCD 7:210 rad=s:To three significant figures, and in vector form we have

E!BCD 7:21Ok rad=s:Solving Eq. (3) for!AB we have

!ABD vCsin C !BCL sin

Hsin :

Plugging in known values, we have

E!ABD 8:58Ok rad=s:

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Dynamics 1e 823

Solution to 6.65


L sin C Hsin D HC R sin and L sin C Hcos D R.1 cos / C L: (5)

Taking the first derivative with respect to time of Eqs. (5) we have

L P cos C HPcos D R Pcos (6)L P sin HPsin D R Psin : (7)

Eqs. (6) and (7) can be viewed as two equations in P andP. Realizing that PD !BC andPD !ABtells us that solving these two equations for P andPwill give us the angular velocities of the two bars as afunction of the specified terms. Solving these equations we have

!BCD PD R PL cos

cos cos tan C sin

tan tan

and

!ABD PD RP

H

sin C tan cos cos tan sin :

Simplifying these we end up with

E!BCD RP

L csc . / sin .C /Ok

and

E!ABD RP

H csc . / sin . C /Ok:

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Problem 6.66

At the instant shown, the arm O Crotates counterclockwise with an angular

velocity of35 rpmabout the fixed sun gear Sof radiusRSD 3:5 in:The planetgearPwith radiusRPD 1:2 in:rolls without slip over both the fixed sun gearand the outer ring gear. Finally, notice that the ring gear is not fixed and it rollswithout slip over the sun gear. Determine the angular velocity of the ring gear

and the velocity of the center of the ring gear at the instant shown.

Solution

PointsC andO are both on the barO C, so from rigid body kinematics

we must have

EvCD EvOC E! COCErC=OD !OCOk .RSC RP/ O{D !OC.RSC RP/O| : (1)

PointC is also on the planetary gear with pointQ, so we must have

EvCD EvQ C E!P ErC=QD !POk RPO{D !PRPO| : (2)


.RSC RP/!OCD RP!P ) !PD RSC RPRP

!OC:

PointA is also on the planetary gear, so we must have

EvAD EvQ C E!P ErA=QD RSC RP

RP!OCOk 2RPO{

D 2.RSC RP/!OCO| : (3)

PointsAand B are both points on the ring gear, so we must have

EvAD EvBC E!R ErA=BD !ROk 2.RSC RP/ O{D 2!R.RSC RP/O| : (4)


2.RSC RP/!OCD 2!R.RSC RP/ ) !RD !OC:

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Dynamics 1e 825

In vector form this is

E!RD 35:0Ok rpm:Calling the center of the ring pointD we must have

EvDD EvBC E!R ErD=BD !Ok .RSC RP/ O{D .RSC RP/!OCO| : (5)Substituting known values into Eq. (5) we have

EvDD 17:2O|in.=s:

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Dynamics 1e 827


For the slider-crank mechanism shown, let RD 20 mm, LD 80 mm, andHD 38 mm.

Problem 6.68 IfPD 1700 rpm, determine the angular velocity of the con-necting rodAB and the speed of the sliderB forD 90.Problem 6.69 Determine the angular velocity of the crankOA when D 20and the slider is moving downward at15 m=s.

Problem 6.70 Determine the general expression for the velocity of the slider

B as a function of,P, and the geometrical parametersR,H, andL, using thevector approach.

Problem 6.71 Determine the general expression for the velocity of the slider

B as a function of,P, and the geometrical parametersR,H, andL, usingdifferentiation of constraints.

Problem 6.72 Plot the velocity of the sliderCas a function of, for

0 360, and forPD 1000 rpm,PD 3000 rpm, andPD 5000 rpm.

