Solutions Manual
Engineering Mechanics: Dynamics 1st Edition
Gary L. Gray The Pennsylvania State University
Francesco Costanzo The Pennsylvania State University
Michael E. Plesha University of Wisconsin–Madison With the assistance of: Chris Punshon Andrew J. Miller Justin High Chris O’Brien Chandan Kumar Joseph Wyne
Version: August 10, 2009
The McGraw-Hill Companies, Inc.
Copyright © 2002–2010 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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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 versions should 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 6–10 are in color. Color will be added to the figures in Chapters 1–5 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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Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.
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Chapter 6 Solutions Problem 6.1 Letting RA D 7:2 in: and RB D 4:6 in:, and assuming that the belt does not slip relative to pulleys A and B, determine the angular velocity and angular acceleration of pulley B when pulley A rotates at 340 rad=s while accelerating at 120 rad=s2 .
Solution Let Q be a point on the periphery of pulley A and let P be a point on the periphery of pulley B. The no slip condition between the pulleys and the belt requires vQ D vP
)
RA !A D RB !B
)
!B D
!EB D 532 kO rad=s
:
RA !A : RB
(1)
Substituting in given data, we obtain:
Differentiating Eq. (1) with respect to time, we obtain ˛B D
RA ˛A ; RB
which, upon substituting in given data gives ˛EB D 188 kO rad=s2
:
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Problem 6.2 Letting RA D 8 in:, RB D 4:2 in:, and RC D 6:5 in:, determine the angular velocity of gears B and C when gear A has an angular speed j!A j D 945 rpm in the direction shown.
Solution Using a vector approach with the component system shown in the figure below, we define points G and F as the points on gears A and B, respectively, that are in contact at this instant. We also define points Q and P as the points on gears B and C , respectively, that are in contact at this instant.
For vEG and vEF we have vEG D !EA rEG=A D !A kO RA . {O/ D !A RA |O; vEF D !EB rEF =B D !B kO RB {O D !B RB |O: Since vEG D vEF , !A RA D !B RB
)
!B D
RA !A RB
)
!EB D
1800 kO rpm:
Similarly, we can find vEQ and vEP using RA !A kO RB . {O/ D RA !A |O; RB D !C kO RC {O D !C RC |O:
vEQ D !EB rEQ=B D vEP D !E C rEP =C Since vEQ D vEP , we have RA !A D !C RC
)
!C D
RA !A RC
)
!EB D 1160 kO rpm:
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Problem 6.3 Letting RA D 203 mm, RB D 107 mm, RC D 165 mm, and RD D 140 mm, determine the angular acceleration of gears B, C , and D when gear A has an angular acceleration with magnitude j˛A j D 47 rad=s2 in the direction shown. Note that gears B and C are mounted on the same shaft and they rotate together as a unit.
Solution We let O and H be the centers of gears A and B respectively. Because the gears can not slip relative to each other, we can enforce the no slip condition vEP =Q D 0E between the contact points P and Q. Using the rigid body velocity equation for gear A we have vEP D vEO C !A kO rEP =O
)
vEP D
!A RA |O:
!B D
RA !A : RB
Doing the same for gear B we have vEQ D vEH C !B kO rEQ=H
)
vEQ D !B RB |O:
Enforcing the no slip condition we must have vEP D vEQ C vEP =Q
)
vEP D vEQ
)
!A RA D !B RB
)
(1)
Differentiating Eq. (1) with respect to time we get ˛B D
RA ˛A RB
Plugging in known values we get ˛EB D
89:2 kO rad=s2 :
Because gears B and C are mounted on the same shaft, we have ˛EC D
89:2 kO rad=s2 :
Next, by considering gears C and D with contact points F and G and proceeding as we did for gears A and B, we get RC ˛D D ˛C : (2) RD Plugging in known values we get ˛ED D 105 kO rad=s2 : August 10, 2009
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Problem 6.4 The bevel gears A and B have nominal radii RA D 20 mm and RB D 5 mm, respectively, and their axes of rotation are mutually perpendicular. If the angular speed of gear A is !A D 150 rad=s, determine the angular speed of gear B.
Solution Since the gears do not slip over their nominal circles relative to one another, we can write !A RA D !B RB
)
!B D
RA !A D 600 rad=s. RB
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Problem 6.5 A rotor with a fixed spin axis identified by point O is accelerated from rest with an angular acceleration ˛O D 0:5 rad=s2 . If the rotor’s diameter is d D 15 ft, determine the time it takes for the points at the outer edge of the rotor to reach a speed v0 D 300 ft=s. Finally, determine the magnitude of the acceleration of these points when the speed v0 is achieved.
Solution Points on the outer edge of the rotor have a speed v D d2 !, where ! is the angular speed of the rotor. Letting !0 be the angular speed corresponding to v0 , we must have v0 D 12 d!0
)
!0 D
2v0 : d
If the angular acceleration is constant, and t0 is the time taken to go from rest to !0 , we have ˛O t0 D !0 D
The angular acceleration of points a distance aEd D ˛EO rEd=O
!02 rEd=O
)
d 2
2v0 d
)
s
2v0 . ˛O d
from O is
aEd D ˛O kO 21 d uO r
ˇ ˇ ˇaEd ˇ D
t0 D
2v02 uO r d
)
aEd D
2v02 uO r C 12 ˛O d uO ; d
2 2 ˛O d 4v04 C D 12;000 ft=s2 : 2 d 4
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Problem 6.6 Do points A and B 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 between A and B change?
Solution Points A and B move relative to one another in the sense that the velocity of A is different from the velocity of B. The difference between the velocity of A and the velocity of B arises from the fact that points A and B are at different distances from the axis of rotation. As far as the distance between A and B is concerned, if we model the gear as a rigid body, then that distance must remain constant.
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Problem 6.7 A Pelton turbine (a type of turbine used in hydroelectric power generation) is spinning at 1100 rpm when the water jets acting on it are shut off, thus causing the turbine to slow down. Assuming that the angular deceleration rate is constant and equal to 1:31 rad=s2 , determine the time it takes for the turbine to stop. In addition, determine the number of revolutions of the turbine during the spin-down.
Solution Letting ! and ˛ be the angular velocity and angular acceleration of the turbine respectively, we have ˛ D 1:31 rad=s2 , and at t D 0, !.0/ D 1100 rpm: Since the angular acceleration is constant we must have !.t / D !.0/ C t ˛:
(1)
At the stop time ! D 0, so we have !.tstop / D 0: Plugging this into Eq. (1) we have tstop D
!.0/ D 87:9 s: ˛
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Problem 6.8 The velocities of points A and B on a disk, which is undergoing planar motion, are such that jE vA j D jE vB j. Is it possible for the disk to be rotating about a fixed axis going through the center of the disk at O? 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 of B would have been perpendicular to the segment OB. However, the figure shows that the velocity vector of B 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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Problem 6.9 The velocity of a point A and the acceleration of a point B 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 at O? Explain.
Solution Yes, it is possible that the motion of the disk is a fixed-axis rotation about O. The velocity vector of point A is 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 disk’s 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 point B would be directed toward the center O. 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 vector aEB has a component that is opposite to the vector vEB D !E O rEB=O . This in turn indicates that the speed of point B is decreasing and therefore that the spin rate of the disk as a whole is slowing down.
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Problem 6.11 The sprinkler shown consists of a pipe AB mounted on a hollow vertical shaft. The water comes in the horizontal pipe at O and goes out the nozzles at A and B, causing the pipe to rotate. Letting d D 7 in:, determine the angular velocity of the sprinkler !s , and jE aB j, the magnitude of the acceleration of B, if B is moving with a constant speed vB D 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, we know that vEB D !S kO rEB=O : Defining the position vector from O to B as shown to the right, we also have vEB D !S kO .d uO R C h uO ´ / D d!S uO
)
d!S D vB :
Plugging in known values and solving, we have !E S D
vB D 34:3 uO ´ rad=s: d
Since O is not in the plane of motion of B, we have aEB D !E S .!E S rEB=O / D
!S2 d
uO r
)
2 vB : jE aB j D d
Plugging in known values we have jE aB 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 the maximum constant angular speed of the gondolas if the magnitude of the acceleration of point A is not to exceed 2:5g.
Solution Under the stated conditions, point A would be moving in uniform circular motion. Therefore, we would have p ˇ ˇ ˇaEA ˇ D ! 2 ` 2:5g ) !OA 2:5g=`: OA Using the given numerical value of `, we therefore have .!OA /max D 2:48 rad=s:
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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 sprocket A be d D 2:5 ft and the radius of sprocket B be ` D 2 ft, determine the angular speed of wheel B if the sprocket A is rotating at 1 rpm.
Solution Because the track does not slip relative to the sprocket A we must have !A
d ` D !B 2 2
)
d !B D !A : `
Plugging in known values we have !B D 1:25 rpm:
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Problem 6.14 A battering ram is suspended in its frame via bars AD and BC , which are identical and pinned at their endpoints. At the instant shown, point E on the ram moves with a speed v0 D 15 m=s. Letting H D 1:75 m and D 20ı , determine the magnitude of the angular velocity of the ram at this instant.
Solution Since bars AD and BC are identical and parallel to one another, we must conclude that the velocities of points A and B are equal to one another. Since point A and B are coplanar and distinct, then we must conclude that the ram is simply translating, which implies that j!E ram j D 0:
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Problem 6.15 A battering ram is suspended in its frame via bars OA and OB, which are pinned at O. At the instant shown, point G on the ram is moving to the right with a speed v0 D 15 m=s. Letting H D 1:75 m, determine the angular velocity of the ram at this instant.
Solution The ram and its supports OA and OB form a single rigid body, therefore we must have vEG D !ram kO vEG=O D ! kO . H / |O D !ram H {O: We are told that jE vG j D v0 ; which means that !ram H D v0 : Plugging in known values we get !E ram D
v0 O k D 8:57 kO rad=s H
:
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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. 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 D 8:06 rad=s2 . Problem 6.16
Determine the velocity and acceleration of G as a function of Problem 6.17 P and . R the geometric parameters shown, ,
Solution to 6.16 Modeling the system as rigid and observing that point O is fixed, we have vEH D vEO C !Ep rEH=O D !Ep rEH=O ;
(1)
where the vectors !Ep and rEH=O are given by !Ep D P kO D
5:98 kO rad=s
and
rEH=O D .` C h/ uO r
d uO D .0:565 m/ uO r
.0:025/ uO :
(2)
Substituting the expressions in Eqs. (2) into Eq. (1) and carrying out the cross product, we have vEH D . 0:15 uO r
3:38 uO / m=s:
Again recalling that O is a fixed point, for the acceleration, we have aEH D ˛Ep rEH=O
!p2 rEH=O ;
(3)
where ˛Ep D R kO D 8:06 kO rad=s is the angular acceleration of the pendulum. Using this expression for ˛Ep and the expression for rEH=O in the last of Eqs. (2), after simplifying, we have aEH D . 20:4 uO r
3:66 uO / m=s2 :
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Solution to 6.17 Modeling the system as rigid and observing that point O is fixed, we have vEH D vEO C !Ep rEH=O D !Ep rEH=O ;
(4)
where the vectors !Ep and rEH=O are given by !Ep D P kO
and rEH=O D .` C h/ uO r
d uO :
(5)
Substituting the expressions in Eqs. (5) into Eq. (4) and carrying out the cross product, we have vEH D P d uO r C P .` C h/ uO : Again recalling that O is a fixed point, for the acceleration, we have aEH D ˛Ep rEH=O
!p2 rEH=O ;
(6)
where ˛Ep D R kO is the angular acceleration of the pendulum. Using this expression for ˛Ep and the expression for rEH=O in the second of Eqs. (5), after simplifying, we have h aEH D R d
i h i P 2 .` C h/ uO r C R .` C h/ C P 2 d uO :
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Problem 6.18 At the instant shown the paper is being unrolled with a speed vp D 7:5 m=s and an acceleration ap D 1 m=s2 . If at this instant the outer radius of the roll is r D 0:75 m, determine the angular velocity !s and acceleration ˛s of the roll.
Solution Referring to the figure shown to the right, the spool is in a fixed axis rotation about its center O. 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. Let P be a point on the outmost layer of paper that is just coming off the spool and let Q 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 have vEP D vEQ , i.e., vEP D
vP {O D vEQ D !E s rEQ=O ;
(1)
where !E s D !s kO is the angular velocity of the spool and rEQ=O D Eq. (1) and solving for !s we have !E s D
.vP =r/ kO
r |O. Substituting these expressions in
:
Enforcing the no slip condition for the acceleration gives that aEQ {O D aEP {O, that is, the components of the accelerations of Q and P are the equal in the direction tangent to the contact. Now, observe that aEP D
aP {O
and
aEQ D ˛Es . r |O/
!s2 . r |O/ D ˛s r {O
Using the expressions in Eq. (2) to enforce the no slip condition we have ˛s r D for ˛s to obtain ˛Es D
.ap =r/ kO
vP2 |O: r
(2)
aP , which can be solved
:
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Problem 6.19 At the instant shown, the propeller is rotating with an angular velocity !p D 400 rpm in the positive ´ direction and an angular acceleration ˛p D 2 rad=s2 in the negative ´ direction, where the ´ axis is also the spin axis of the propeller. Consider the cylindrical coordinate system shown, with origin O on the ´ axis and unit vector uO R pointing toward point Q on the propeller which is 14 ft away from the spin axis. Compute the velocity and acceleration of Q. Express your answer using the component system shown.
Solution The propeller is a rigid body, so we must have vEQ D !EP rEQ=O D !P uO ´ RQ uO R D !P RQ uO where !P D 400 2 60 D 41:89 rad=s and RQ = 14 ft. Plugging in known values we get vEQ D 586 uO ft=s: Similarly for aEQ we get aEQ D D D
!P2 rEQ=O C ˛P kO rEQ=O ! 2 RQ uO R C ˛P kO RQ P !P2 RQ
uO R C ˛P RQ uO :
Plugging in known values we get aEQ D Π24:6103 uO R
28:0 uO ft=s2 :
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Problem 6.20 The wheel A, with diameter d D 5 cm, is mounted on the shaft of the motor shown and is rotating with a constant angular speed !A D 250 rpm. The wheel B, with center at the fixed point O, is connected to A with a belt, which does not slip relative to A or B. The radius of B is R D 12:5 cm. At point C the wheel B is connected to a saw. If point C is at distance ` D 10 cm from O, determine the velocity and acceleration of C when D 20ı . Express your answers using the component system shown.
Solution Since the belt does not slip relative to A or B, we have !A
d D !B R 2
)
!B D
d !A : 2R
(1)
Next, observing that the crank is in a fixed axis rotation about O, we have vEC D !EB rEC =O ;
(2)
where the
d!A O k and rEC =O D `.cos {O C sin |O/: 2R Substituting Eqs. (3) into Eq. (2), after simplifying we obtain !EB D
vEC D
(3)
d `!A . sin {O C cos |O/ D . 0:179 {O C 0:492 |O/ m=s: 2R
Recalling that the angular velocity of the wheel B is constant and that O is a fixed point, we have aEC D
!B2 rEC =O :
(4)
Hence we have aEC D
d 2 `!A2 .cos {O C sin |O/ D 4R2
.2:58 {O C 0:938 |O/ m=s2 :
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Problem 6.21 In a contraption built by a fraternity, a person is sitting at the center of a swinging platform with length L D 12 ft that is suspended via two identical arms each of length H D 10 ft. Determine the angle and the angular speed of the arms if the person is moving upward and to the left with a speed vp D 25 ft=s at the angle D 33ı .
Solution The platform and its rider are undergoing a curvilinear translation. Therefore, the velocity vector of the person, vEP , is equal to the velocity vectors of points C and D. vEC D vEP D vED
)
D
)
D 33:0ı :
Also, we must have vP D j!AB jH D j!CD jH: Plugging in known values we get j!AB j D j!CD j D
vP D 2:50 rad=s: H
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Problem 6.22 A geosynchronous equatorial orbit is a circular orbit above the Earth’s equator that has a period of 1 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 rotation rate 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 satellite’s orbit. For the satellite to be geosynchronous, the satellite’s angular velocity must be such that it moves through 2 radians every day !s D
2 rad 1 h D 72:710 24 h 3600 s
6
rad=s:
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Problem 6.23 The bucket of a backhoe is being operated while holding the arm OA fixed. At the instant shown, point B has a horizontal component of velocity v0 D 0:25 ft=s and is vertically aligned with point A. Letting ` D 0:9 ft, w D 2:65 ft, and h D 1:95 ft, determine the velocity of point C . In addition, assuming that, at the instant shown, point B is not accelerating in the horizontal direction, compute the acceleration of point C . Express your answers using the component system shown.
