dynamics solns ch06
TRANSCRIPT
-
8/13/2019 Dynamics Solns Ch06
1/187
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.
-
8/13/2019 Dynamics Solns Ch06
2/187
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.
-
8/13/2019 Dynamics Solns Ch06
3/187
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:
We welcome your input.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
4/187
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
5/187
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 :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
6/187
-
8/13/2019 Dynamics Solns Ch06
7/187
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)
Plugging in known values we get
EDD 105Ok rad=s2:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
8/187
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
9/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
10/187
760 Solutions Manual
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
11/187
-
8/13/2019 Dynamics Solns Ch06
12/187
762 Solutions Manual
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
13/187
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
14/187
764 Solutions Manual
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
15/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
16/187
766 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
17/187
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
`:
Plugging in known values we have
!BD 1:25 rpm:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
18/187
768 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
19/187
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:
Plugging in known values we get
E!ramD v0H
OkD 8:57Ok rad=s :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
20/187
770 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
21/187
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 :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
22/187
772 Solutions Manual
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 :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
23/187
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 :
Plugging in known values we get
EaQ
D
24:6
103
OuR
28:0
Ou ft=s
2:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
24/187
774 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
25/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
26/187
776 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
27/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
28/187
778 Solutions Manual
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 :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
29/187
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 :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
30/187
780 Solutions Manual
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 :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
31/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
32/187
-
8/13/2019 Dynamics Solns Ch06
33/187
-
8/13/2019 Dynamics Solns Ch06
34/187
784 Solutions Manual
Plugging in known values we get
vcyclistD RCRS
RW!CD 17:6 m=s:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
35/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
36/187
-
8/13/2019 Dynamics Solns Ch06
37/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
38/187
788 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
39/187
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)
Plugging in known values we get
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
40/187
790 Solutions Manual
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| :
Plugging in known values we get
EvCD .5:42 O{ C 0:272O| / m=s:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
41/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
42/187
792 Solutions Manual
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
43/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
44/187
794 Solutions Manual
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:
Plugging in known values we get
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 :
Plugging in known values we get
EvQD 0:154Oum=s:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
45/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
46/187
796 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
47/187
Dynamics 1e 797
Problems 6.42 and 6.43
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
48/187
798 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
49/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
50/187
800 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
51/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
52/187
-
8/13/2019 Dynamics Solns Ch06
53/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
54/187
804 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
55/187
-
8/13/2019 Dynamics Solns Ch06
56/187
806 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
57/187
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:
From rigid body kinematics we have
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)
Substituting Eq. (11) into Eq. (10) we have
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
-
8/13/2019 Dynamics Solns Ch06
58/187
808 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
59/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
60/187
810 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
61/187
-
8/13/2019 Dynamics Solns Ch06
62/187
812 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
63/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
64/187
-
8/13/2019 Dynamics Solns Ch06
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
66/187
-
8/13/2019 Dynamics Solns Ch06
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
68/187
818 Solutions Manual
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:
From rigid body kinematics we have
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)
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
69/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
70/187
820 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
71/187
Dynamics 1e 821
Problems 6.63 through 6.65
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
72/187
822 Solutions Manual
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
73/187
Dynamics 1e 823
Solution to 6.65
From the geometry of the problem, we have
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
74/187
824 Solutions Manual
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)
Equating Eqs. (1) and (2) we have
.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)
Equating Eqs. (3) and (4) we have
2.RSC RP/!OCD 2!R.RSC RP/ ) !RD !OC:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
75/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
76/187
-
8/13/2019 Dynamics Solns Ch06
77/187
Dynamics 1e 827
Problems 6.68 through 6.72
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
-
8/13/2019 Dynamics Solns Ch06
78/187
828 Solutions Manual
Plugging in known values we have
EvBD 1:92O|m=s:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
79/187
Dynamics 1e 829
Solution to 6.69
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 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)
Substituting Eq. (2) into Eq. (1) we have
PD15p
L2 .H R cos /2R.R cos H / :
Plugging in known values we have
E!ODPOkD 877Ok rad=s:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
80/187
830 Solutions Manual
Solution to 6.70
From the rigid body kinematics of the wheel, and letting!ODPwe have
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
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 sin !AB
qL2 .H R cos /2
O{ C !AB.H R cos / C R Pcos O| : (3)
BecauseB is confined to move vertically in the slot,vBxD 0. From this we have
!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| :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
81/187
Dynamics 1e 831
Solution to 6.71
From the geometry of the problem, we have
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)
Substituting Eq. (6) into Eq. (7) we have
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)
Substituting Eq. (11) into Eq. (8) we have
vBD R Pcos R Psin H Rcos
Lp
L2 .H R cos /2:
Knowing thatB is confined to theO| direction, we have
EvBD
R Pcos RPsin .H R cos /p
L2 .H R cos /2
!O| :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
82/187
832 Solutions Manual
Solution to 6.72
Converting units on known values we have
RD 0:020 m;L
D0:080 m;
HD 0:038 m:
From the rigid body kinematics of the wheel, and letting!ODPwe have
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
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 sin !AB
qL2 .H R cos /2
O{ C !AB.H R cos / C R Pcos O| : (12)
BecauseB is confined to move vertically in the slot,vBxD 0. From this we have
!ABD RPsin
pL2 .H R cos /2
: (13)
Plugging Eq. (13) into Eq. (12) we have
EvBD
R Pcos RPsin .H R cos /p
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
83/187
-
8/13/2019 Dynamics Solns Ch06
84/187
834 Solutions Manual
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| :
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
85/187
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
86/187
836 Solutions Manual
Problems 6.75 through 6.77
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| ;
which can be evaluated to obtain
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
87/187
-
8/13/2019 Dynamics Solns Ch06
88/187
838 Solutions Manual
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
89/187
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 ;
which can be evaluated to obtain
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.
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
90/187
840 Solutions Manual
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
:
Plugging in known values we have
RD 307103Ok rad=s2:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
91/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
92/187
842 Solutions Manual
Problems 6.81 and 6.82
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
93/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
94/187
844 Solutions Manual
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{
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
95/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
96/187
846 Solutions Manual
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)
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
97/187
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:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
98/187
-
8/13/2019 Dynamics Solns Ch06
99/187
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
!ABD 0:4577 rad=s and ABD 0:2582 rad=s:August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
100/187
850 Solutions Manual
In vector form, and to three significant figure we have
E!ABD 0:458Ok rad=s EABD 0:258Ok rad=s2:
August 10, 2009
-
8/13/2019 Dynamics Solns Ch06
101/187
Dynamics 1e 851
Problems 6.87 through 6.89
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