Solution to 6.68

Converting units on known values we have

RD 20 mm D 0:020 m; L D 80 mm D 0:080 m; HD 38 mm D 0:038 m:

From the rigid body kinematics of the wheel, and letting!ODPwe have

EvAD EvOC E!O ErA=ODPOk RO|D P R O{:From the rigid body kinematics of arm ABwhere ErB=AD HO{C

pL2 H2 O| ,

we have

EvBD EvA C E!AB ErB=AD R PO{ C !ABOk .HO{ C

pL2 H2 O| /

D . P R !ABp

L2 H2/ O{ C !ABHO| :BecauseB is confined to move vertically in the slot,vBxD 0. From this we have

!ABD R

pL2 H2 P :In vector form, and plugging in known values we have

E!ABD 50:6Ok rad=s:

From they component ofEvB we have

EvBD vByO|D !ABHO|DPH RpL2 H2

O| :August 10, 2009


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EvBD 1:92O|m=s:

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Dynamics 1e 829

Solution to 6.69


RD 20 mm D 0:020 m; L D 80 mm D 0:080 m; HD 38 mm D 0:038 m:

From the rigid body kinematics of the wheel, and letting!ODPwe haveEvAD EvOC E!O ErA=ODPOk R.cos O{ C sin O| / DPR. sin O{ C cos O| /:

From the rigid body kinematics of arm AB where ErB=AD .H R cos / O{ Cp

L2 .H R cos /2 O| , wehave

EvBD EvA C E!AB ErB=AD R PO{ C !ABOk

.H R cos / O{ C

qL2 .H R cos /2 O|

D P R !ABqL2

.H

R cos /2O{ C !AB.H R cos /O| :

BecauseB is confined to move vertically in the slot,vBxD 0. From this we have

PD!ABp

L2 .H R cos /2R

: (1)

We also know thatEvBD 15O|m=s. From this we have

!AB.H R cos / D 15 ) !ABD 15R cos H: (2)


PD15p

L2 .H R cos /2R.R cos H / :


E!ODPOkD 877Ok rad=s:

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Solution to 6.70


EvAD EvOC E!O ErA=ODPOk R.cos O{ C sin O| / DPR. sin O{ C cos O| /:

From the rigid body kinematics of arm AB , where ErB=AD .HR cos / O{ CpL2 .H R cos /2 O| , wehave


.H R cos / O{ C

qL2 .H R cos /2 O|

D P R sin !AB

qL2 .H R cos /2

O{ C !AB.H R cos / C R Pcos O| : (3)


!ABD R Psin pL2 .H R cos /2 : (4)

Plugging Eq. (4) into Eq. (3) we have

EvBD

R Pcos RPsin .H R cos /p

L2 .H R cos /2

!O| :

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Dynamics 1e 831

Solution to 6.71


xBD R cos C L cos D H (5)yBD

R sin C

L sin :

Taking a time derivative of these two equations we have

R Psin L Psin D 0 ) PD RPsin

L sin (6)

vBD R Pcos C L Pcos : (7)


vBD R Pcos R Psin cot : (8)

From Eq. (5) we know thatcos D H R cos

L ; (9)

and from the Pythagorean theorem, we know that

sin Dq

L2 .H R cos /2: (10)

Combining Eqs. (9) and (10) we have

cot D H R cos L

pL2 .H R cos /2

: (11)


vBD R Pcos R Psin H Rcos

Lp

L2 .H R cos /2:

Knowing thatB is confined to theO| direction, we have

EvBD


L2 .H R cos /2

!O| :

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Solution to 6.72


RD 0:020 m;L

D0:080 m;

HD 0:038 m:


EvAD EvOC E!O ErA=ODPOk R.cos O{ C sin O| /DP R. sin O{ C cos O| /:

From the rigid body kinematics of arm AB , where ErB=AD .HR cos / O{ Cp

L2 .H R cos /2 O| , wehave


.H R cos / O{ C

qL2 .H R cos /2 O|

D P R sin !AB

qL2 .H R cos /2

O{ C !AB.H R cos / C R Pcos O| : (12)


!ABD RPsin

pL2 .H R cos /2

: (13)

Plugging Eq. (13) into Eq. (12) we have

EvBD


L2 .H R cos /2

!O| :