Solution Since the bucket is in a fixed axis rotation about point A, we have vEB D !AB kO rEB=A D v0 {O
!AB kO ` |O D v0 {O
)
)
`!AB {O D v0 {O v0 D ) !AB D `
0:2778 rad=s;
where !AB is the angular velocity of the bucket. Since B is not accelerating in the horizontal direction, ˛AB D 0 and we have vEC D !AB kO rEC =A D !AB Œw {O
.h
`/ |O
)
vEC D .h
`/!AB {O C !AB w |O:
Upon substituting in given quantities, we have vEC D . 0:292 {O Since ˛AB D 0, aEC D
2 !AB rEC =A D
v02 Œw {O
.h
0:736 |O/ ft=s: `/ |O, which means that
aEC D . 0:204 {O C 0:0810 |O/ ft=s2 :
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Problem 6.24 Wheels A and C are mounted on the same shaft and rotate together. Wheels A and B are connected via a belt and so are wheels C and D. 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 A has an angular velocity !A D 2 rad=s and an angular acceleration ˛A D 0:5 rad=s2 , determine the angular velocity and acceleration of wheels B and D. The radii of the wheels are RA D 1 ft, RB D 0:25 ft, RC D 0:6 ft, and RD D 0:75 ft.
Solution Since the belts do not slip relative to the wheels they connect, we must have that !A RA D !B RB ;
˛A RA D ˛B RB ;
!C RC D !D RD ;
˛C RC D ˛D RD ;
where !A D !C ;
and
˛A D ˛C :
Using the given quantities, we have RA !A kO D 8:00 kO rad=s ; RB RA ˛EB D ˛A kO D 2:00 kO rad=s2 ; RB RC !ED D !C kO D 2:50 kO rad=s ; RD RC ˛ED D ˛C kO D 0:0625 kO rad=s2 : RD !EB D
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Problem 6.25 An acrobat lands at the end A of a board and, at the instant shown, point A has a downward vertical component of velocity v0 D 5:5 m=s. Letting D 15ı , ` D 1 m, and d D 2:5 m, determine the vertical component of velocity of point B 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 vEA D !EAB rEA=O D !AB kO `.cos {O C sin |O/ D !AB `. sin {O C cos |O/: Looking at the y component of vEA we have vAy D
v0 D !AB ` cos
)
!AB D
v0 : ` cos
Similarly, for point B we know that vEB D !AB kO rEB=O D
v0 O k . d cos {O ` cos
d sin |O/ D
v0 d v0 d tan {O C |O: ` `
Plugging known values into the y component of vEB we have vBy D
v0 d D 13:8 m=s `
:
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Problem 6.26 At the instant shown, A is moving upward with a speed v0 D 5 ft=s and acceleration a0 D 0:65 ft=s2 . Assuming that the rope that connects the pulleys does not slip relative to the pulleys and letting ` D 6 in: and d D 4 in:, determine the angular velocity and angular acceleration of pulley C .
Solution We can define the length of the rope as L D yD C 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 yPD C 2yPA
)
yPD D
2yPA D 2v0
0 D yRD C 2yRA
)
yRD D
2yRA D 2a0
Because the point Q is contacting the rope, and the rope is not slipping, point Q has the same velocity and vertical component of acceleration as D. vQ D vD D 2v0 D !C
` 2
)
!E C D 4
v0 D 40:0 kO rad=s `
aQy D aD D 2a0 D ˛C
` 2
)
˛EC D 4
a0 D 5:20 kO rad=s2 `
;
:
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Problem 6.27 At the instant shown, the angle D 30ı , jE vA j D 292 ft=s, and the turbine is rotating clockwise. Letting OA D R, OB D R=2, R D 182 ft, and treating the blades as being equally spaced, determine the velocity of point B at the given instant and express it using the component system shown.
Solution At the instant shown D 30ı , so the positions of A and B in the given cartesian coordinate system are rEA D rEA=O D R.sin {O C cos |O/ R rEB D rEB=O D {O: 2 Because the windmill is being treated as a rigid body, we also know vEA D !E T rEA=O D !T kO rEA=O D !T R. cos {O C sin |O/
)
vA D j!T jR D 292 ft=s
where !E T is the angular velocity of the turbine and vA is the speed of A. Solving for !T we have !T D
vA R
because the turbine is rotating clockwise. The velocity of point B is then vEB D !E T rEB=O D vEB D
vA O k R
R {O 2
vA |O D 146 |O ft=s: 2
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Problem 6.28 At the instant shown, the angle D 30ı , the turbine is rotating clockwise, and aEB D .70:8 {O 12:8 |O/ m=s2 . Letting OA D R, OB D R=2, R D 55:5 m, and assuming the blades are equally spaced, determine the angular velocity and angular acceleration of the turbine blades as well as the acceleration of point A at the given instant.
Solution At the instant shown D 30ı , so the positions of A and B in the given cartesian coordinate system are rEA D rEA=O D R.sin {O C cos |O/ R rEB D rEB=O D {O: 2 Because the windmill is being treated as a rigid body, we also know !T2 rEB=O C ˛T kO rEB=O R R 2 O {O C ˛T k {O D !T 2 2 R R D !T2 {O ˛T |O D aBx {O C aBy |O: 2 2
aEB D
Therefore, r !E T D ˛ET D
2aBx O k D 1:60 kO rad=s; R 2aBy O k D 0:461 kO rad=s2 : R
The negative root is chosen for !T because the turbine is rotating clockwise. The acceleration of A is aEA D
!T2 rEA=O C ˛T kO rEA=O
D
!T2 R.sin {O C cos |O/ C ˛T kO R.sin {O C cos |O/
D
R.!T2 sin C ˛T cos / {O C R.˛T sin
!T2 cos / |O:
Plugging in the known values of R; !T ; ˛T ; and we get aEA D . 93:0 {O
110 |O/ m=s2 :
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Problems 6.29 through 6.31 A bicycle has wheels 700 mm in diameter and a gear set with the dimensions given in the table below. Crank Sprocket No. of Cogs Radius (mm)
C1 26 52.6
C2 36 72.8
C3 48 97.0
Cassette (9 speeds) Sprocket No. of Cogs Radius (mm)
S1 11 22.2
S2 12 24.3
S3 14 28.3
S4 16 32.3
S5 18 36.4
S6 21 42.4
S7 24 48.5
S8 28 56.6
S9 34 68.7
If a cyclist has a cadence of 1 Hz, determine the angular speed of the rear wheel in rpm when using the combination of C3 and S2. In addition, knowing that the speed of the cyclist is equal to the speed of a point on the tire relative to the wheel’s center, determine the cyclist’s speed in m=s. Problem 6.29
Problem 6.30 If a cyclist has a cadence of 68 rpm, determine which combination of chain ring (a sprocket mounted on the crank) and (rear) sprocket would most closely make the rear wheel rotate with an angular speed of 127 rpm. Having found a chain ring/sprocket combination, determine the wheel’s exact angular speed corresponding to the chosen chain ring/sprocket combination and the given cadence.
If a cyclist is pedaling so that the rear wheel rotates with an angular speed of 16 rad=s, determine all possible (rear) sprocket/chain ring (crank-mounted sprocket) combinations that would allow him or her to pedal with a frequency within the range 1:00–1:25 Hz. Problem 6.31
Solution to 6.29 Using C3 and S2, RC D 97:0 mm; RS D 24:3 mm; and RW D 0:7 m where the subscripts C; S; and W denote the crank, sprocket and wheel respectively. We let !C ; !S ; !W be the angular velocities of the crank, sprocket, and wheel. The sprocket is fixed to the wheel, so !S D !W : Since the chain does not slip relative to the crank or sprocket, we have !S RS D !C RC
)
!S D !W D
RC !C : RS
Because the cyclist’s cadence is 1 Hz; !C D 2 rad=s: Plugging in known values, we get !W D 25:1 rad=s D 240 rpm: We are also told that vcyclist D vP =O where vEP =O is the velocity of a point on the rim of the wheel relative to the wheel’s center O. From this, we have vEP =O D !E W rEP =O
)
vcyclist D vP =O D !W RW : August 10, 2009
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Plugging in known values we get vcyclist D
RC RW !C D 17:6 m=s: RS
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Solution to 6.30 We let !C ; !W ; and !S be the angular velocities and RC ; RW and RS 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 !C RC D !S RS : Given !C D 68 rpm, we are asked to pick a sprocket and crank combination such that !S is close to 128 rpm. C In other words, R RS must be close to !S 127 D D 1:868: !C 68 By trial and error we find that pairing C1 with S3 gives RC 52:6 mm D D 1:86; RS 28:3 mm which is the closest pairing to the desired ratio. For this choice of gears and the given cadence, !W D !S D
RC 52:6 mm !C D 68 rpm D 126 rpm: RS 28:3 mm
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Solution to 6.31 We let !C ; !W ; and !S be the angular velocities and RC ; RW ; and RS be the radii of the crank, wheel, and sprocket respectively. The chain does not slip with respect to the crank or sprocket, and !S D !W so we must have RS !S RS D !W RS D !C RC ) !C D !W : (1) RC We need to find the combinations of crank and sprocket such that 2 rad=s !C 2:5 rad=s
(2)
Plugging Eq. (1) into Eq. (2) we must have that 0:393
RS 0:491: RC
By trial and error we find that all possible combinations include: RS RC RS S2 and C1: RC RS S4 and C2: RC RS S6 and C3: RC S1 and C1:
D 0:422 D 0:460 D 0:444 D 0:437:
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Problem 6.32 A carrier is maneuvering so that, at the instant shown, jE vA j D 25 knots (1 kn is exactly equal to 1:852 km=h) and D 33ı . Letting the distance between A and B be 220 m and D 22ı , determine vEB at the given instant if the ship’s turning rate at this instant is P D 2ı =s clockwise.
Solution Since A and B are on the same rigid body, we must have vEB D vEA C !E carrier rEB=A
(1)
vEA D vA .sin {O C cos |O/; O !E carrier D ! kO D . 2 ı =s/ kO D . 0:03491 rad=s/ k;
(2)
rEB=A D AB.cos {O C sin |O/:
(4)
where (3)
and
Substituting Eqs. (2)–(4) into Eq. (1) we have vEB D vA .sin {O C cos |O/ C ! kO AB.cos {O C sin |O/ D .vA sin
!AB sin / {O C .vA cos C !AB cos / |O:
Substituting in known values we get vEB D .9:88 {O C 3:67 |O/ m=s:
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Problem 6.33 At the instant shown, the pinion is rotating between two racks with an angular velocity !P D 55 rad=s. If the nominal radius of the pinion is R D 4 cm and if the lower rack is moving to the right with a speed vL D 1:2 m=s, determine the velocity of the upper rack.
Solution We define P and Q to be the contact points between the pinion and the top and bottom racks, respectively. Then we have vEQ D vEL D .1:2 m=s/ {O: Because Q and P are points on the same rigid body, we must also have vEP D vEQ C !EP rEP =Q D vL {O C !P kO .2R/ |O D .vL
2R!P / {O:
Substituting known values we get vEP D
3:2 {O m=s:
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Problem 6.34 At the instant shown the lower rack is moving to the right with a speed of vL D 4 ft=s, while the upper rack is fixed. If the nominal radius of the pinion is R D 2:5 in:, determine !P , the angular velocity of the pinion, as well as the velocity of point O, i.e., the center of the pinion.
Solution We define P and Q to be the contact points between the pinion and the top and bottom racks, respectively. Then we have vEQ D vEL D .1:2 m=s/ {O
and
E vEP D 0:
Because Q and P are points on the same rigid body, we must also have vEQ D !EP rEQ=P D vEL
)
!P kO . 2R/ |O D vEL
)
2R!P {O D vEL
)
!P D
vL : (1) 2R
Plugging in known values we get !EP D 9:60 kO rad=s: Points O and P are also on the same rigid body, so we also must have vEO D !EP rEO=P D !P kO . R/ |O D !P R {O:
(2)
Substituting Eq. (1) into Eq. (2) we have vEO D 2:00 {O ft=s:
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Problem 6.35 A bar of length L D 2:5 m is pin-connected to a roller at A. The roller is moving along a horizontal rail as shown with vA D 5 m=s. If at a certain instant D 33ı and P D 0:4 rad=s, compute the velocity of the bar’s midpoint C .
Solution Because C and A are points on the same rigid body, we must have vEC D vEA C !EAC rEC =A ; where vEA D vA {O;
O !EAC D P k;
rEC =A D
L .sin {O 2
(1)
cos |O/:
(2)
Plugging Eqs. (2) into Eq. (1) we have L vEC D vA {O C !AC kO .sin {O cos |O/ 2 L L D .vA C !AC cos / {O C !AC sin |O: 2 2 Plugging in known values we get vEC D .5:42 {O C 0:272 |O/ m=s:
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Problem 6.36 If the motion of the bar is planar, what would the speed of A need to be for vEC to be perpendicular to the bar AB? Why?
Solution The kinematic relation between points C and A is vEC D vEA C !EAC rEA=C : For vEC to be perpendicular to rEA=C in planar motion, we must have vEA D 0E
)
vA D 0:
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Problem 6.37 Points A and B are both on the trailer part of the truck. If the relative velocity of point B with respect to A is as shown, is the body undergoing a planar rigid body motion?
Solution For the motion to be a planar rigid body motion, vEB=A must be perpendicular to the vector rEB=A . The figure indicates that vEB=A and rEB=A are not perpendicular. Therefore, the motion of the trailer is not a planar rigid body motion.
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Problem 6.38 A truck is moving to the right with a speed v0 D 12 km=h while the pipe section with radius R D 1:25 m and center at C rolls without slipping over the truck’s bed. The center of the pipe section C is moving to the right at 2 m=s relative to the truck. Determine the angular velocity of the pipe section and the absolute velocity of C .
Solution We let !EP be the angular velocity of the pipe. Because C and Q can be considered points on the same rigid body, we must have vEC D vEQ C !EP rEC =Q
(1)
where O !EP D !P k;
vEQ D vET D v0 {O;
and
rEC =Q D R |O:
(2)
Substituting Eqs. (2) into Eq. (1) we have vEC D .v0
!P R/ {O:
(3)
We are also told that vEC =Q D !EP rEC =Q D vpipe/truck {O Plugging in known values we have !EP D
)
!P R {O D 2:000 {O m=s:
1:600; which to three significant figures is !EP D
1:60 kO rad=s:
Plugging the unrounded value for !EP and all known values into Eq. (3), and rounding to three significant figures we have vEC D 5:33 {O m=s:
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Problem 6.39 A wheel W of radius RW D 7 mm is connected to point O via the rotating arm OC, and it rolls without slip over the stationary cylinder S of radius RS D 15 mm. If, at the instant shown, D 47ı and !OC D 3:5 rad=s, determine the angular velocity of the wheel and the velocity of point Q, where point Q lies on the edge of W and along the extension of the line OC.
Solution Points C and O are two points on the arm, which is a rigid body, so we must have vEC D vEO C !E OC rEC =O ;
(1)
where !E OC D !OC kO
and
rEC =O D .RS C RW / uO r :
(2)
Substituting Eqs. (2) into Eq. (1) we have vEC D .RS C RW /!OC uO :
(3)
E and We let E be the contact point between cylinder S and wheel W , so vEE D 0, vEC D vEE C !E W rEC =E D !W kO RW uO r D !W RW uO :
(4)
Equating Eqs. (3) and (4) we have !W RW uO D !OC .RS C RW /uO
)
!W D
.RS C RW / !OC : RW
Plugging in known values we get !E W D 11:0 kO rad=s: Because Q is a point on the same rigid body as E, we also must have vEQ D vEE C !W kO 2RW uO r D 2RW !W uO D 2.RS C RW /!OC uO : Plugging in known values we get vEQ D 0:154 uO m=s: August 10, 2009
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Problem 6.40 For the slider-crank mechanism shown, let R D 20 mm, L D 80 mm, and H D 38 mm. Use the concept of instantaneous center of rotation to determine the values of , with 0ı 360ı , for which vB D 0. Also, determine the angular velocity of the connecting rod at these values of .
Solution When vB D 0, B is the instantaneous center of rotation of AB, and vEA must be perpendicular to the line BA. However, vEA must also be perpendicular to the line OA, since O is the center of rotation of the crank. Therefore, O, A, and B 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 1 D H
)
1 D cos
1
H LCR
1 D 67:7ı :
)
In the second case we have that .L
R/ cos 2 D H
)
2 D 180ı C cos
H
1
L )
R 2 D 231ı :
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Problem 6.41 At the instant shown bars AB and BC are perpendicular to each other, and bar BC is rotating counterclockwise at 20 rad=s. Letting L D 2:5 ft and D 45ı , determine the angular velocity of bar AB as well as the velocity of the slider C .
Solution Because D 45ı , the instantaneous center of rotation of bar BC is point A. Therefore, p vC D 2L!BC D 70:7 ft=s Because C is constrained to move along the guide with vector .O{ C |O/, and is moving up and to the right, we have vEC D .50:0 {O C 50:0 {O/ ft=s: A is also the center of rotation of bar AB. Therefore, in the present configuration vB
D
!AB L
D
!BC L
)
!AB
D
!BC
)
!EAB D 20:0 kO rad=s:
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Problems 6.42 and 6.43 A ball of radius RA D 3 in: is rolling without slip in a stationary spherical bowl of radius RB D 8 in: Assume that the ball’s motion is planar. Problem 6.42 If the speed of the center of the ball is vA D 1:75 ft=s and if the ball is moving down and to the right, determine the angular velocity of the ball.