The expression forEvB just derived can be reduced to a function of andP by substituting in thegiven parameters. For each of the required values ofP, the function can be plotted with software such as

Mathematicaor MATLAB. The plots presented below were obtained using Mathematicawith the following

code:

Parameters R 0.02,L 0.08,H 0.038;

Velocity R dotCosR dotSinHRCos

L2 HRCos2. Parameters;

dotval 1

301000, 3000, 5000;

TablePlotVelocity.dot i,, 0, 2, AxesLabel "", "vBy ",

Ticks 0,, 2, 3, Automatic,i, dotval

Using the above code we obtain the following plots:

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84/187


Problem 6.73

Complete the velocity analysis of the slider-crank mechanism, using differenti-

ation of constraints that was outlined beginning on p. 478. That is, determine

the velocity of the pistonC and the angular velocity of the connecting rod as a

function of the given quantities,!AB ,R, andL. Use the component systemshown for your answers.

Solution

One way to solve this problem is to write yC as

yCD R cos C L cos ; (1)

where for cos and sin we willalwaysuse

sin D RL

sin and cos Dq

1 sin2 Ds

1 R2

L2sin2 : (2)

Now we differentiate Eq. (1) with respect to time:

PyCD R Psin L Psin ; (3)

wherePis found by differentiating the first of Eq. (2) with respect to time:Pcos D R

LPcos ) PD R

LPcos

cos : (4)

Realizing thatPD !BC andPD !AB , we have

!BCD !AB RL

cos

cos ) E!BCD !AB R

L

cos

cos Ok:

Also, noting thatEvCD PyCO| , and substituting Eq. (4) forPin Eq. (3) we have

PyCD R Psin L

R Pcos L cos

!R sin

L

D R Psin R

2 Psin cos L cos

: (5)

After substituting!AB into Eq. (5), and putting it in vector form we have

EvCDR!ABsin R

2!ABsin cos

L cos

O| :

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Dynamics 1e 835

Problem 6.74

A truck on an exit ramp is moving in such a way that, at the instant shown, jEaAj D 17 ft=s2,PD 0:3 rad=s,andRD 0:1 rad=s2. If the distance between points A and B isdABD 12 ft,D 57, andD 13,determineEaB .

Solution

Based on the picture shown, we letEaAD jEaAj.cos O{ sin O| /;E!ABD POk; EABD ROk; andErB=ADdAB.cos O{ C sin O| /:From the vector approach to acceleration analysis we have

EaBD EaA C EAB ErB=A !2ABErB=AD jEaAj.cos O{ sin / CROk dAB.cos O{ C sin O| / P2dAB.cos O{ C sin O| /D .jEaAj cos R dABsin P2dABcos / O{ C .jEaAj sin CR dABcos P2dABsin /O| ;

which can be evaluated to obtain

EaBD .17:0 O{ 5:38O| / ft=s2

;

using the given valuesjEaAj D 17 ft=s2,PD 0:3 rad=s,RD 0:1 rad=s2, dABD 12 ft, D 57, andD 13.

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LetLD4 ft, let point A travel parallel to the guide shown, and let Cbe themidpoint of the bar.

Problem 6.75 If pointA is accelerating to the right with aAD 27 ft=s2

andPD 7 rad=sD constant, determine the acceleration of point C whenD 24.Problem 6.76 If pointA is accelerating to the right with aAD 27 ft=s2,PD 7 rad=s, andRD 0:45 rad=s2, determine the acceleration of point CwhenD 26.Problem 6.77 If, whenD 0, A is accelerating to the right with aAD27 ft=s2 andEaCDE0, determineP andR.

Solution to 6.75

We haveEaAD aA O{;E!ABD POk; EABD ROkDE0; andErC=AD L2 .sin O{ cos O| /: From the vectorapproach to acceleration analysis we have

EaCD EaA C EAB ErC=A !2AB ErC=AD aA O{ L

P22

.sin O{ C cos O| /

D

aA L2

P2 sin O{ CP2L

2 cos O| ;


EaCD .12:9 O{ C 89:5O| / ft=s2:

using the given valuesL D 4 ft,aAD 27 ft=s2,PD 7 rad=s, andD 24.

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Solution to 6.77

We haveEaAD aA O{,E!ABD POk,EABD ROkDE0, andErC=A D L2 O| : From the vector approach toacceleration analysis we have

EaC

DEaA

C EAB

ErC=A

!2AB

ErC=A

E0 D aA O{ CROk L

2

O|CL P

2

2 O|

E0 D

aA CRL2

O{ CP2L

2 O| : (1)

Separating the components of Eq. (1) we have

0 D aA CRL2

) RD aA 2L

D 13:5 rad=s2;

and

0 DP2L2

) PD 0;

using the given valuesL D 4 ft,aAD 27 ft=s2,aCD 0 ft=s2, andD 0.

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Dynamics 1e 839

Problem 6.78

A wheelWof radiusRWD 5 cmrolls without slip over the stationary cylinderSof radiusRSD 12 cm, and the wheel is connected to pointO via the armOC. If!OCD constantD3:5 rad=s, determine the acceleration of point Q,which lies on the edge ofW and along the extension of the lineO C.

Solution

Let E be the contact point betweenW andS. From the no slip condition,

EvE DE0. We let E!W D !W Ok and from the geometry of the problemErC=ED RW.cos O{ C cos O| /. From rigid body kinematics of the wheel wehave

EvCD EvEC E!W ErC=ED !W Ok RWOurD !WRWOu : (1)

PointsC andO are also points on the arm where

ErC=O D .RS C RW/Our . From rigid body kinematics we must have

EvCD EvOC E!OC ErC=OD !OCOk .RSC RW/ OurD !OC.RSC RW/ Ou : (2)Equating Eqs. (1) and (2) we have

!WRWD !OC.RSC RW/ ) !WD RSC RWRW

!OC:

BecauseCis undergoing a uniform circular motion with radius .RSC RW/aboutO , we haveEaCD !2OC.RSC RW/Our :

From the vector approach to acceleration analysis we have

EaQD EaCC EW ErQ=C !2WErQ=CD !2OC.RSC RW/Our

RSC RWRW

2

!2OCRWOur

D !2OC.RSC RW/

1 C RSC RWRW

Our

D !2OC.RSC RW/

2 C RSRW

Our ;


9:16Ourm=s2:using the given valuesRWD 5 cm D 0:05000 m,RSD 12 cm D 0:1200 m, and!OCD 3:5 rad=s.

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Problem 6.79

One way to convert rotational motion into linear motion and vice versa is via

the use of a mechanism called the Scotch yoke, which consists of a crankC

that is connected to a sliderB via a pin A. The pin rotates with the crank while

sliding within the yoke, which, in turn, rigidly translates with the slider. Thismechanism has been used, for example, to control the opening and closing of

valves in pipelines. Letting the radius of the crank be RD25 cm, determinethe angular velocity !C and the angular accelerationCof the crank at the

instant shown ifD 25and the slider is moving to the right with a constantspeedvBD 40 m=s.

Solution

Choosing pointO as the origin of the reference frame, from the geometry of the

problem we have

ErAD R. cos O{ C sin O| /: (1)By construction, the velocity of the slider is equal to the horizontal component of

the velocity of point A. We can getEvA by taking the derivative of Eq. (1) withrespect to time. Doing so we have

EvAD R P.sin O{ C cos O| / ) EvBD R Psin O{ ) PD vBR sin

D 378:6 rad=s;

where we have used the given valuesRD25 cmD0:2500 m,vBD 40 m=s, andD 25. Realizing that!CD Pwe have

E!CD 379Ok rad=s:

Because we are toldvBD 0we must haved

dtvBD R sin RC R cos P2 D 0 ) RD

P2tan

:


RD 307103Ok rad=s2:

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Dynamics 1e 841

Problem 6.80

CollarC moves along a circular guide with radiusRD 2 ftwith a constantspeedvCD 18 ft=s. At the instant shown, the bars AB andB Care verticaland horizontal, respectively. LettingLD 4 ft and HD 5 m, determine theangular accelerations of the barsAB and B Cat this instant.