If the angular speed of the ball j!A j D 4 rad=s is counterclockwise, determine the velocity of the center of the ball. Problem 6.43
Solution to 6.42 Converting RA to feet we have RA D 0:2500 ft: 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, vA D j!EA jRA
)
!A D
vA RA
)
!A D 7:000 rad=s
Observing that the ball will be rotating counterclockwise and given that kO D uO r uO , to three significant figures we have !EA D
7:00 kO 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, vA D j!EA jRA D 4RA
)
vEA D
12 uO in=s:
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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 crank C that is connected to a slider B 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 the radius of the crank be R D 1:5 ft, determine the angular velocity !C of the crank so that the maximum speed of the slider is vB D 90 ft=s.
Solution Calling the center of the wheel the origin, from the geometry of the problem we have rEA D 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 get vEA by taking the derivative of Eq. 1 with respect to time. Doing so we have vEA D RP . sin {O C cos |O/ ) vEB D RP sin {O: P for .vB /max we have Because RP sin varies as a function of , with its maximum value equal to Rjj, .vB /max D RjP j D R!C
)
!C D
.vB /max R
)
!E C D 60:0 kO rad=s:
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Problems 6.45 through 6.47 The system shown consists of a wheel of radius R D 14 in: rolling on a horizontal surface. A bar AB of length L D 40 in: is pin-connected to the center of the wheel and to a slider A that is constrained to move along a vertical guide. Point C is the bar’s midpoint. If, when D 72ı , the wheel is moving to the right so that vB D 7 ft=s, determine the angular velocity of the bar as well as the velocity of the slider A. Problem 6.45
If, when D 53ı , the slider is moving downward with a speed vA D 8 ft=s, determine the velocity of points B and C . Problem 6.46
If the wheel rolls without slip with a constant counterclockwise angular velocity of 10 rad=s, determine the velocity of the slider A when D 45ı .
Problem 6.47
Solution to 6.45 Converting units on known values, we have R D 1:167 ft
L D 3:333 ft:
We are told the wheel is moving to the right. Therefore, using the component system shown vEB D 7 {O ft=s:
(1)
In addition, the slider must be moving down, so we have vEA D vA |O; and the angular velocity of the bar is O Also, the position vector of B with respect to A is rEB=A D L.cos {O sin |O/: From rigid !EAB D !AB k: body kinematics, we must have vEB D vEA C !EAB rEB=A D vA |O C !AB kO L.cos {O D !AB L cos {O C .!AB L sin
sin |O/ vA / |O:
(2)
vB ; L sin
(3)
Equating Eq. (1) and Eq. (2) we have vB D !AB L sin
)
!AB D
vA D !AB L cos :
(4)
Plugging known values into Eq. (3) we get !AB D 2:208 rad=s: To three significant figures this is !EAB D 2:21 kO rad=s: Plugging the unrounded value for !AB into Eq. (4) we get vA D 2:27 ft=s
)
vEA D
2:27 |O ft=s: August 10, 2009
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Solution to 6.46 Converting units on known values, we have R D 1:167 ft L D 3:333 ft: We are told the slider is moving down. Therefore, using the component system shown, vEA D 8 |O ft=s: In addition, because the slider is moving down, the the wheel must be O and !E W D moving to the right, so we have vEB D vB {O; !EAB D !AB k, O Also, the position vector of B relative to A is rEB=A D L.cos {O !W k: sin |O/; and of B relative to Q is rEB=Q D R |O: From rigid body kinematics we must have vEB D vEA C !EAB rEB=A D vA |O C !AB kO L.cos {O D !AB L cos {O C .!AB L sin
sin |O/ vA / |O:
(5)
We must also have vEB D vEQ C !E W rEB=Q D !W kO R |O D
!W R {O:
(6)
Equating components of Eq. (5) with those of Eq. (6) we have !W R D !AB L sin ;
(7)
vA D !AB L cos :
(8)
Solving Eq. (8) gives !AB D 3:988 rad=s: Plugging !AB into Eq. (7) gives !W D !W into Eq. (6) gives
9:100 rad=s: Plugging
vEB D 10:6 {O ft=s: Realizing that the position vector from A to C is rEC =A D
rEB=A 2 ,
from rigid body kinematics we have
vEC D vEA C !EAB rEC =A L vA |O C !AB kO .cos {O 2 D !AB L cos {O C .!AB L sin D
sin |O/ vA / |O:
(9)
Plugging all known values into Eq. (9) gives vEC D .5:31 {O
4 |O/ ft=s:
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Solution to 6.47 Converting units on known values, we have R D 1:167 ft L D 3:333 ft: We are told that the wheel spins counterclockwise at 10 rad=s, so we have !E W D 10 kO rad=s:
(10)
In addition, because the wheel is turning counterclockwise, the slider is moving up, and B is moving to the left, we must have vEA D vA |O;
vEB D
vB {O;
and
O !EAB D !AB k:
(11)
Also, the position vector of B relative to A is rEB=A D L.cos {O rEB=Q D R |O: From rigid body kinematics we must have vEB D vEA C !EAB rEB=A D vA |O C !AB kO L.cos {O
sin |O/; and of B relative to Q is
sin |O/
D !AB L cos {O C .!AB L sin C vA / |O:
(12)
We must also have vEB D vEQ C !E W rEB=Q D !W kO R |O D
!W R {O D
11:67 {O ft=s:
(13)
Equating components of Eq. (12) with those of Eq. (13) we have 11:67 D !AB L sin ; vA D Solving Eq. (14) gives !AB D
(14)
!AB L cos :
(15)
4:952 rad=s: Substituting this value for !AB into Eq. (15), we have vEA D 11:7 |O ft=s:
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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 of 1:7 m=s. If the nominal radius of the pinion O is R D 0:25 m, determine the angular velocity of the pinion, as well as the position of the pinion’s instantaneous center of rotation relative to point O.
Solution From the no slip condition at points U and Q we know that vEU D
1:7 {O m=s
(1)
and vEQ D 2:7 {O m=s:
(2)
From rigid body kinematics we have vEU D vEQ C !EP rEU=Q D vL {O C !P kO 2R |O D vL {O
2R!P {O;
(3)
where !P is the angular speed of the pinion. Comparing Eq. (3) with Eq. (1) component by component, we have vL C 1:7 ) !P D 8:800 rad=s: 1:7 D vL 2R!P ) !P D 2R In vector form, and to three significant figures this is !EP D 8:80 kO rad=s: We also know that vEIC D vEQ C !EP rEIC =Q 0E D vL {O C !P kO Œ.dIC =Q /x {O C .dIC =Q /y |O 0E D ŒvL
.dIC =Q /y !P {O C .dIC =Q /x !P |O:
(4)
Breaking Eq. (4) into two scalar equations by components we have 0 D vL
.dIC =Q /y !P
0 D .dIC =Q /x !P
)
)
.dIC =Q /y D
vL ; !P
.dIC =Q /x D 0:
Substituting known values we have rEIC =Q D 0:307 |O rad=s:
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Problem 6.49 A carrier is maneuvering so that, at the instant shown, jE vA j D 22 kn, D 35ı , jE vB j D 24 kn (1 kn is equal to 1 nautical mile (nml) per hour or 6076 ft=h). Letting D 19ı and the distance between A and B be 720 ft, determine the ship’s turning rate at the given instant if the ship is rotating clockwise.
Solution O and rEB=A D dAB .cos {O C sin |O/. From rigid body We let vEA D vA .sin {O C cos |O/, !EA D !A k, kinematics we know vEB D vEA C !EAB rEB=A D vA .sin {O C cos |O/ C !AB kO dAB .cos {O C sin |O/ D .vA sin
!AB dAB sin / {O C .vA cos C !AB dAB cos / |O:
(1)
Taking the magnitude of both sides of Eq. (1) we have 2 vB D .vA sin
!AB dAB sin /2 C .vA cos C !AB dAB cos /2 :
(2)
All values in Eq. (2) are known except for !AB . Solving, we have !AB D ˙0:06805 rad=s: Since we are told that the ship is turning clockwise, using the component system shown we must have !EAB D
0:0681 kO rad=s:
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Problems 6.50 and 6.51 For the slider-crank mechanism shown, let R D 1:9 in:, L D 6:1 in:, and H D 1:2 in: Also, at the instant shown, let D 27ı and !AB D 4850 rpm. Problem 6.50
Determine the velocity of the piston at the instant shown.
Problem 6.51
Determine P and the velocity of point D at the instant shown.
Solution to 6.50 O Converting all given values to ft or rad=s we have We let rEB=A D R.cos {O C sin |O/ and !EAB D !AB k. R D 0:1583 ft;
L D 0:5083 ft;
H D 0:1000 ft;
and !AB D 507:9 rad=s:
From rigid body kinematics we have vEB D vEA C !EAB rEB=A D !AB kO R.cos {O C sin |O/ D
!AB R sin {O C !AB R cos |O:
(1)
O rEC =B D L.sin {O C cos |O/, and vEC D vC |O, then we have If we let !EBC D !BC k, vEC D vC |O D vEB C !EBC rEC =B : D vEB C !BC kO L.sin {O C cos |O/ D vEB
!BC L cos {O C !BC L sin |O:
(2)
Substituting Eq. (1) for vEB in Eq. (2) we get vC |O D
.!AB R sin C !BC L cos / {O C .!AB R cos C !BC L sin / |O:
(3)
From the geometry of the problem we also have L sin D R cos
)
Substituting Eq. (4) for in Eq. (3) we have vC |O D !AB R sin C !BC L cos sin
1
D sin
R cos L
1
R cos : L
(4)
{O C 2R cos .!AB C !BC / |O:
(5)
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Breaking Eq. (5) into two scalar equations by components we have R 0 D !AB R sin !BC cos sin 1 cos L vC D 2R cos .!AB C !BC /:
(6) (7)
Eqs. (6) and (7) are two equations in vC , !BC , and known values. Solving these equations for vC we get vEC D 61:1 |O ft=s:
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Solution to 6.51 O Converting all given values to ft=s or rad=s we have We let rEB=A D R.cos {O C sin |O/ and !EAB D !AB k. R D 0:1583 ft;
L D 0:5083 ft;
H D 0:1000 ft;
and !AB D 507:9 rad=s:
From rigid body kinematics we have vEB D !EAB rEB=A D !AB kO R.cos {O C sin |O/ D
!AB R sin {O C !AB R cos |O:
(8)
O rEC =B D L.sin {O C cos |O/, and vEC D vC |O, then from rigid body kinematics we If we let !EBC D !BC k, have vEC D vC |O D vEB C !EBC rEC =B : D vEB C !BC kO L.sin {O C cos |O/ D vEB
!BC L cos {O C !BC L sin |O:
(9)
Substituting Eq. (8) into Eq. (9) we get vC |O D
.!AB R sin C !BC L cos / {O C .!AB R cos C !BC L sin / |O:
(10)
From the geometry of the problem we also have L sin D R cos
)
Substituting Eq. (11) into Eq. (10) we have vC |O D !AB R sin C !BC L cos sin
1
D sin
R cos L
1
R cos : L
(11)
{O C 2R cos .!AB C !BC / |O:
(12)
Equating components of Eq. (12) we have 0D
!AB R sin
!BC
cos sin
1
R cos L
(13)
vC D 2R cos .!AB C !BC /:
(14)
Eqs. (13) and (14) are two equations in vC , !BC , and known values. Solving these equations for !BC , we get !BC D 74:76 rad=s: Because measures the angle of BC with respect to the horizontal axis, P D !BC . Therefore, P D
74:8 rad=s:
We let rED=B D H.sin {O C cos |O/. Therefore, from rigid body kinematics we have vED D vEB C !EBC rED=B D vEB C !BC kO H.sin {O C cos |O/ D vEB
!BC H cos {O C !BC H sin |O:
(15)
Substituting Eq. (8) into Eq. (15) we get vED D
.!AB R sin C !BC H cos / {O C .!AB R cos C !BC H sin / |O:
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Combining Eq. (11) with Eq. (16) to eliminate gives R H vED D !AB R sin C !BC H cos sin 1 cos {O C R cos !AB C !BC |O: L L
(17)
Substituting all known values into Eq. (17) we get vED D . 29:3 {O C 69:6 |O/ ft=s:
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Dynamics 1e
Problem 6.52 A wheel W of radius RW D 7 mm is connected to point O via the rotating arm OC, and it rolls without slip over the stationary cylinder S of radius RS D 15 mm. If, at the instant shown, D 63ı and !W D 9 rad=s, determine the angular velocity of the arm OC and the velocity of point P , where point P lies on the edge of W and is vertically aligned with point C .
Solution Let E be the contact point between W and S. From the no slip condition, E We let !E W D !W kO and from the geometry of the problem vEE D 0. rEC =E D RW .cos {O C cos |O/. From rigid body kinematics we have vEC D vEE C !E W rEC =E D !W kO RW .cos {O C sin |O/ D !W RW . sin {O C cos |O/:
(1)
Points C and O are also points on the arm where rEC =O D .RS C RW /.cos {O C sin |O/. Therefore, we must have vEC D vEO C !E OC rEC =O D !OC kO .RS C RW /.cos {O C sin |O/ D !OC .RS C RW /. sin |O C cos |O/:
(2)
Equating Eqs. (1) and (2) we have !W RW D !OC .RS C RW /
)
!E OC D
RW !W kO D 2:86 kO rad=s: RS C RW
Points P and C are both points on the wheel with rEP =C D RW |O. Therefore, we must have vEP D vEC C !E W rEP =C : Substituting in Eq. (1) for vEC we have vEP D !W RW . sin {O C cos |O/ C !W kO RW |O D !W RW Œ. sin
1/ {O C cos |O:
(3)
Substituting known values into Eq. (3) we get vEP D . 0:119 {O C 0:0286 |O/ m=s:
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Solutions Manual
Problem 6.53 At the instant shown, the center O of a spool with inner and outer radii r D 1 m and R D 2:2 m, respectively, is moving up the incline at speed vO D 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 length of rope being wound or unwound per unit time.
Solution E We also know vEO D vO {O , From the no slip condition, vEQ D 0. O rEO=Q D R |O, and !E S D !S k. From rigid body kinematics we have vEO D vEQ C !E S rEO=Q
vO {O D !S kO R |O
)
)
!S D
vO : R
We also know that rED=O D r |O, and vEC D vED . From rigid body kinematics we have vED D vEO C !E S rED=O vEC D vO {O C !S kO r |O vEC D .vO
!S r/ {O
)
r vEC D vO 1 C {O: R
Since vEC and vEO are in the same direction, and vC > vO , the cord is being unwound at a rate of jE vC
vEO j D
r vO D 1:36 m=s: R
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Dynamics 1e
Problem 6.54 At the instant shown, the center O of a spool with inner and outer radii r D 3 ft and R D 7 ft, respectively, is moving down the incline at a speed vO D 12:2 ft=s. If the spool does not slip relative to the rope and if the rope is fixed at one end, determine the velocity of point C (the point on the spool that is in contact with the incline) as well as the rope’s unwinding rate, that is, the length of rope being unwound per unit time.
Solution We let point A be the point where the rope is losing contact with the spool. Because the end of the rope not around the spool is fixed, the speed of the rope is 0, and therefore vA D 0. We also know that vEO D 12:2 {O ft=s. From rigid body kinematics we have vEO D vEA C !E O rEO=A vO {O D !O kO r |O vO {O D
r!O {O
)
!O D
vO : r
We must also have vEC D vEA C !E O rEC =A vO O D k .r R/ |O r vO .r R/ D : r
Plugging in known values we have vEC D
16:3 {O ft=s:
E the rope is being unwound, and its rate is Since vEA D 0, jE vO
vEA j D 12:2 ft=s:
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Problem 6.55 The bucket of a backhoe is the element AB of the four-bar linkage system ABCD. Assume that the points A and D are fixed and that, at the instant shown, point B is vertically 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 C at the instant shown, along with the angular velocities of elements BC and CD. Let h D 0:66 ft, e D 0:46 ft, l D 0:9 ft, and w D 1:0 ft.
Solution Point C is a point on both bar BC and bar CD, and therefore can be related to both points B and D through rigid body kinematics. We have that vEC D vED C !E CD rEC =D D !CD kO Π.w
h/ {O C .e C `/ |O D
!CD .e C `/ {O
!CD .w
h/ |O (1)
Since vEB D vB {O, we ahve vEC D vEB C !EBC rEC =B D vB {O C !BC . w/ {O D vB {O
!BC w |O:
(2)
Equating Eqs. (1) and (2) component by component we have !CD .e C `/ {O D vB {O
h/ |O D
!CD .w
and
!BC w |O:
(3)
Solving Eqs. (3) we have !CD D
vB eC`
and !BC D
!CD .w w
h/
)
!BC D
vB w h : eC` w
(4)
Substituting known values into Eq. (4), we have !E CD D
0:882 kO rad=s;
!EBC D
0:300 kO rad=s:
Substituting Eqs. (4) into Eq. (2) we have vEC D vB {O C
vB .w eC`
h/ |O:
(5)
Substituting known values into Eq. (5) we have vEC D .1:20 {O C 0:300 |O/ ft=s:
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Dynamics 1e
Problem 6.56 Bar AB is rotating counterclockwise with an angular velocity of 15 rad=s. Letting L D 1:25 m, determine the angular velocity of bar CD when D 45ı .
Solution Because bar AB is rotating about the fixed point A, at the given instant point B is only moving in the {O direction, and we have vEB D !AB L {O: Points B and C are both on the bar BC so we have vEC D vEB C !EBC rEC =B D !AB L {O C !BC kO L {O D !AB L {O C !BC L |O:
(1)
Points C and D are also both on the bar CD so we have vEC D vED C !E CD rEC =D D !CD kO
L L .cos {O C sin |O/ D !CD . sin {O C cos |O/: 2 2
(2)
Equating the {O components of Eqs. (1) and (2) we have !AB L D
L !CD sin 2
)
!CD D
2!AB : sin
Substituting in known values we have !E CD D
42:4 kO rad=s:
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Solutions Manual
Problem 6.57 At the instant shown, bars AB and CD are vertical and point C is moving to the left with a speed of 35 ft=s. Letting L D 1:5 ft and H D 0:6 ft, determine the velocity of point B.
Solution From the concept of instantaneous center of rotation, we see that lines ABA and CD are parallel, so the instantaneous center of rrotation of bar BC is at infinity, so !BC D 0. Therefore, from rigid body kinematics we have vEB D vEC , which gives us vEC D
35 {O ft=s:
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Dynamics 1e
Problem 6.58 Collars A and B are constrained to slide along the guides shown and are connected by a bar with length L D 0:75 m. Letting D 45ı , determine the angular velocity of the bar AB at the instant shown if, at this instant, vB D 2:7 m=s.
Solution From rigid body kinematics, we must have vEA D vEB C !EAB rEA=B ; where vEA D
vA |O, vEB D vB {O, and rEA=B D
L p 2
{O C
L p 2
|O. Plugging these values in we have L L O vA |O D vB {O C !AB k p {O C p |O 2 2 L L vA |O D vB {O !AB p {O !AB p |O: 2 2
(1)
From the {O component of Eq. (1) we have L vB D !AB p 2
)
!AB D
vB p 2: L
Substituting known values we have !EAB D 5:09 kO rad=s:
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Solutions Manual
Problem 6.59 At the instant shown, an overhead garage door is being shut with point B moving to the left within the horizontal part of the door guide at a speed of 5 ft=s, while point A is moving vertically downward. Determine the angular velocity of the door and the velocity of the counterweight C at this instant if L D 6 ft and d D 1:5 ft.
Solution Using the concept of instantaneous center of rotation we have vB D !AB d; p vA D !AB L2
(1) d 2:
(2)
Solving Eq. (1) for !AB we have !AB D
vB d
)
!AB D 3:333 rad=s:
In vector form and to three significant figures we have !EAB D 3:33 kO rad=s: Substituting all known values into Eq. 2 we have vA D
vB p 2 L d
Given the geometry of the pulley system, vC D
d2 D
19:35 ft=s:
vA . In vector form, we have
vEC D 19:4 |O ft=s:
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Dynamics 1e
Problem 6.60 A spool with inner radius R D 1:5 m rolls without slip over a horizontal rail as shown. If the cable on the spool is unwound at a rate vA D 5 m=s in such a way that the unwound cable remains perpendicular to the rail, determine the angular velocity of the spool and the velocity of the spool’s center O.
Solution We let P be the contact point between the rope and the spool. By the no slip condition, the |O component of vEP is equal to vA . Also, vQ D 0 because it is the contact point between the spool and the rail. From rigid body kinematics we then have vEP D vEQ C!E spool rEP =Q D !spool kO . R {O C R |O/ D
R!spool . {O C |O/:
Equating vP y to vA we have vA D
R!spool
)
!spool D
vA D R
3:33 rad=s:
In vector form this is !E spool D
3:33 kO rad=s:
We must also have vEO D vEQ C !E spool rEO=Q D !spool kO R |O D
!spool R {O
)
vEO D 5:00 {O rad=s:
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Solutions Manual
Problem 6.61 In the four-bar linkage system shown, the lengths of the bars AB and CD are LAB D 46 mm and LCD D 25 mm, respectively. In addition, the distance between points A and D is dAD D 43 mm. The dimensions of the mechanism are such that when the angle D 132ı , the angle D 69ı . For D 132ı and P D 27 rad=s, determine the angular velocity of bars BC and CD as well as the velocity of the point E, the midpoint of bar BC . Note that the figure is drawn to scale and that bars BC and CD are not collinear.
Solution We let point D be the origin. From the geometry of the problem we have rEA D dAD {O D 0:04300 {O m; rEB=A D LAB .cos {O C sin |O/ D . 0:03078 {O C 0:03418 |O/ m; rEB D rEA C rEB=A D .0:01222 {O C 0:03418 |O/ m; rEC D LCD .cos {O C sin |O/ D .0:008959 {O C 0:02334 |O/ m; and rEC =B D rEC
rEB D .LCD cos
C .LCD sin
LAB cos / {O
dAD
LAB sin / |O D . 0:003261 {O
0:01085 |O/ m:
From rigid body kinematics we have vEB D vEA C !EAB rEB=A : D !AB kO LAB .cos {O C sin |O/ D !AB LAB . sin {O C cos |O/ D . 0:9230 {O
0:8311 |O/ m=s:
We must also have vEC D vEB C !EBC rEC =B D !AB LAB . sin {O C cos |O/ C !BC kO .LCD cos C .LCD sin D Œ!BC .LAB sin
dAD
LAB cos / {O
LAB sin / |O LCD sin /
C Œ!BC .LCD cos
dAD
!AB LAB sin {O LAB cos / C !AB LAB cos |O:
(1)
Point C is also on bar CD where rEC =D D LCD .cos {O C sin |O/. From rigid body kinematics we then have vEC D vED C !E CD rEC =D D !CD LCD . sin {O C cos |O/:
(2)
Equating Eqs. (1) and (2) component by component we have !BC .LAB sin
LCD sin /
!AB LAB sin D
!CD LCD sin
(3)
LAB cos / C !AB LAB cos D !CD LCD cos :
(4)
and !BC .LCD cos
dAD
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Dynamics 1e
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 !BC .LAB sin LCD sin / !AB LAB sin !BC .LCD cos dAD LAB cos / C !AB LAB cos tan cos / D !BC Œ.LAB sin LCD sin / tan D
!AB LAB .sin
)
!BC
C tan .LCD cos dAD LAB cos / !AB LAB .sin tan cos / D LAB sin LCD sin C tan .LCD cos dAD
)
LAB cos /
: (5)
Substituting known values into Eq. (5) we get !EBC D 1310 kO rad=s: Solving Eq. (3) for !CD we have !CD D
!AB LAB sin
!BC .LAB sin LCD sin
LCD sin /
:
(6)
Substituting Eq. (5) for !BC in Eq. (6) we have !CD D
!AB LAB sin LCD sin !AB LAB .sin tan cos /.LAB sin LAB sin LCD sin C tan .LCD cos dAD
LCD sin / LAB cos /LCD sin
(7)
Substituting known values into Eq. (7) we have !E CD D
571 kO rad=s:
We can calculate vEE from the rigid body kinematics of bar BC , where rEE=B D 21 rEC =D . We have vEE D vEB C !BC kO rEE=B 1 D vBx {O C vBy |O C !BC kO rEC =D 2 !BC O D vBx {O C vBy |O C k Œ.rC =B /x {O C .rC =B /y |O 2 !AB LAB .sin tan cos / .rC =B /y {O D ŒvBx 2ŒLAB sin LCD sin C tan .LCD cos dAD LAB cos / !AB LAB .sin tan cos / C ŒvBy C .rC =B /x |O: (8) 2ŒLAB sin LCD sin C tan .LCD cos dAD LAB cos / Substituting all known values into Eq. (8) we have vEE D .6:20 {O
2:97 |O/ m=s:
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Solutions Manual
Problem 6.62 A person is closing a heavy gate with rusty hinges by pushing the gate with car A. If w D 24 m and vA D 1:2 m=s, determine the angular velocity of the gate when D 15ı .
Solution From the geometry of the problem, we have yH D w tan :
(1)
Taking the first derivative of Eq. (1) we have yPH D Realizing that yPH D
w P : cos2
vA we have vA D
w P : cos2
Solving for P we are left with cos2 : P D vA w Finally, substituting known values into Eq. (2) we have P D
(2)
0:0467 rad=s:
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Dynamics 1e
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 0ı , the bars AB and BC are vertical and horizontal, respectively. In addition, let R D 2 ft, L D 3 ft, and H D 3:5 ft. When D 0ı , the collar at C is sliding downward with a speed of 23 ft=s. Determine the angular velocities of the bars AB and BC at this instant.
Problem 6.63
When D 37ı , ˇ D 25:07ı , D 78:71ı , and the collar is sliding clockwise with a speed vC D 23 ft=s. Determine the angular velocities of the bars AB and BC . Problem 6.64
Determine the general expression for the angular velocities of bars AB and BC as a P function of , ˇ, , R, L, H , and . Problem 6.65
Solution to 6.63 Because point B is the instantaneous center of rotation of bar CB, it has zero velocity. Point A is pinned and cannot move, so bar AB cannot be moving. Therefore we have E !EAB D 0: From the instantaneous center of rotation method, we then have vC D L!BC
)
!BC D
vC D 7:667 rad=s: L
Following the given component system, in vector form, and to three significant figures we have !EBC D 7:67 kO rad=s:
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Solutions Manual
Solution to 6.64 From the geometry of the problem, we have the following position vectors: rEC =O D R.cos {O C sin |O/; rEB=C D L.cos ˇ {O
sin ˇ |O;
rEB=A D H. cos {O C sin |O/; and the velocity vector vEC D vC .sin {O
cos |O/:
O Points B and C are both on bar BC so from rigid body We also define !EBC D !BC kO and !EAB D !AB k. kinematics we have vEB D vEC C !BC kO rEB=C D vC .sin {O
cos |O/ C !BC L.sin ˇ {O C cos ˇ |O/
D .vC sin C !BC L sin ˇ/ {O C .!BC L cos ˇ
vC cos / |O:
(1)
Also, because points A and B are both on bar AB, we have vEB D vEA C !EAB rEB=A D
!AB H.sin {O C cos |O/:
(2)
Equating Eq. (1) with Eq. (2) component by component we have vC sin C !BC L sin ˇ D !BC L cos ˇ
vC cos D
!AB H sin
(3)
!AB H cos :
(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 vC sin C !BC L sin ˇ D tan !BC L cos ˇ vC cos
)
vC sin C !BC L sin ˇ D tan .!BC L cos ˇ )
!BC D vC
vC cos /
sin C tan cos : L.tan cos ˇ sin ˇ/
Substituting in known values, we have !BC D
7:210 rad=s:
To three significant figures, and in vector form we have !EBC D
7:21 kO rad=s:
Solving Eq. (3) for !AB we have !AB D
vC sin C !BC L sin ˇ : H sin
Plugging in known values, we have !EAB D 8:58 kO rad=s: August 10, 2009
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Dynamics 1e
Solution to 6.65 From the geometry of the problem, we have L sin ˇ C H sin D H C R sin
and L sin ˇ C H cos D R.1
cos / C L:
(5)
Taking the first derivative with respect to time of Eqs. (5) we have LˇP cos ˇ C H P cos D RP cos LˇP sin ˇ H P sin D RP sin :
(6) (7)
Eqs. (6) and (7) can be viewed as two equations in ˇP and . P Realizing that ˇP D !BC and P D !AB P tells us that solving these two equations for ˇ and P will give us the angular velocities of the two bars as a function of the specified terms. Solving these equations we have !BC D
ˇP D
!AB D
P D
RP cos L cos ˇ
cos tan ˇ C sin tan ˇ tan
and RP sin C tan ˇ cos : H cos tan ˇ sin
Simplifying these we end up with
!EBC D
RP csc .ˇ L
/ sin . C / kO
!EAB D
RP csc .ˇ H
O
/ sin .ˇ C / k:
and
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Solutions Manual
Problem 6.66 At the instant shown, the arm OC rotates counterclockwise with an angular velocity of 35 rpm about the fixed sun gear S of radius RS D 3:5 in: The planet gear P with radius RP D 1:2 in: rolls without slip over both the fixed sun gear and the outer ring gear. Finally, notice that the ring gear is not fixed and it rolls without 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 Points C and O are both on the bar OC , so from rigid body kinematics we must have vEC D vEO C !E COC ErC =O D !OC kO .RS C RP / {O D !OC .RS C RP / |O:
(1)
Point C is also on the planetary gear with point Q, so we must have vEC D vEQ C !EP rEC =Q D !P kO RP {O D !P RP |O:
(2)
Equating Eqs. (1) and (2) we have .RS C RP /!OC D RP !P
)
!P D
RS C RP !OC : RP
Point A is also on the planetary gear, so we must have vEA D vEQ C !EP rEA=Q RS C RP D !OC kO 2RP {O RP D 2.RS C RP /!OC |O:
(3)
Points A and B are both points on the ring gear, so we must have vEA D vEB C !E R rEA=B D !R kO 2.RS C RP / {O D 2!R .RS C RP / |O:
(4)
Equating Eqs. (3) and (4) we have 2.RS C RP /!OC D 2!R .RS C RP /
)
!R D !OC : August 10, 2009
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Dynamics 1e In vector form this is !E R D 35:0 kO rpm: Calling the center of the ring point D we must have vED D vEB C !E R rED=B D ! kO .RS C RP / {O D .RS C RP /!OC |O:
(5)
Substituting known values into Eq. (5) we have vED D 17:2 |O in.=s:
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Problem 6.67 The crank AB is rotating counterclockwise at a constant angular velocity of 12 rad=s while the pin B slides within the slot in the bar CD, which is pinned at C . Letting R D 0:5 m, h D 1 m, and d D 0:25 m, determine the angular velocity of CD at the instant shown (with points A, B, and C vertically aligned) as well as the velocity of the horizontal bar to which bar CD is connected.
Solution If we let ` be the distance between B and C , then from the geometry of the problem we have the constraint equation R sin D ` sin : Differentiating this constraint with respect to time we have RP cos D `P sin C ` P cos :
(1)
At the instant given in the problem, we have D 0ı ; D 0ı , and ` D h R ) `P D 0. Substituting these values into Eq. (1) we have RP D .h
R/ P
)
Making the observation that !AB D P and !CD D !CD D
R h
R
!AB D
12:00 rad=s
P D
R h
R
P :
P we have , )
!E CD D
12:0 kO rad=s:
Points Q and C are both on the arm CD so from rigid body kinematics we must have vEQ D vEC C !E CD rEQ=C D !CD kO . d / |O D !CD d {O: Substituting known values we have vEQ D 3:00 {O m=s:
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Dynamics 1e
Problems 6.68 through 6.72 For the slider-crank mechanism shown, let R D 20 mm, L D 80 mm, and H D 38 mm. If P D 1700 rpm, determine the angular velocity of the connecting rod AB and the speed of the slider B for D 90ı .
Problem 6.68
Determine the angular velocity of the crank OA when D 20ı and the slider is moving downward at 15 m=s.
Problem 6.69
Determine the general expression for the velocity of the slider Problem 6.70 P and the geometrical parameters R, H , and L, using the B as a function of , , vector approach. Determine the general expression for the velocity of the slider B as a function of , P , and the geometrical parameters R, H , and L, using differentiation of constraints. Problem 6.71
Problem 6.72 Plot the velocity of the slider C as a function of , for 0 360ı , and for P D 1000 rpm, P D 3000 rpm, and P D 5000 rpm.