Solution

Note: The problem statement givesHD 5 m, which is incorrect. The solution will useHD 5 ft.Realizing thatB is the instantaneous center of rotation of bar AB andA is not moving, we must have

!ABD 0:BecauseB is also the instantaneous center of rotation of barB C, we must have

!BC

D

vC

L ) E!BC

D

vB

L

Ok:

SinceCmoves in a uniform circular motion about O at constant speed, we have

EaCD v2CR

O{;

and from the vector approach to acceleration analysis where,ErB=CD L O{ andEBCD BCOk, we have

EaBDEaCC EBC ErB=C !2BCErB=CD v

2C

RO{ C BCOk L O{

v2CR2

L O{

D v2

CR

1 C L

R

O{ C BCLO| : (1)Also from the vector approach to acceleration analysis whereErB=AD HO| andEABD ABOk we have

EaBD EaA C EAB ErB=A !2AB ErB=AD ABOk HO|D ABHO{: (2)

Equating components of Eqs. (1) and (2) we have

v2C

R

1 C L

R

D ABH and BCL D 0: (3)

Solving Eq. (3) forAB andBC, we have

ABD v2C

R

1 C L

R

) EABD 97:2Ok rad=s2 and EBCDE0:

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A ball of radiusRAD 5 in:is rolling without slip inside a stationary sphericalbowl of radiusRBD 17 in:Assume that the motion of the ball is planar.Problem 6.81 If, at the instant shown, the center of the ball is traveling

counterclockwise with a speedvAD 32 ft=sand such thatPvAD 0, determinethe acceleration of the center of the ball as well as the acceleration of the point

on the ball that is contact with the bowl.

Problem 6.82 If, at the instant shown, the center of the ball is traveling

counterclockwise with a speed vAD 32 ft=s and such thatPvAD 24 ft=s2,determine the acceleration of the center of the ball as well as the acceleration of

the point on the ball that is contact with the bowl.

Solution to 6.81

The center of the ball is making a uniform circular motion of radiusRB RAaboutO at constant speed, so we must have

EaAD v2A

RB RAOur (1)

Since the ball rolls without slip over a stationary surface, the point of

contact between the ball and the bowl, pointQ, is the instantaneous

center of rotation of the ball. Therefore,

vAD j E!AjRA ) !AD vARA

:

Notice that AD 0 because !A is constant. From the vector approach to acceleration analysis whereErQ=AD RAOur we must have

EaQD EaA C EA ErQ=A !2AErQ=AD v2A

RB RAOur

v2A

R2ARAOurD v2A

1

RB RAC 1

RA

Our (2)

Substituting given values vA D 32 ft=s, RA D 5 in:D 0:4167 ft, and RB D 17 in: D 1:417 ft intoEqs. (1) and (2) we have

EaAD 1020Ourft=s2;EaQD 3480Ourft=s2:

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Dynamics 1e 843

Solution to 6.82

The center of the ball is moving in a uniform circular motion of

radiusRB RAaboutO , as well as accelerating in the direction,so we must have

EaAD v2ARB RA

OurC PvAOu (3)

Substituting given values vA D 32 ft=s, PvA D 24 ft=s2, RA D0:4167 ft, andRBD 1:417 ft into Eq. (3), we have

EaAD .1020OurC 24:0Ou / ft=s2

Since the ball rolls without slip over a stationary surface, the point of contact between the ball and the bowl,

pointQ, is the instantaneous center of rotation of the ball. Therefore,

vAD j E!AjRA ) !AD vA

RA :