Solution to 6.68 Converting units on known values we have R D 20 mm D 0:020 m;
L D 80 mm D 0:080 m;
H D 38 mm D 0:038 m:
From the rigid body kinematics of the wheel, and letting !O D P we have vEA D vEO C !E O rEA=O D P kO R |O D
P R {O:
p From the rigid body kinematics of arm AB where rEB=A D H {OC L2 we have
H 2 |O,
vEB D vEA C !EAB rEB=A
p RP {O C !AB kO .H {O C L2 H 2 |O/ p D . P R !AB L2 H 2 / {O C !AB H |O:
D
Because B is confined to move vertically in the slot, vBx D 0. From this we have !AB D
p
R L2
H2
P :
In vector form, and plugging in known values we have !EAB D
50:6 kO rad=s:
From the y component of vEB we have vEB D vBy |O D !AB H |O D P H p
R L2
H2
|O: August 10, 2009
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Solutions Manual
Plugging in known values we have vEB D 1:92 |O m=s:
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Dynamics 1e
Solution to 6.69 Converting units on known values we have R D 20 mm D 0:020 m;
L D 80 mm D 0:080 m;
H D 38 mm D 0:038 m:
From the rigid body kinematics of the wheel, and letting !O D P we have vEA D vEO C !E O rEA=O D P kO R.cos {O C sin |O/ D P R. sin {O C cos |O/: From the rigid body kinematics of arm AB where rEB=A D .H have
R cos / {O C
p
L2
.H
R cos /2 |O, we
vEB D vEA C !EAB rEB=A P O R {O C !AB k .H
D D
P R
!AB
q
L2
q 2 2 R cos / {O C L .H R cos / |O 2 .H R cos / {O C !AB .H R cos / |O:
Because B is confined to move vertically in the slot, vBx D 0. From this we have p !AB L2 .H R cos /2 P D : R We also know that vEB D
(1)
15 |O m=s. From this we have
!AB .H
R cos / D
15
)
!AB D
15 R cos
H
:
(2)
Substituting Eq. (2) into Eq. (1) we have P D
p 15 L2 .H R.R cos
R cos /2 : H/
Plugging in known values we have !E O D P kO D
877 kO rad=s:
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Solutions Manual
Solution to 6.70 From the rigid body kinematics of the wheel, and letting !O D P we have vEA D vEO C !E O rEA=O D P kO R.cos {O C sin |O/ D P R. sin {O C cos |O/: From the rigid body kinematics of arm AB, where rEB=A D .H have
R cos / {O C
p
L2
.H
R cos /2 |O, we
vEB D vEA C !EAB rEB=A RP {O C !AB kO .H
D D
P R sin
!AB
q
L2
L2 .H R cos /2 |O 2 R cos / {O C Œ!AB .H R cos / C RP cos |O:
R cos / {O C .H
q
(3)
Because B is confined to move vertically in the slot, vBx D 0. From this we have !AB D
RP sin p
L2
.H
R cos /2
:
(4)
Plugging Eq. (4) into Eq. (3) we have vEB D RP cos
RP sin .H p L2 .H
R cos / R cos /2
! |O:
August 10, 2009
831
Dynamics 1e
Solution to 6.71 From the geometry of the problem, we have xB D R cos C L cos D H
(5)
yB D R sin C L sin : Taking a time derivative of these two equations we have RP sin
RP sin L sin vB D RP cos C L P cos :
L P sin D 0
)
P D
(6) (7)
Substituting Eq. (6) into Eq. (7) we have vB D RP cos
RP sin cot :
(8)
From Eq. (5) we know that cos D and from the Pythagorean theorem, we know that q sin D L2
H
R cos ; L
.H
(9)
R cos /2 :
(10)
Combining Eqs. (9) and (10) we have cot D
H R cos : p 2 L L .H R cos /2
(11)
Substituting Eq. (11) into Eq. (8) we have vB D RP cos
H R cos RP sin p : L L2 .H R cos /2
Knowing that B is confined to the |O direction, we have vEB D RP cos
RP sin .H p L2 .H
R cos / R cos /2
! |O:
August 10, 2009
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Solutions Manual
Solution to 6.72 Converting units on known values we have R D 0:020 m; L D 0:080 m; H D 0:038 m: From the rigid body kinematics of the wheel, and letting !O D P we have vEA D vEO C !E O rEA=O D P kO R.cos {O C sin |O/ D P R. sin {O C cos |O/: From the rigid body kinematics of arm AB, where rEB=A D .H have
R cos / {O C
p
L2
.H
R cos /2 |O, we
vEB D vEA C !EAB rEB=A P O R {O C !AB k .H
D D
P R sin
R cos / {O C
q !AB L2
q
L2
R cos /2
.H
.H
R cos /2 |O
R cos / C RP cos |O:
{O C Œ!AB .H
(12)
Because B is confined to move vertically in the slot, vBx D 0. From this we have RP sin
!AB D
p
L2
.H
R cos /2
:
(13)
Plugging Eq. (13) into Eq. (12) we have vEB D RP cos
RP sin .H p L2 .H
R cos / R cos /2
! |O:
The expression for vEB just derived can be reduced to a function of and P by substituting in the given parameters. For each of the required values of P , the function can be plotted with software such as Mathematica or M ATLAB. The plots presented below were obtained using Mathematica with the following code: Parameters ! !R " 0.02, L " 0.08, H " 0.038"; Velocity ! R Θdot Cos#Θ$ $
R Θdot Sin#Θ$ %H $ R Cos#Θ$& L2
Θdotval !
1 30
$ %H $ R
'. Parameters;
Cos#Θ$&2
Π !1000, 3000, 5000";
Table(Plot(Velocity '. !Θdot " i", !Θ, 0, 2 Π", AxesLabel " )"Θ", "vBy "*, Ticks " !!0, Π, 2 Π, 3 Π", Automatic"+, !i, Θdotval"+
Using the above code we obtain the following plots: August 10, 2009
833
Dynamics 1e vBy 2
1
Θ Π
2Π
Π
2Π
Π
2Π
-1
-2
vBy 6 4 2 Θ -2 -4 -6 -8 vBy 10
5
Θ
-5
-10
August 10, 2009
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Solutions Manual
Problem 6.73 Complete the velocity analysis of the slider-crank mechanism, using differentiation of constraints that was outlined beginning on p. 478. That is, determine the velocity of the piston C and the angular velocity of the connecting rod as a function of the given quantities , !AB , R, and L. Use the component system shown for your answers.
Solution One way to solve this problem is to write yC as yC D R cos C L cos ;
(1)
where for cos and sin we will always use s
R sin D sin L
and
cos D
q
1
sin2 D
R2 2 sin : L2
1
(2)
Now we differentiate Eq. (1) with respect to time: yPC D
RP sin
L P sin ;
(3)
where P is found by differentiating the first of Eq. (2) with respect to time: R P cos D P cos L
)
R cos P D P : L cos
(4)
Realizing that P D !BC and P D !AB , we have !BC D !AB
R cos L cos
)
!EBC D !AB
R cos O k: L cos
Also, noting that vEC D yPC |O, and substituting Eq. (4) for P in Eq. (3) we have ! P cos R sin R R2 P sin cos yPC D RP sin L D RP sin : L cos L L cos
(5)
After substituting !AB into Eq. (5), and putting it in vector form we have vEC D
R!AB sin
R2 !AB sin cos L cos
|O:
August 10, 2009
835
Dynamics 1e
Problem 6.74 A truck on an exit ramp is moving in such a way that, at the instant shown, jE aA j D 17 ft=s2 , P D 0:3 rad=s, and R D 0:1 rad=s2 . If the distance between points A and B is dAB D 12 ft, D 57ı , and D 13ı , determine aEB .
Solution O ˛EAB D R k; O and rEB=A D Based on the picture shown, we let aEA D jE aA j.cos {O sin |O/; !EAB D P k; dAB .cos {O C sin |O/: From the vector approach to acceleration analysis we have 2 !AB rEB=A sin / C R kO dAB .cos {O C sin |O/
aEB D aEA C ˛EAB rEB=A D jE aA j.cos {O D .jE aA j cos
R dAB sin
P 2 dAB .cos {O C sin |O/ P 2 dAB cos / {O C . jE aA j sin C R dAB cos P 2 dAB sin / |O;
which can be evaluated to obtain aEB D .17:0 {O using the given values jE aA j D 17 ft=s2 , P D ı D 13 .
5:38 |O/ ft=s2 ;
0:3 rad=s, R D
0:1 rad=s2 , dAB D 12 ft, D 57ı , and
August 10, 2009
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Solutions Manual
Problems 6.75 through 6.77 Let L D 4 ft, let point A travel parallel to the guide shown, and let C be the midpoint of the bar. If point A is accelerating to the right with aA D 27 ft=s2 and P D 7 rad=s D constant, determine the acceleration of point C when D 24ı . Problem 6.75
If point A is accelerating to the right with aA D 27 ft=s2 , P D 7 rad=s, and R D 0:45 rad=s2 , determine the acceleration of point C when D 26ı . Problem 6.76
If, when D 0ı , A is accelerating to the right with aA D R E determine P and . and aEC D 0,
Problem 6.77
27 ft=s2
Solution to 6.75 O ˛EAB D R kO D 0; E and rEC =A D We have aEA D aA {O; !EAB D P k; approach to acceleration analysis we have
L 2 .sin
{O
cos |O/: From the vector
2 aEC D aEA C ˛EAB rEC =A !AB rEC =A LP 2 D aA {O .sin {O C cos |O/ 2 L P2 L D aA sin {O C P 2 cos |O; 2 2
which can be evaluated to obtain aEC D . 12:9 {O C 89:5 |O/ ft=s2 : using the given values L D 4 ft, aA D 27 ft=s2 , P D 7 rad=s, and D 24ı .
August 10, 2009
837
Dynamics 1e
Solution to 6.76 O ˛EAB D R kO D 0, E and rEC =A D We have aEA D aA {O; !EAB D P k, approach to acceleration analysis we have aEC DE aA C ˛EAB rEC =A
L 2 .sin
{O
cos |O/: From the vector
2 !AB rEC =A
L LP 2 D aA {O C R kO .sin {O C cos |O/ .sin {O C cos |O/ 2 2 L L P2 L 2L R R P D aA C cos sin {O C sin C cos |O; 2 2 2 2 which can be evaluated to obtain .16:8 {O C 87:7 |O/ ft=s2 : using the given values L D 4 ft, aA D 27 ft=s2 , P D 7 rad=s, R D
0:45 rad=s2 , and D 26ı .
August 10, 2009
838
Solutions Manual
Solution to 6.77 O ˛EAB D R kO D 0, E and rEC =A D We have aEA D aA {O, !EAB D P k, acceleration analysis we have
L 2
|O: From the vector approach to
2 aEC DE aA C ˛EAB rEC =A !AB rEC =A L LP 2 0E D aA {O C R kO |O C |O 2 2 E0 D aA C R L {O C P 2 L |O: 2 2
(1)
Separating the components of Eq. (1) we have L 0 D aA C R 2
)
R D
aA
2 D L
13:5 rad=s2 ;
and L 0 D P 2 2
)
P D 0;
using the given values L D 4 ft, aA D 27 ft=s2 , aC D 0 ft=s2 , and D 0ı .
August 10, 2009
839
Dynamics 1e
Problem 6.78 A wheel W of radius RW D 5 cm rolls without slip over the stationary cylinder S of radius RS D 12 cm, and the wheel is connected to point O via the arm OC . If !OC D constant D 3:5 rad=s, determine the acceleration of point Q, which lies on the edge of W and along the extension of the line OC .
Solution Let E be the contact point between W and S. From the no slip condition, E We let !E W D !W kO and from the geometry of the problem vEE D 0. rEC =E D RW .cos {O C cos |O/. From rigid body kinematics of the wheel we have vEC D vEE C !E W rEC =E D !W kO RW uO r D !W RW uO :
(1)
Points C and O are also points on the arm where rEC =O D .RS C RW / uO r . From rigid body kinematics we must have vEC D vEO C !E OC rEC =O D !OC kO .RS C RW /uO r D !OC .RS C RW /uO :
(2)
Equating Eqs. (1) and (2) we have !W RW D !OC .RS C RW /
)
!W D
RS C RW !OC : RW
Because C is undergoing a uniform circular motion with radius .RS C RW / about O, we have aEC D
2 !OC .RS C RW / uO r :
From the vector approach to acceleration analysis we have 2 !W rEQ=C RS C RW 2 2 2 !OC RW uO r !OC .RS C RW / uO r RW RS C RW 2 !OC .RS C RW / 1 C uO r RW RS 2 !OC .RS C RW / 2 C uO r ; RW
aEQ D aEC C ˛EW rEQ=C D D D
which can be evaluated to obtain 9:16 uO r m=s2 : using the given values RW D 5 cm D 0:05000 m, RS D 12 cm D 0:1200 m, and !OC D 3:5 rad=s. August 10, 2009
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 crank C that is connected to a slider B 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 of valves in pipelines. Letting the radius of the crank be R D 25 cm, determine the angular velocity !C and the angular acceleration ˛C of the crank at the instant shown if D 25ı and the slider is moving to the right with a constant speed vB D 40 m=s.
Solution Choosing point O as the origin of the reference frame, from the geometry of the problem we have rEA D 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 get vEA by taking the derivative of Eq. (1) with respect to time. Doing so we have vEA D RP .sin {O C cos |O/
vEB D RP sin {O
)
)
P D
vB D 378:6 rad=s; R sin
where we have used the given values R D 25 cm D 0:2500 m, vB D 40 m=s, and D 25ı . Realizing that !C D P we have !E C D
379 kO rad=s:
Because we are told vB D 0 we must have d vB D R sin R C R cos P 2 D 0 dt
)
R D
P 2 : tan
Plugging in known values we have R D
307103 kO rad=s2 :
August 10, 2009
841
Dynamics 1e
Problem 6.80 Collar C moves along a circular guide with radius R D 2 ft with a constant speed vC D 18 ft=s. At the instant shown, the bars AB and BC are vertical and horizontal, respectively. Letting L D 4 ft and H D 5 m, determine the angular accelerations of the bars AB and BC at this instant.
Solution Note: The problem statement gives H D 5 m, which is incorrect. The solution will use H D 5 ft. Realizing that B is the instantaneous center of rotation of bar AB and A is not moving, we must have !AB D 0: Because B is also the instantaneous center of rotation of bar BC , we must have vC L
!BC D
)
!EBC D
vB O k: L
Since C moves in a uniform circular motion about O at constant speed, we have aEC D
2 vC {O; R
O we have and from the vector approach to acceleration analysis where, rEB=C D L {O and ˛EBC D ˛BC k, aEB DE aC C ˛EBC rEB=C
2 !BC rEB=C
2 2 vC vC {O C ˛BC kO L {O L {O R R2 2 vC L 1C {O C ˛BC L |O: R R
D D
(1)
Also from the vector approach to acceleration analysis where rEB=A D H |O and ˛EAB D ˛AB kO we have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A D ˛AB kO H |O D
˛AB H {O:
(2)
Equating components of Eqs. (1) and (2) we have 2 vC R
L 1C R
D
˛AB H
and
˛BC L D 0:
(3)
Solving Eq. (3) for ˛AB and ˛BC , we have ˛AB D
2 vC R
L 1C R
)
˛EAB D 97:2 kO rad=s2
and
E ˛EBC D 0:
August 10, 2009
842
Solutions Manual
Problems 6.81 and 6.82 A ball of radius RA D 5 in: is rolling without slip inside a stationary spherical bowl of radius RB D 17 in: Assume that the motion of the ball is planar. If, at the instant shown, the center of the ball is traveling counterclockwise with a speed vA D 32 ft=s and such that vPA D 0, 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. Problem 6.81
If, at the instant shown, the center of the ball is traveling counterclockwise with a speed vA D 32 ft=s and such that vPA D 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. Problem 6.82
Solution to 6.81 The center of the ball is making a uniform circular motion of radius RB RA about O at constant speed, so we must have aEA D
vA2 RB
RA
uO r
(1)
Since the ball rolls without slip over a stationary surface, the point of contact between the ball and the bowl, point Q, is the instantaneous center of rotation of the ball. Therefore, vA D j!EA jRA
)
!A D
vA : RA
Notice that ˛A D 0 because !A is constant. From the vector approach to acceleration analysis where rEQ=A D RA uO r we must have aEQ D aEA C ˛EA rEQ=A
!A2 rEQ=A
D
vA2 RB
RA
uO r
vA2 RA2
RA uO r D
vA2
1 RB
1 C uO r RA RA
(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 into Eqs. (1) and (2) we have aEA D
1020 uO r ft=s2 ;
aEQ D
3480 uO r ft=s2 :
August 10, 2009
843
Dynamics 1e
Solution to 6.82 The center of the ball is moving in a uniform circular motion of radius RB RA about O, as well as accelerating in the direction, so we must have aEA D
vA2 RB
RA
uO r C vPA uO
(3)
Substituting given values vA D 32 ft=s, vPA D 24 ft=s2 , RA D 0:4167 ft, and RB D 1:417 ft into Eq. (3), we have aEA D . 1020 uO r C 24:0 uO / ft=s2 Since the ball rolls without slip over a stationary surface, the point of contact between the ball and the bowl, point Q, is the instantaneous center of rotation of the ball. Therefore, vA D j!EA jRA
)
!A D
vA : RA
From the no slip condition, aQ D 0. From the vector approach to acceleration analysis where rEQ=A D RA uO r and ˛EA D ˛A kO we must have aEQ D aEA C ˛EA rEQ=A D D
vA2
!A2 rEQ=A vA2
uO r C vPA uO C ˛A RA uO
RA uO r RB RA RA2 1 1 2 vA C uO r C .vPA C ˛A RA / uO : RB RA RA
(4)
However, vPA C ˛A RA D 0, so we are left with aEQ D
vA2
1 RB
1 C RA RA
uO r :
(5)
Substituting given values vA D 32 ft=s, vPA D 24 ft=s2 , RA D 0:4167 ft, and RB D 1:417 ft into Eq. (5), we have aEQ D
3480 uO r ft=s2 :
August 10, 2009
844
Solutions Manual
Problem 6.83 A bar of length L D 2:5 m is falling so that, when D 34ı , vA D 3 m=s and aA D 8:7 m=s2 . At this instant, determine the angular acceleration of the bar AB and the acceleration of point D, where D is the midpoint of the bar.
Solution We have rEB=A D L.sin {O cos |O/ and rED=A D L 2 .sin {O analysis where vEA D vA |O and vEB D vB {O, we have vEB D vEA C !EAB rEB=A
)
vB {O D
cos |O/. From the vector approach to velocity
vA |O C !AB kO L.sin {O )
cos |O/
vB {O D .L!AB cos / {O C .L!AB sin
vA / |O: (1)
Equating the two sides of Eq. (1) component by component, we have 0 D L!AB sin
vA
)
!AB D
vA ; L sin
and vB D L!AB cos :
(2)
Substituting given values vA D 3 m=s, L D 2:5 m, and D 34ı into Eqs. (2), we have !AB D 2:146 rad=s
vB D 4:448 m=s:
and
From the vector approach to acceleration analysis where aEA D 2 !AB rEB=A
aEB D aEA C ˛EAB rEB=A )
aA |O and aEB D aB {O, we have
aB {O D
aA |O C ˛AB kO L.sin {O
D .˛AB L cos
2 !AB L sin / {O
cos |O/
2 !AB L.sin {O
C .˛AB L sin C
cos |O/
2 !AB L cos
aA / |O:
(3)
From the |O component of Eq. (3) we have 2 ˛AB L sin C !AB L cos
aA D 0
)
˛AB D
aA
2 !AB L cos : L sin
(4)
Substituting known values aA D 8:7 m=s2 , L D 2:5 m, !AB D 2:146 rad=s, and D 34ı into Eq. (4), we have ˛EAB D 0:604 kO rad=s2 : For D we have aED D aEA C ˛EAB rED=A
" D
L 2 L .sin {O cos |O/ !AB .sin {O 2 ! #2 2 !AB L cos L 2 L cos !AB sin {O L sin 2 2
aA |O C ˛AB kO
D
aA
2 !AB rED=A
cos |O/
August 10, 2009
845
Dynamics 1e "
aA
2 L cos !AB L sin
!
L 2 L C sin C !AB cos 2 2 1 2 2 L .aA !AB L cos / cot !AB sin {O D 2 2 1 2 2 L C .aA !AB L cos / C !AB cos aA |O 2 2
# aA |O
(5)
Substituting known values aA D 8:7 m=s2 , L D 2:5 m, !AB D 2:146 rad=s, and D 34ı into Eq. (5), we have aED D . 3:84 {O
4:35 |O/ m=s2 :
August 10, 2009
846
Solutions Manual
Problem 6.84 A bar of length L D 8 ft and midpoint D is falling so that when D 27ı , jE vD j D 18 ft=s, and the vertical acceleration of point D is 23 ft=s2 downward. At this instant, compute the angular acceleration of the bar and the acceleration of point B.
Solution We have rEB=A D L.sin {O cos |O/ and rED=A D L 2 .sin {O cos |O/. Realizing that point H is the instantaneous center of rotation of the line HD, where the distance between H and D is dHD D L 2 , we have !AB D
vD D 4:500 rad=s: L=2
O From the vector approach to acceleration analysis where ˛EAB D ˛AB k, O aEA D aA |O and !EAB D !AB k we have 2 aED D aEA C ˛EAB rED=A !AB rED=A L L D aA |O C ˛AB cos {O C ˛AB sin |O 2 2
D .˛AB L cos
L .sin {O cos |O/ 2 L 2 2 L !AB L sin / {O C aA C ˛AB sin C !AB cos |O 2 2 2 !AB
(1)
We also have 2 aEB D aEA C ˛EAB rEB=A !AB rEB=A D aA |O C ˛AB kO L.sin {O cos |O/
D .˛AB L cos
2 !AB L.sin {O
cos |O/
2 2 !AB L sin / {O C .aA C ˛AB L sin C !AB L cos / |O
(2)
Recalling that aDy is given and aBy D 0, we have 2 aBy D 0 D aA C ˛AB L sin C !AB L cos L 2 L cos aDy D 23:00 ft=s2 D aA C ˛AB sin C !AB 2 2
(3) (4)
Solving Eq. (3) for aA and substituting into Eq. (4) we have 23:00 ft=s2 D D
L 2 2 L cos ˛AB L sin C !AB L cos C ˛AB sin C !AB 2 2 L 46:00 ft=s2 2 L 2 ˛AB sin !AB cos ) ˛AB D !AB cot 2 2 L sin
(5)
August 10, 2009
847
Dynamics 1e Substituting given values L D 8 ft, D 27ı , and !AB D 4:500 rad=s into Eq. (3), we have ˛AB D
27:08 rad=s2
)
˛EAB D
27:1 kO rad=s2 :
Substituting known values L D 8 ft, D 27ı , ˛AB D 27:08 rad=s2 , and !AB D 4:500 rad=s into Eq. (3), we have aA D 45:99 ft=s2 : Substituting known values L D 8 ft, D 27ı , ˛AB D 45:99 ft=s2 into Eq. (1) we have aED D . 133 {O
27:08 rad=s2 , !AB D 4:500 rad=s, and aA D
23:0 |O/ ft=s2 :
August 10, 2009
848
Solutions Manual
Problem 6.85 Assuming that, for 0ı 90ı , vA is constant, compute the expression for the acceleration of point D, the midpoint of the bar, as a function of and vA .
Solution Calling O the origin of the reference frame, from the geometry of the problem we have xD D
L sin ; 2
yD D L
L cos 2
; and
yA D L cos :
(1)
Observing that aED D xRD {O yRD |O, by taking time derivatives of Eqs. (1) we have L P2 L L P2 L aED D sin C R cos {O cos C R sin |O; (2) 2 2 2 2
yPA D
LP sin D vA
)
P D
vA : L sin
)
R D
vA2
vA cos P D L sin2
L2 sin3
cos
(3)
Substituting Eqs. (3) for P and R in Eq. (2) we have aED D
vA2 2L sin
! 1 vA2 cos2 {O 2 L sin3
vA2 2L sin2
cos
!
vA2 2L sin2
cos |O
)
aED D
vA2 2L sin3
{O:
August 10, 2009
849
Dynamics 1e
Problem 6.86 A truck on an exit ramp is moving in such a way that, at the instant shown, jE aA j D 6 m=s2 and D 13ı . Let the distance between points A and B be dAB D 4 m. If, at this instant, the truck is turning clockwise, D 59ı , aBx D 6:3 m=s2 , and aBy D 2:6 m=s2 , determine the angular velocity and angular acceleration of the truck.
Solution From the information given in the problem we have O ˛EAB D ˛AB k;
rEB=A D dAB .cos {O C sin |O/;
and aEA D jE aA j.cos {O
sin |O/:
(1)
From rigid body kinematics we have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A
(2)
Substituting Eqs. (1) into Eq. (2) we have aEB D jE aA j.cos {O D .jE aA j cos
sin |O/ C ˛AB kO dAB .cos {O C sin |O/ ˛AB dAB sin
2 !AB dAB .cos {O C sin |O/
2 !AB dAB cos / {O
C . jE aA j sin C ˛AB dAB cos
2 !AB dAB sin / |O
(3)
Breaking Eq. (3) into components we have aBx D jE aA j cos aBy D
˛AB dAB sin
2 !AB dAB cos
jE aA j sin C ˛AB dAB cos
2 !AB dAB
sin
Eqs. (4) and (5) are two equations in ˛AB and !AB . Solving them we have s jE aA j.cos cot sin / aBx cot aBy !AB D ; dAB .cos cot sin / and ŒaBy cos jE aA j.sin cos C sin cos / aBx sin ˛AB D : dAB Substituting known values dAB D 4 m, D 13ı , D 59ı , aBx D 6:3 m=s2 , aBy D jaA j D 6 m=s2 into Eqs. (6) and (7) we have !AB D
0:4577 rad=s
and ˛AB D
(4) (5)
(6)
(7) 2:6 m=s2 , and
0:2582 rad=s: August 10, 2009
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Solutions Manual
In vector form, and to three significant figure we have !EAB D
0:458 kO rad=s ˛EAB D
0:258 kO rad=s2 :
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Dynamics 1e
Problems 6.87 through 6.89 The system shown consists of a wheel of radius R D 1:4 m rolling without slip on a horizontal surface. A bar AB of length L D 3:7 m is pin-connected to the center of the wheel and to a slider A constrained to move along a vertical guide. Point C is the bar’s midpoint. If the wheel is rolling clockwise with a constant angular speed of 2 rad=s, determine the angular acceleration of the bar when D 72ı . Problem 6.87
If the slider A is moving downward with a constant speed 3 m=s, determine the angular acceleration of the wheel when D 53ı .
Problem 6.88
Determine the general relation expressing the acceleration of Problem 6.89 the slider A as a function of , L, R, the angular velocity of the wheel !W , and the angular acceleration of the wheel ˛W .
Solution to 6.87 We assume the wheel does not leave the ground, hence B is constrained to move in the horizontal direction, and from the rolling without slip condition we have vEB D vB {O D R!W {O: (1) Also, calling point O the origin of the reference frame, from the geometry of the problem we have xB D L cos : (2) Differentiating Eq. (2) with respect to time, and observing that vB D xPB we have xPB D
L sin P D
R!W
)
R!W P D L sin
(3)
Taking a time derivative of Eq. (3) we have R D
R!W cos P L sin2
)
2 R 2 !W cos R D D 0:2057 rad=s2 ; 3 2 L sin
where we have substituted the known values !AB D 2 rad=s, R D 1:4 m, L D 3:7 m, and D 72ı . R in vector form and to three significant figures, we have Observing that ˛AB D , ˛EAB D 0:206 kO rad=s2 :
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Solution to 6.88 We assume the wheel does not leave the ground, hence B is constrained to move in the horizontal direction, and from the rolling without slip condition, we have vEB D vB {O D
R!W {O:
(4)
Also, calling point O the origin of the reference frame, from the geometry of the problem we have xB D L cos
and yA D R C L sin :
(5)
Differentiating Eqs. (5) with respect to time, and observing that vB D xPB and vA D yPA , we have
yA D R C L sin
)
yPA D
vA D LP cos
P D
)
vA L cos
)
R D
vA2 sin vA2 sin P D L2 cos L2 cos3
(6)
and xPB D
L sin P D
R!W
)
L L vA !W D P sin D sin D R R L cos
vA tan : R
(7)
Taking a time derivative of Eq. (7), we have ˛W D
L cos P 2 R
L sin R : R
Substituting the expressions for P and R in Eqs. (6) into Eq. (8), we have 2 vA sin 2 L sin L2 cos3 A L cos L vcos vA2 vA2 sin2 ˛W D D D R R LR cos LR cos3
(8)
vA2 .1 C tan2 / : (9) LR cos
Substituting known values vA D 3 m=s, R D 1:4 m, L D 3:7 m, and D 53ı into Eq. (9), we have ˛W D
7:971
)
˛EW D
7:97 kO rad=s2
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Dynamics 1e
Solution to 6.89 We assume the wheel does not leave the ground, hence B is constrained to move in the horizontal direction, and from the rolling without slip condition we have vEB D vB {O D
R!W {O
and aEB D
R˛W {O:
(10)
From rigid body kinematics we have vEA D vEB C !EAB rEA=B D D . R!W
R!W C !AB kO L. cos {O C sin |O/ L!AB sin / {O C . L!AB cos / |O: (11)
Because the slider is constrained to move in the slot, vAx D 0: Therefore, setting the x component of the last of Eqs. (11), we have !AB D
!W
R : L sin
We must also have aEA D aEB C ˛EAB rEA=B D D
2 !AB rEA=B
R2 2 R˛W {O C ˛AB kO L. cos {O C sin |O/ !W L. cos {O C sin |O/ L2 sin2 ! 2 R2 !W R2 2 cos {O C ˛AB L cos !W R˛W ˛AB L sin C |O: L sin L sin2
(12)
Because the slider is constrained to move in the slot, aAx D 0: Therefore, setting the x component of the last of Eqs. (11), we have ! 2 R 2 !W 1 ˛AB D cos R˛W : (13) L sin L sin2 Substituting Eq. (13) for ˛AB in Eq. (12) we have aEA D
R˛W tan
2 R2 !W cot2 L sin
2 !W
R2 L sin
! |O D
! 2 R 2 !W cot2 C 1 |O L sin
R˛W tan )
aEA D
R˛W tan
2 R 2 !W
L sin3
! |O:
(14)
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Solutions Manual
Problems 6.90 through 6.93 For the slider-crank mechanism shown, let R D 0:75 m and H D 2 m, and let the length of bar BC be LBC D 3:25 m. Problem 6.90
for D
27ı .
Assume that P D 50 rad=s D constant and compute the angular acceleration of the slider
Assume that, at the instant shown, D 27ı , P D 50 rad=s, and R D 15 rad=s2 . Compute the angular acceleration of the slider at this instant as well as the acceleration of point C .
Problem 6.91
Assuming that P is constant, determine the expression of the angular acceleration of the slider as a function of and P (and the accompanying geometrical parameters).
Problem 6.92
Problem 6.93
function of for
0ı
Letting P D 300 rpm D constant, plot the angular acceleration of the slider as a 360ı . In addition, plot the speed of point C for the same range of .
Solution to 6.90 R we can use differentiation of constraints to find an expression for ˛BC . From the Observing that ˛BC D , geometry of the problem we have the constraint equations dBO sin D R sin
and
Solving Eqs. (1) for dBO and , we have R sin 1 D tan H R cos
dBO cos C R cos D H:
and
dBO D
(1)
H sin : cos sin C cos sin
(2)
Using Eqs. (2) for D 27ı , we have D 0:2503 rad
and
dBO D 1:375 m:
Taking time derivatives of Eqs. (1) we have dPBO sin C dBO P cos D RP cos
and
dPBO cos
dBO P sin
RP sin D 0:
(3)
Solving Eqs. (3) for dPBO and P we have dPBO D R sin C P
and
R cos. C /P P D : dBO
(4)
Substituting known values R D 0:75 m, D 0:2503ı D 27ı , P D 50 rad=s, and dBO D 1:375 m into Eqs. (4), we have dPBO D 24:77 m=s and P D 20:48 rad=s: August 10, 2009
855
Dynamics 1e Taking a time derivative of P from Eq. (4) and letting R D 0, we have P R sin . C /P .P C / R cos . C /dPBO P C : R D 2 dBO dBO
(5)
Substituting known values R D 0:75 m, D 0:2503ı , P D 20:48 rad=s, D 27ı , P D 50 rad=s, dBO D 1:375 m, and dPBO D 24:77 m=s into Eq. (5), we have R D
1639 rad=s2
)
˛EBO D 1640 kO rad=s2 :
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Solutions Manual
Solution to 6.91 R we can use differentiation of constraints to find an expression for ˛BC . From the Observing that ˛BC D , geometry of the problem we have the constraint equations dBO sin D R sin
and
Solving Eqs. (6) for dBO and we have R sin 1 D tan H R cos
dBO cos C R cos D H:
and
dBO D
(6)
H sin : cos sin C cos sin
(7)
Using Eqs. (7) for D 27ı , we have D 0:2503 rad
dBO D 1:375 m:
and
Taking time derivatives of Eqs. (6) we have dPBO sin C dBO P cos D RP cos
and
dPBO cos
dBO P sin
RP sin D 0:
(8)
Solving Eqs. (8) for dPBO and P we have dPBO D R sin C P
R cos. C /P : P D dBO
and
(9)
Substituting known values R D 0:75 m, D 0:2503ı D 27ı , P D 50 rad=s, and dBO D 1:375 m into Eqs. (9), we have dPBO D 24:77 m=s and P D 20:48 rad=s: Taking a time derivative of P from Eq. (9), we have P R cos . C /dPBO P R sin . C /P .P C / R D C 2 dBO dBO
R cos . C /R : dBO
Substituting known values R D 0:75 m, D 0:2503ı , P D 20:48 rad=s, D 27ı , P D 50 rad=s, R D 15 rad=s2 , dBO D 1:375 m, and dPBO D 24:77 m=s into Eq. (5), we have R D
1632 rad=s2
)
˛EBO D 1630 kO rad=s2 :
The acceleration of C can be found by applying rigid body kinematics. Observing that rEB=A D R.sin {O R we have cos |O/, !AB D P , and ˛AB D , aEB D aEA C ˛EAB rEB=A D ˛AB kO R.sin {O
2 !AB rEB=A
2 !AB R.sin {O cos |O/ D . RP 2 sin C RR cos / {O C .RP 2 cos C RR sin / |O
cos |O/
D . 841:2 {O C 1676s |O/ m=s2 ;
(10)
where we have used the known values R D 0:75 m, D 27ı , P D 50 rad=s, and R D 15 rad=s2 . Observing R and rEC =B D L. sin {O cos |O/ we must also have P ˛BC D , that !BC D , aEC D aEB C ˛EOB rEC =B
2 !OB rEC =B
August 10, 2009
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Dynamics 1e D . 841:2 {O C 1676s |O/ m=s2 C .1632 kO rad=s2 / L. sin {O .20:48 rad=s/L. sin {O
cos |O/
cos |O/
D . 848:2 {O C 7153 |O/ m=s; where we have used the expression for aB from Eq. (), and known values R D 0:75 m, D 0:2503ı , P D 20:48 rad=s, R D 1632 rad=s2 . To three significant figures we have aEC D . 848 {O C 7150 |O/ m=s:
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Solutions Manual
Solution to 6.92 R we can use differentiation of constraints to find an expression for ˛BC . From the Observing that ˛BC D , geometry of the problem we have the constraint equations dBO sin D R sin
and
dBO cos C R cos D H:
(11)
Solving Eqs. (11) for we have D tan
1
R sin H R cos
:
(12)
Taking time derivatives of in Eq. (11) we have RP .H cos R/ H 2 C R2 2HR cos 2HR2 . R C H cos / sin P R D .H 2 C R2 2HR cos /2 P D
HRP sin H 2 C R2 2HR cos
(13)
For ˛EOB , we then have
˛EOB D
HRP sin 2HR2 . R C H cos / sin P C .H 2 C R2 2HR cos /2 H 2 C R2 2HR cos
! O k:
August 10, 2009
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Dynamics 1e
Solution to 6.93 R we can use differentiation of constraints to find an expression for ˛BC . From the Observing that ˛BC D , geometry of the problem we have the constraint equations dBO sin D R sin
dBO cos C R cos D H:
and
(14)
Solving Eqs. (14) for we have D tan
1
R sin H R cos
:
(15)
Taking time derivatives of in Eq. (14), we have RP .H cos R/ H 2 C R2 2HR cos 2HR2 . R C H cos / sin P R D .H 2 C R2 2HR cos /2
P D
(16) HRP sin : H 2 C R2 2HR cos
For ˛EOB , we then have
˛EOB D
HRP sin 2HR2 . R C H cos / sin P C .H 2 C R2 2HR cos /2 H 2 C R2 2HR cos
! O k:
We know the position of C can be expressed as rEC D .R sin
L sin / {O
.R cos C L cos / |O:
(17)
For the velocity of C , we take a time derivative of Eq. (17), which gives us vEC D .RP cos
L P cos / {O C .RP sin C L P sin / |O:
(18)
Substituting from Eq. (15) and P from Eq. (16) into Eq. (18), we have ( " # ) P .H cos R/ R R sin vEC D RP cos L cos tan 1 {O H 2 C R2 2HR/ H R cos ( " # ) P .H cos R/ R sin R sin tan 1 |O: C RP sin C L H 2 C R2 2HR H R cos The expressions for ˛EOB and vEC just derived are reduced to functions of and P . The function can be plotted with software such as Mathematica or M ATLAB. The plots presented below were obtained using Mathematica with the following code:
August 10, 2009
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Solutions Manual
Parameters ! !R " 0.75, LBC " 3.25, H " 2, ΘDot " 300 $
ΑOB !