From the no slip condition, aQD 0. From the vector approach to acceleration analysis where ErQ=AD RAOurandEAD AOkwe must have

EaQD EaA C EA ErQ=A !2AErQ=AD v

2A

RB RAOurC PvAOuC ARAOu

v2A

R2ARAOur

D v2A

1

RB RAC 1

RA

OurC . PvA C ARA/Ou : (4)

However,PvA C ARAD 0, so we are left with

EaQD v2A

1

RB RAC 1

RA

Our : (5)

Substituting given valuesvAD 32 ft=s,PvAD 24 ft=s2,RAD 0:4167 ft, andRBD 1:417 ftinto Eq. (5), wehave

EaQD 3480Ourft=s2:

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Problem 6.83

A bar of length LD2:5 m is falling so that, whenD 34,vAD3 m=sandaAD 8:7 m=s2. At this instant, determine the angular acceleration of the barABand the acceleration of pointD, whereD is the midpoint of the bar.

Solution

We have ErB=AD L.sin O{ cos O| /and ErD=AD L2 .sin O{ cos O| /. From the vector approach to velocityanalysis whereEvAD vAO| andEvBD vBO{, we have

EvBD EvA C E!AB ErB=A ) vBO{D vAO|C !ABOk L.sin O{ cos O| /) vBO{D .L!ABcos / O{ C .L!ABsin vA/O| : (1)

Equating the two sides of Eq. (1) component by component, we have

0 D L!ABsin vA ) !ABD vAL sin

; and vBD L!ABcos : (2)

Substituting given valuesvAD 3 m=s,L D 2:5 m, andD 34 into Eqs. (2), we have!ABD 2:146 rad=s and vBD 4:448 m=s:

From the vector approach to acceleration analysis where

EaA

D aA

O| and

EaB

DaB

O{, we have

EaBD EaA C EAB ErB=A !2ABErB=A) aBO{D aAO|C ABOk L.sin O{ cos O| / !2ABL.sin O{ cos O| /

D . ABL cos !2ABL sin / O{ C . ABL sin C !2ABL cos aA/O| : (3)From theO| component of Eq. (3) we have

ABL sin C !2ABL cos aAD 0 ) ABD aA !2ABL cos

L sin : (4)

Substituting known valuesaAD8:7 m=s2,LD2:5 m,!ABD 2:146 rad=s, andD 34 into Eq. (4), wehave

EABD 0:604Ok rad=s2:ForD we have

EaDD EaA C EAB ErD=A !2ABErD=AD aAO|C ABOk L

2.sin O{ cos O| / !2AB

L

2.sin O{ cos O| /

D"

aA !2ABL cos L sin

!L

2 cos !2AB

L

2 sin

#O{

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Dynamics 1e 845

C"

aA !2ABL cos L sin

!L

2 sin C !2AB

L

2 cos aA

#O|

D

1

2.aA !2ABL cos / cot !2AB

L

2 sin

O{

C 12 .aA !2ABL cos / C !2AB L2 cos aAO| (5)Substituting known valuesaAD8:7 m=s2,LD2:5 m,!ABD 2:146 rad=s, andD 34 into Eq. (5), wehave

EaDD .3:84 O{ 4:35O| / m=s2:

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Problem 6.84

A bar of length LD 8 ft and midpoint D is falling so that when D 27,jEvDj D 18 ft=s, and the vertical acceleration of pointD is 23 ft=s2 downward.At this instant, compute the angular acceleration of the bar and the acceleration

of pointB .

Solution

We have ErB=AD L.sin O{ cos O| /and ErD=AD L2 .sin O{ cos O| /. Realizingthat pointH is the instantaneous center of rotation of the lineHD, where the

distance betweenH

andD

isdHDD

L

2, we have

!ABD vDL=2

D 4:500 rad=s:

From the vector approach to acceleration analysis whereEABD ABOk,EaAD aAO| andE!ABD !ABOk we have

EaDD EaA C EAB ErD=A !2ABErD=AD aAO|C AB L

2 cos O{ C AB L

2 sin O| !2AB

L

2.sin O{ cos O| /

D . ABL cos !2ABL sin / O{ C

aA C AB L2

sin C !2AB L2 cos O| (1)

We also have

EaBD EaA C EAB ErB=A !2ABErB=AD aAO|C ABOk L.sin O{ cos O| / !2ABL.sin O{ cos O| /D . ABL cos !2ABL sin / O{ C .aA C ABL sin C !2ABL cos /O| (2)

Recalling thataDy is given and aByD 0, we have

aByD

0

DaA

CABL sin

C!2

AB

L cos (3)

aDyD 23:00 ft=s2 D aA C AB L2

sin C !2ABL

2 cos (4)

Solving Eq. (3) foraAand substituting into Eq. (4) we have

23:00 ft=s2 D ABL sin C !2ABL cos C AB L2 sin C !2AB L2 cos D AB L

2 sin !2AB

L

2 cos ) ABD 46:00 ft=s

2

L sin !2ABcot (5)

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Dynamics 1e 847

Substituting given valuesL D 8 ft,D 27, and!ABD 4:500 rad=s into Eq. (3), we have

ABD 27:08 rad=s2 ) EABD 27:1Ok rad=s2:

Substituting known valuesL

D8 ft,

D27,AB

D 27:08 rad=s2, and!AB

D4:500 rad=sinto Eq. (3),

we have

aAD 45:99 ft=s2:Substituting known values LD 8 ft, D 27, ABD 27:08 rad=s2, !ABD 4:500 rad=s, and aAD45:99 ft=s2 into Eq. (1) we have

EaDD .133 O{ 23:0O| / ft=s2:

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Dynamics 1e 849

Problem 6.86

A truck on an exit ramp is moving in such a way that, at the instant shown, jEaAj D 6 m=s2 andD 13.Let the distance between pointsAand B bedABD 4 m. If, at this instant, the truck is turning clockwise,D 59, aBxD 6:3 m=s2, and aByD 2:6 m=s2, determine the angular velocity and angular accelerationof the truck.

Solution

From the information given in the problem we have

ErB=AD dAB.cos O{ C sin O| /; EABD ABOk; and EaAD jEaAj.cos O{ sin O| /: (1)From rigid body kinematics we have

EaBD EaA C EAB ErB=A !2ABErB=A (2)Substituting Eqs. (1) into Eq. (2) we have

EaBD jEaAj.cos O{ sin O| / C ABOk dAB.cos O{ C sin O| / !2

ABdAB.cos O{ C sin O| /D .jEaAj cos ABdABsin !2ABdABcos / O{C .jEaAj sin C ABdABcos !2ABdABsin /O| (3)

Breaking Eq. (3) into components we have

aBxD jEaAj cos ABdABsin !2ABdABcos (4)aByD jEaAj sin C ABdABcos !2ABdABsin (5)

Eqs. (4) and (5) are two equations inAB and!AB . Solving them we have

!ABD sjEaAj.cos cot sin / aBx cot aBydAB.cos cot sin / ;

(6)

and

ABD aBycos jEaAj.sin cos C sin cos / aBx sin

dAB: (7)

Substituting known values dABD 4 m, D 13, D 59, aBx D 6:3 m=s2, aByD 2:6 m=s2, andjaAj D 6 m=s2 into Eqs. (6) and (7) we have

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In vector form, and to three significant figure we have

E!ABD 0:458Ok rad=s EABD 0:258Ok rad=s2:

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Dynamics 1e 851


The system shown consists of a wheel of radius RD 1:4 m rolling without slipon a horizontal surface. A barAB of lengthL D 3:7 m is pin-connected to thecenter of the wheel and to a slider Aconstrained to move along a vertical guide.

PointCis the bars midpoint.

Problem 6.87 If the wheel is rolling clockwise with a constant angular speed

of2 rad=s, determine the angular acceleration of the bar

dynamics solns ch06

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