2 H R 2 #H Cos$Θ% ' R& Sin$Θ% ΘDot 'H2 ( R 2 ' 2 H R Cos$Θ%(
2
vC ! Sqrt* R ΘDot Cos$Θ% ' LBC
30
"; ). Parameters;
H R ΘDot Sin$Θ%
(
H2 ( R 2 ' 2 H R Cos$Θ%
R ΘDot #H Cos$Θ% ' R& H2 ( R 2 ' 2 H R
R ΘDot #H Cos$Θ% ' R&
R ΘDot Sin$Θ% ( LBC
Π
H2 ( R 2 ' 2 H R
Cos*ArcTan*
Sin*ArcTan*
R Sin$Θ% H ' R Cos$Θ% R Sin$Θ%
H ' R Cos$Θ%
Plot*ΑOB , ,Θ, 0, 2 Π-, AxesLabel " ,"Θ", "ΑOB "-, Ticks " !!0,
Π
++
2
2
( + ). Parameters;
3Π
, 2 Π", Automatic"+ 2 Π 3Π Plot*vC , ,Θ, 0, 2 Π-, AxesLabel " ,"Θ", "vC "-, Ticks " !!0, , Π, , 2 Π", Automatic"+ 2 2 2
, Π,
++
Using the above code we obtain the following plots: ΑOB 20
10
Θ Π 2
Π
3Π 2
2Π
-10
-20 vC
100
80
60
Θ Π 2
Π
3Π 2
2Π
August 10, 2009
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Dynamics 1e
Problem 6.94 A spool with inner radius R D 5 ft is made to roll without slip over a horizontal rail as shown. If the cable on the spool is unwound in such a way that the free or vertical portion of cable remains perpendicular to the rail, determine the angular acceleration of the spool and the acceleration of the spool’s center O. The vertical component of the velcocity of point A is vA D 12 ft=s, and the vertical component of its acceleration is aA D 2 ft=s2 .
Solution We let point P be the contact point between the spool and rail, and point Q be the contact point between the spool and rope. From the no slip condition, we then have vEP D 0E and
vQy D vA :
(1)
From rigid body kinematics, we must have vEQ D vEP C !E spool rEQ=P D !spool kO . R {O C R |O/ D
!spool R. {O C |O/;
(2)
Equating the y component of vEQ in Eq. (2) with vQy in Eq. (1), we have !spool R D vA
)
!spool D
vA : R
Now that we have found the angular velocity of the spool, we look at the acceleration. From the no slip condition we have aP x D 0 ) aEP D aP y |O: Also, because O can only move in the horizontal direction, we must have aEO D aOx {O: From rigid body kinematics, we must have 2 !spool rEO=P v 2 A aOx {O D aP y |O C ˛spool kO R |O R |O R
aEO D aEP C ˛Espool rEO=P
August 10, 2009
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Solutions Manual " aOx {O D
˛spool R {O C aP y
vA2 R
# |O:
(3)
Writing Eq. (3) as two scalar equations we have aOx D
˛spool R
aP y D
and
vA2 R
)
aEP D
vA2 |O: R
We now know the acceleration of two points on the spool, and we can relate them to find the angular acceleration. We have 2 !spool rEQ=P v 2 v2 A D A |O C ˛spool kO R. {O C |O/ R. {O C |O/ R R ! vA2 ˛spool R {O ˛spool R |O: D R
aEQ D aEP C ˛Espool rEQ=P
(4)
From the rolling without slip condition at Q we have aQy D aA :
(5)
Equating Eq. (4) with the expression for aQy in Eq. (5), we have ˛spool R D aA
)
˛spool D
aA R
)
˛Espool D
0:400 kO rad=s2 :
where we have used the given values R D 5 ft and aA D 2 ft=s2 . Substituting this expression for ˛spool into Eq. (3) we have aEO D aA {O
)
aEO D 2:00 {O m=s2 ;
where we have used the given value aA D 2 ft=s2 .
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Dynamics 1e
Problem 6.95 At the instant shown, bars AB and BC are perpendicular to each other while the slider C has a velocity vC D 24 m=s and an acceleration aC D 2:5 m=s2 in the directions shown. Letting L D 1:75 m and D 45ı , determine the angular acceleration of bars AB and BC .
Solution Because D 45ı , the instantaneous center of rotation of bar BC is point A. Therefore, p vC vC D 2L!BC ) !BC D p D 9:697 rad=s; 2L where we have used the given values vC D 24 m=s and L D 1:75 m. Because C is constrained to move along the guide oriented by the unit vector p uO t D .O{ C |O/= 2, and is moving up and to the right, we have vEC D vC .cos 45ı {O C sin 45ı |O/ D .16:97 {O C 16:97 {O/ ft=s
(1)
and aEC D aC .cos 45ı {O C sin 45ı |O/ D . 1:786 {O
1:786 |O/ ft=s2 ;
(2)
where we have used the given values vC D 24 m=s and aC D 2:5 m=s2 . Point A is also the center of rotation of bar AB. Therefore, in the present configuration vB D !AB L D !BC L
)
!AB D !BC D 9:697 rad=s:
From rigid body kinematics we have aEB D aEA C ˛EAB rEB=A
2 2 2 !AB rEB=A D ˛AB kO . L |O/ C !AB L |O D L˛AB {O C !AB L |O:
We must also have 2 !BC rEC =B L |O/ C ˛BC kO L {O
aEC D aEB C ˛EBC rEC =B 2 D .L˛AB {O C !AB
D .˛AB L
!AB L {O
2 2 !AB L/ {O C .˛BC L C !AB L/ |O
(3)
Equating the expressions for aEC in Eqs. (2) and (3), on a component by component basis, we have 1:786 m=s2 D ˛AB L
2 !AB L and
2 1:786 m=s2 D ˛BC L C !AB L:
(4) August 10, 2009
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Solutions Manual
Eqs. (4) are two equations in the two unknowns ˛AB and ˛BC . Solving them we have ˛AB D
2 L !AB
1:786 m=s2 L
and ˛BC D
2 L/ .1:786 m=s2 C !AB : L
Substituting in known values L D 1:75 m and !AB D 9:697 rad=s, and putting the angular accelerations in vector form, we have ˛EAB D 93:0 kO rad=s2
and ˛EBC D
95:1 kO rad=s2 :
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Dynamics 1e
Problem 6.96 A flood gate is controlled via the hydraulic cylinder AB. If the length of the cylinder is increased with a constant time rate of 2:5 ft=s, determine the angular acceleration of the gate when D 0ı . Let ` D 10 ft, h D 2:5 ft, and d D 5 ft.
Solution At any position of the gate, we have the constraint equations `
LAB cos D d cos
and LAB sin
h D d sin :
(1)
where LAB is the distance between points A and B. For D 0, we then have q h D tan 1 and LAB D h2 C .` d /2 : (2) ` d Substituting known values h D 2:5 ft, d D 5 ft, and ` D 10 ft into Eqs. (2), we have D 0:4636 rad
and LAB D 5:590 ft:
Taking time derivatives of Eqs. (1) we have LAB P sin D d P cos
and LAB P cos D d P sin :
(3)
P Solving them, we have Eqs. (3) are two equations in P and . P AB cot . C /L P D LAB
P AB csc . C /L and P D : d
(4)
P AB D 2:5 ft=s into Substituting known values D 0:4636 rad, D 0 rad, d D 5 ft, LAB D 5:590 ft, and L Eqs. (4) we have P D 0:8946 rad=s and P D 1:118 rad=s: R we take a time derivative of P from Eqs. (4), and remembering that LR AB D 0, Observing that ˛gate D , we have cot . C / csc . C /LP AB P C P ˛gate D R D d P AB D Substituting in known values D 0:4636 rad, P D 0:8946 rad=s, D 0 rad, P D 1:118 rad=s, L 2:5 ft=s, and d D 5 ft, and expressing in vector form with three significant figures, we have ˛Egate D 4:50 kO rad=s2 :
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Solutions Manual
Problem 6.97 The bucket of a backhoe is the element AB of the four-bar linkage system ABCD. Assume that the points A and D are fixed and that the bucket rotates with a constant angular velocity !AB D 0:25 rad=s. In addition, suppose that, at the instant shown, point B is aligned vertically with point A, and C is aligned horizontally with B. Determine the acceleration of point C at the instant shown along with the angular accelerations of the elements BC and CD. Let h D 0:66 ft, e D 0:46 ft, l D 0:9 ft, and w D 1:0 ft.
Solution Point C is a point on both bar BC and bar CD, and therefore can be related to both points B and D through rigid body kinematics. We have that vEC D vED C !E CD rEC =D D !CD kO Π.w
h/ {O C .e C `/ |O D
!CD .e C `/ {O
!CD .w
h/ |O: (1)
Observing that A is the instantaneous enter of rotation of AB, vBy D 0, and vEB D vBx {O. We then have vEC D vEB C !EBC rEC =B D vBx {O C !BC . w/ {O D vBx {O
!BC w |O:
(2)
Equating Eqs. (1) and (2) component by component we have !CD .e C `/ {O D vBx {O
!CD .w
and
h/ |O D
!BC w |O:
(3)
Observing that A is the instantaneous center of rotation of AB, we also have vB D j!AB j`. Solving Eqs. (3), for !CD and !BC , we have !CD .e C `/ D j!AB j`
)
!CD D
j!AB j` eC`
(4)
)
!BC D
j!AB j` w h : eC` w
(5)
and !BC D
!CD .w w
h/
Substituting known values j!AB j D 0:25 rad=s, e D 0:46 ft, ` D 0:9 ft, h D 0:66 ft, and w D 1:0 ft into Eqs. (4) and (5), we have !E CD D 0:1654 kO rad=s;
!EBC D 0:05625 kO rad=s:
From rigid body kinematics we must also have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A D
aEC D aEB C ˛EBC rEC =B
2 !BC rEC =B
D
2 !AB ` |O
C ˛BC
kO . w {O/
2 !AB ` |O;
!AB ` w h eC` w
(6) 2 . w {O/ August 10, 2009
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Dynamics 1e
!AB ` w h Dw eC` w
2 {O
2 .!AB ` C ˛BC `/ |O;
(7)
and 2 !CD rEC =D !AB ` 2 O D ˛CD k .e C `/ |O .e C `/ |O eC` .!AB `/2 D ˛CD .e C `/ {O |O eC`
aEC D aED C ˛ECD rEC =D
(8)
Equating Eqs. (7) and (8) component by component we have
!AB ` w h w eC` w
2 .!AB `
2 D
.!AB `/2 C ˛BC `/ D eC`
˛CD .e C `/
)
˛BC
)
˛CD D
!2 ` D AB eC`
Œ!AB `.w h/2 D .e C `/w
2 !AB
)
˛BC D
0:01749 rad=s2
2 !AB
` eC`
(9)
1
D 0:004516 rad=s2 : (10) In vector form, we have ˛EBC D
0:0175 |O rad=s2
and
˛ECD D 0:00452 kO rad=s2 :
Substituting known values w D 1:0 ft, ` D 0:9 ft, e D 0:46 ft, h D 0:66 ft, j!AB j D 0:25 rad=s, and ˛AB D 0:004516 rad=s2 into Eq. (7) we have aEC D .0:00316 {O
0:0388 |O/ ft=s2 :
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Solutions Manual
Problem 6.98 At the instant shown, bar CD is rotating with an angular velocity 20 rad=s and with angular acceleration 2 rad=s2 in the directions shown. Furthermore, at this instant D 45ı . Letting L D 2:25 ft, determine the angular accelerations of bars AB and BC .
Solution Because bar AB is rotating about the fixed point A, at the given instant point B is only moving in the {O direction, and we have vEB D !AB L {O: Points B and C are both on the bar BC so we have vEC D vEB C !EBC rEC =B D !AB L {O C !BC kO L {O D !AB L {O C !BC L |O:
(1)
Points C and D are also both on the bar CD so we have L L vEC D vED C !E CD rEC =D D !CD kO .cos {O C sin |O/ D !CD . sin {O C cos |O/: 2 2
(2)
Equating Eqs. (1) and (2) component by component, we have L !CD sin !CD sin ) !AB D 2 2 L !CD cos !BC L D !CD cos ) !BC D : 2 2 !AB L D
(3) (4)
Similarly for the accelerations, we have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A
2 !CD sin2 . L |O/ 4 2 L!CD sin2 D L˛AB {O C |O; 4
D ˛AB kO . L |O/
as well as 2 !BC rEC =B ! 2 L!CD sin2 D L˛AB {O C |O C ˛BC kO L {O 4
aEC D aEB C ˛EBC rEC =B
2 !CD cos2 L {O 4
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869
Dynamics 1e D L˛AB
2 cos2 L!CD 4
! {O C
2 sin2 L!CD C L˛BC 4
! |O;
(5)
and 2 aEC D aED C ˛ECD rEC =D !CD rEC =D L 2 L .cos {O C sin |O/ D ˛CD kO .cos {O C sin |O/ !CD 2 2 L L 2 L 2 L D ˛CD sin C !CD cos {O C ˛CD cos !CD sin |O: 2 2 2 2
(6)
Equatings Eqs. (5) and (6) component by component we have L˛AB
2 cos2 L!CD D 4
2 L!CD sin2 C L˛BC 4
L 2 L ˛CD sin C !CD cos 2 2 2 sin 2 cos 2 cos ) ˛AB D !CD ˛CD !CD 4 2 2 L 2 L D ˛CD cos !CD sin 2 2 2 !CD sin2 cos 2 sin C ˛CD !CD : ) ˛BC D 4 2 2
Substituting known values !CD D 20 rad=s; ˛CD D 2 rad=s2 ; D 45ı ; and L D 2:25 ft; in vector form we have ˛EAB D 92:1 kO rad=s2 and ˛EBC D 191 kO rad=s2 :
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Solutions Manual
Problems 6.99 through 6.101 A wheel W of radius RW D 2 in: rolls without slip over the stationary cylinder S of radius RS D 5 in:, and the wheel is connected to point O via the arm OC . Determine the acceleration of the point on the wheel W that is in contact with S for !OC D 7:5 rad=s D constant. Problem 6.99
Determine the acceleration of the point on the wheel W that Problem 6.100 is in contact with S for !OC D 7:5 rad=s and ˛OC D 2 rad=s2 . If, at the instant shown, D 63ı , !W D 9 rad=s, and ˛W D 1:3 rad=s2 , determine the angular acceleration of the arm OC and the acceleration of point P , where P lies on the edge of W and is aligned vertically with point C . Problem 6.101
Solution to 6.99 We call the contact point between the wheel W and cylinder S point Q. From the geometry of the problem, we have rEC =O D .RS C RW / uO r
and
rEC =Q D RW uO r :
From rigid body kinematics we have vEC D vEO C !E OC rEC =O D !OC kO .RS C RW / uO r D !OC .RS C RW / uO :
(1)
E Therefore, we must have Due to rolling without slip, vEQ D 0: vEC D vEQ C !E W rEC =Q D !W kO RW uO r D !W RW uO
(2)
Equating Eqs. (1) and (2) component by component we have !OC .RS C RW / D !W RW
)
!W D
.RS C RW / !OC : RW
(3)
We must also have aEC D aEO C ˛EOC rEC =O D
2 !OC rEC =O
2 !OC .RS C RW / uO r
and 2 !W rEQ=C 2 2 !OC .RS C RW / uO r C ˛W kO . RW / uO r C !W RW uO r
aEQ D aEC C ˛EW rEQ=C D
2 2 D Π!OC .RS C RW / C !W RW uO r
RW ˛W uO
(4)
August 10, 2009
871
Dynamics 1e From the no slip condition, we know that aEQ D 0. Substituting Eq. (3) for !W in Eq. (4) we have aEQ D
2 !OC .RS
.RS C RW /2 2 RS C RW 2 C RW / C !OC uO r D !OC .RS C RW / RW RW
1 uO r :
(5)
Substituting known values RW D 2 in: D 0:1667 ft; RS D 5 in: D 0:4167 ft; and !OC D 7:5 rad=s, we have aEQ D 82:0 uO r ft=s2 :
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Solutions Manual
Solution to 6.100 We call the contact point between the wheel W and cylinder S point Q. From the geometry of the problem, we have rEC =O D .RS C RW / uO r
rEC =Q D RW uO r :
and
From rigid body kinematics we have vEC D vEO C !E OC rEC =O D !OC kO .RS C RW / uO r D !OC .RS C RW / uO :
(6)
E Therefore, we must have Due to rolling without slip, vEQ D 0: vEC D vEQ C !E W rEC =Q D !W kO RW uO r D !W RW uO :
(7)
Equating Eqs. (6) and (7) component by component we have !OC .RS C RW / D !W RW
)
!W D
.RS C RW / !OC : RW
(8)
We must also have 2 aEC D aEO C ˛EOC rEC =O !OC rEC =O 2 D ˛OC kO .RS C RW / uO r !OC .RS C RW / uO r
D
2 !OC .RS C RW / uO r C .RS C RW /˛OC uO
and aEQ D aEC C ˛EW rEQ=C
2 !W rEQ=C
2 2 D Œ !OC .RS C RW / uO r C .RS C RW /˛OC uO C ˛W kO . RW / uO r C !W RW uO r 2 2 D Œ !OC .RS C RW / C !W RW uO r C Œ.RS C RW /˛OC
RW ˛W uO :
(9)
From the no slip condition, we know that aEQ D 0
)
˛W D
RS C RW ˛OC : RW
Substituting Eq. (8) for !W in Eq. (9) we have .RS C RW /2 2 RS C RW 2 2 !OC uO r D !OC .RS C RW / aEQ D !OC .RS C RW / C RW RW
(10)
1 uO r : (11)
Substituting known values RW D 2 in: D 0:1667 ft; RS D 5 in: D 0:4167 ft; and !OC D 7:5 rad=s, we have aEQ D 82:0 uO r ft=s2 :
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Dynamics 1e
Solution to 6.101 We call the contact point between the wheel W and cylinder S point Q. From the geometry of the problem, we have rEC =O D .RS C RW / uO r ; rEP =C D RW Œsin uO r C cos uO ; and rEC =Q D RW uO r : From rigid body kinematics we have vEC D vEO C !E OC rEC =O D !OC kO .RS C RW / uO r D !OC .RS C RW / uO :
(12)
E Therefore, we must have Due to rolling without slip, vEQ D 0: vEC D vEQ C !E W rEC =Q D !W kO RW uO r D !W RW uO
(13)
Equating Eqs. (12) and (13) component by component we have !OC .RS C RW / D !W RW
)
!OC D
RW !W : .RS C RW /
(14)
We must also have 2 aEC D aEO C ˛EOC rEC =O !OC rEC =O 2 D ˛OC kO .RS C RW / uO r !OC .RS C RW / uO r 2 !OC .RS C RW / uO r C .RS C RW /˛OC uO
D and
aEQ D aEC C ˛EW rEQ=C
2 !W rEQ=C
2 2 D Œ !OC .RS C RW / uO r C .RS C RW /˛OC uO C ˛W kO . RW / uO r C !W RW uO r 2 2 D Œ !OC .RS C RW / C !W RW uO r C Œ.RS C RW /˛OC
RW ˛W uO :
From the no slip condition, we know that aEQ D 0
)
˛OC D
RW ˛W : RS C RW
(15)
We must also have aEP D aEC C ˛EW rEP =C
2 !W rEP =C
2 !OC .RS C RW / uO r C .RS C RW /˛OC uO C ˛W kO RW .sin uO r C cos uO /
D
2 !W RW .sin uO r C cos uO /
D
2 2 !OC .RS C RW / RW ˛W cos !W RW sin uO r 2 C .RS C RW /˛OC C RW ˛W sin !W RW cos uO
(16) August 10, 2009
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Solutions Manual
Substituting Eqs. (14) and (15) into Eq. (16) we have " # 2 RW 2 2 aEP D ! RW ˛W cos !W RW sin uO r .RS C RW / W 2 C RW ˛W .1 C sin / !W RW cos uO
(17)
Substituting known values RW D 2 in: D 0:1667 ft; RS D 5 in: D 0:4167 ft; D 63ı ; !W D 9 rad=s; and ˛W D 1:3 rad=s2 , into Eqs. (15) and (17), to three significant figures we have ˛EOC D
0:371 kO rad=s2
and aEP D . 15:8 uO r
6:54 uO / ft=s2 :
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Dynamics 1e
Problem 6.102 At the instant shown, bars AB and CD are vertical. In addition, point C is moving to the left with a speed of 4 m=s, and the magnitude of the acceleration of C is 55 m=s2 . Letting L D 0:5 m and H D 0:2 m, determine the angular accelerations of bars AB and BC .
Solution Realizing that if C is moving left, !CD must be counterclockwise, from the concept of instantaneous center of rotation, we have vC O aC O !E CD D k D 8:000 kO rad=s and ˛ECD D k: (1) L L Similarly, because the center of rotation of bar BC is at infinity, we have E !EBC D 0: Therefore, the velocity of B is the same as the velocity of C , so we must have !EAB D
vC O kD L
8:000 kO rad=s:
(2)
Observing that at the given instant we have the position vectors rEB=A D
L |O;
rEB=C D
.H {O C 2L |O/;
and rEC =D D L |O;
then from rigid body kinematics we must have aEC D aED C ˛ECD rEC =D
2 !CD rEC =D D ˛CD kO L |O
aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A D ˛AB kO . L/ |O
2 vC |O; L 2 2 !AB . L/ |O D L˛AB {O C L!AB |O;
2 !CD .L |O/ D
L˛CD {O
(3)
and 2 !BC rEC =B 2 D L˛AB {O C L!AB |O C ˛BC kO .H {O C 2L |O/
aEC D aEB C ˛EBC rEC =B D .L˛AB
2L˛BC
2 !BC .H {O C 2L |O/
2 2 H !BC / {O C .L!AB C H˛BC
2 2L!BC / |O:
(4)
Equations (3) and (4) are two expressions for aEC , and setting them equal component by component, we have L˛CD D L˛AB
2L˛BC
2 H !BC
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Solutions Manual
and 2 vC 2 D L!AB C H˛BC L
2 2L!BC ;
which are two equations in ˛AB and ˛BC . Solving them, we have 2 2 2 2 2 L H 2vC a C 2L! C AB C H C 4L !BC L ˛AB D HL 2 2 2 2!BC C L2 !AB vC ˛BC D ; HL where we have let ˛CD D aLC . However, closer inspection, only the magnitude of aC is given in the problem, so we must find answers for both the positive and negative values. Choosing aC D 55 m=s2 , and substituting the given values vC D 4 m=s, aC D 55 m=s2 , L D 0:5 m, H D 0:2 m, !AB D 8:000 rad=s, and !BC D 8:000 rad=s, in vector forma and to three significant figures, we have ˛EAB D
84:4 kO rad=s2
and
˛EBC D
315 kO rad=s2 :
Choosing aC D 55 m=s2 and substituting the given values vC D 4 m=s, aC D 55 m=s2 , L D 0:5 m, H D 0:2 m, !AB D 8:000 rad=s, and !BC D 8:000 rad=s, in vector forma and to three significant figures, we have ˛EAB D 136 kO rad=s2 and ˛EBC D 315 kO rad=s2 :
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Dynamics 1e
Problems 6.103 through 6.106 For the slider-crank mechanism shown, let R D 1:9 in:, L D 6:1 in:, and H D 1:2 in: Assuming that !AB D 4850 rpm and is constant, determine the angular acceleration of Problem 6.103 the connecting rod BC and the acceleration of point C at the instant when D 27ı . Assuming that !AB D 4850 rpm and is constant, determine the acceleration of point D at the instant when D 10ı .
Problem 6.104
Assuming that, at the instant shown, D 31ı , !AB D 4850 rpm, and ˛AB D !PAB D 280 rad=s2 , determine the angular acceleration of the connecting rod and the acceleration of point C .
Problem 6.105
Determine the general expression of the acceleration of the piston C as a function of L, P and ˛AB D . R R, , !AB D , Problem 6.106
Solution to 6.103 From the geometry of the problem we have rEB=A D R.cos {O C sin |O/; rEC =B D L. sin {O C cos |O/; and L sin D R cos
)
D sin
1
R cos : L
From rigid body kinematics we have vEB D vEA C !EAB rEB=A D !AB kO R.cos {O C sin |O/ D
R!AB sin {O C R!AB cos |O
and vEC D vEB C !EBC rEC =B D
R!AB sin {O C R!AB cos |O C !BC kO L. sin {O C cos |O/
D . R!AB sin
L!BC cos / {O C .R!AB cos
L!BC sin / |O: August 10, 2009
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Solutions Manual
Because C is constrained to move in the vertical direction, we have L!BC cos D 0
R!AB sin
)
!BC D
R!AB sin : L cos
Remembering that ˛AB D 0, we must also have 2 !AB rEB=A
aEB D aEA D
2 !AB R.cos {O C sin |O/
D
2 R!AB cos {O
2 !AB sin |O
and aEC D aEB C ˛EBC rEC =B
2 !BC rEC =B
2 2 R!AB cos {O !AB sin |O 2 C ˛BC kO L. sin {O C cos |O/ !BC L. sin {O C cos |O/ ! # " 2 ! 2 sin2 R 2 AB sin L˛BC cos {O D R!AB cos C L cos2 " ! # 2 ! 2 sin2 R 2 AB C R!AB sin L˛BC sin cos |O L cos2
D
(1)
Because C is constrained to move in the vertical direction, we have 2 R!AB cos C
2 R2 !AB sin2 L cos2
! sin
L˛BC cos 2 R!AB cos C
)
˛BC D
2 R2 !AB sin2 2 L cos
L cos
sin (2)
Substituting known values R D 1:9 in: D 0:1583 ft; L D 6:1 in: D 0:5083 ft; H D 1:2 in: D 0:1000 ft; D 27ı ; and !AB D 4850 rpm D 507:9 rad=s into Eqs. (1) and (2), rounding to three significant figures, and putting in vector form we have aEC D
11000 |O ft=s2
and ˛EBC D
72900 kO rad=s2 :
(3)
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Dynamics 1e
Solution to 6.104 From the geometry of the problem we have rEB=A D R.cos {O C sin |O/; rEC =B D L. sin {O C cos |O/; rED=C D .L
H /.sin {O
cos |O/;
and L sin D R cos
)
D cos
1
L sin R
D 56:17ı :
From rigid body kinematics we have vEB D vEA C !EAB rEB=A D !AB kO R.cos {O C sin |O/ D
R!AB sin {O C R!AB cos |O
and vEC D vEB C !EBC rEC =B D
R!AB sin {O C R!AB cos |O C !BC kO L. sin {O C cos |O/
D . R!AB sin
L!BC cos / {O C .R!AB cos
L!BC sin / |O:
Because C is constrained to move in the vertical direction, we have L!BC cos D 0
R!AB sin
)
!BC D
R!AB sin : L cos
Remembering that ˛AB D 0, we must also have 2 !AB rEB=A D
aEB D aEA
2 !AB R.cos {O C sin |O/ D
2 R!AB cos {O
2 !AB sin |O
and aEC D aEB C ˛EBC rEC =B
2 !BC rEC =B
2 2 R!AB cos {O !AB sin |O 2 C ˛BC kO L. sin {O C cos |O/ !BC L. sin {O C cos |O/ ! # " 2 2 2 R !AB sin 2 sin L˛BC cos {O D R!AB cos C L cos2 " ! # 2 R2 !AB sin2 2 R!AB sin L˛BC sin C cos |O: L cos2
D
(4)
Because C is constrained to move in the vertical direction, aC x D 0, and we have ! 2 R2 !AB sin2 2 R!AB cos C sin L˛BC cos D 0 L cos2 2 R!AB
)
˛BC D
cos C
2 R2 !AB sin2 2 L cos
sin :
L cos
Substituting known values R D 1:9 in: D 0:1583 ft, L D 6:1 in: D 0:5083 ft, H D 1:2 in: D 0:1000 ft, D 10ı , D 56:12ı , and !AB D 4850 rpm D 507:9 rad=s, we have aEC D
39070 |O ft=s2
and ˛EBC D
42330 kO rad=s2 : August 10, 2009
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Solutions Manual
Finally, we must have aED D aEC C ˛EBC rED=C D aEC C ˛BC kO .L " D
2 R!AB cos
H
2 !BC rED=C
H /.sin {O
cos |O/
R˛AB sin C
2 cos LR!AB
1 H tan L2
2 sin2 HR2 !AB L2 cos
H /.sin {O
2 R!AB sin 2 LR!AB
cos |O/
tan
2 sec sin 2 tan LR˛AB sin C R2 !AB L2
" C R˛AB cos
2 !BC .L
# {O
2 sec sin 2 HR2 !AB L2
cos
LR˛AB sin C R
2
2 !AB
2
sec sin tan
|O
(5)
Substituting known values R D 1:9 in: D 0:1583 ft, L D 6:1 in: D 0:5083 ft, H D 1:2 in: D 0:1000 ft, D 10ı , D 56:12ı , and !AB D 4850 rpm D 507:9 rad=s into Eq. (5), and rounding to three significant figures, we have aED D .31000 {O
46100 |O/ ft=s2 :
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Dynamics 1e
Solution to 6.105 From the geometry of the problem we have rEB=A D R.cos {O C sin |O/;
rEC =B D L. sin {O C cos |O/;
and L sin D R cos
)
D sin
1
R cos : L
From rigid body kinematics we have vEB D vEA C !EAB rEB=A D !AB kO R.cos {O C sin |O/ D
R!AB sin {O C R!AB cos |O
and vEC D vEB C !EBC rEC =B D
R!AB sin {O C R!AB cos |O C !BC kO L. sin {O C cos |O/
D . R!AB sin
L!BC cos / {O C .R!AB cos
L!BC sin / |O:
Because C is constrained to move in the vertical direction, we have L!BC cos D 0
R!AB sin
)
!BC D
R!AB sin : L cos
We must also have 2 aEB D aEA C ˛EAB rEB=A !AB rEB=A 2 D ˛AB kO R.cos {O C sin |O/ !AB R.cos {O C sin |O/
D . R˛AB sin
2 R!AB cos / {O C .R˛AB cos
2 !AB sin / |O
and aEC D aEB C ˛EBC rEC =B
2 !BC rEC =B
2 2 D . R˛AB sin R!AB cos / {O C .R˛AB cos !AB sin / |O 2 C ˛BC kO L. sin {O C cos |O/ !BC L. sin {O C cos |O/ ! # " 2 R2 !AB sin2 2 sin L˛BC cos {O D R˛AB sin R!AB cos C L cos2 " ! # 2 R2 !AB sin2 2 C R˛AB cos R!AB sin L˛BC sin cos |O: L cos2
(6)
Because C is constrained to move in the vertical direction we have ! 2 R2 !AB sin2 2 R˛AB sin R!AB cos C sin L˛BC cos D 0 L cos2 2 R2 !AB sin2 2 R!AB cos C sin L˛BC cos L cos2 ) ˛AB D (7) R sin Substituting known values R D 1:9 in: D 0:1583 ft; L D 6:1 in: D 0:5083 ft; H D 1:2 in: D 0:1000 ft; D 31ı ; ˛AB D 280 rad=s2 and !AB D 4850 rpm D 507:9 rad=s into Eqs. (6) and (7), rounding to three significant figures, and putting it in vector form, we have ˛EBC D
69400 kO rad=s2
and
aEC D
15200 |O ft=s2 : August 10, 2009
882
Solutions Manual
Solution to 6.106 Letting the origin of the reference frame be at A, at the indicated instant, we have that the vertical distance from A to C , yC , is yC D R sin C L cos : We also know that L sin D R cos : Hence, yC can be rewritten as s
yC D R sin C L 1
R cos L
2 :
Because C is constrained to move vertically, have 3
2 6 vEC D yPC |O D RP cos 41 C
R sin 7 q 5 |O 2 2 R cos L 1 L2
and differentiating again with respect to time, we have 2 0 6 B aEC D 4L @
R2 cos2 P 2
q L2 1 0 B @
R2 cos 2 L2
R2 sin 2 P 2 q L2 1
R2 cos2 L2
1 2 P sin C 3=2 A 2 2
R4 cos2 L4 1 C
R cos L2
R2 cos q L2 1
2
3 1 R sin C 7 sin P 2 C cos R 5 |O: ACR 2 2
R cos L2
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Dynamics 1e
Problems 6.107 and 6.108 In the four-bar linkage system shown, let the circular guide with center at O be fixed and such that, for D 0ı , the bars AB and BC are vertical and horizontal, respectively. In addition, let R D 0:6 m, L D 1 m, and H D 1:25 m. When D 37ı , ˇ D 25:07ı , and D 78:71ı , assume collar C is sliding clockwise with a speed 7 m=s. Assuming that, at the instant in question, the speed is increasing and that jE aC j D 93 m=s2 , determine the angular accelerations of the bars AB and BC . Problem 6.107
Problem 6.108 Use the method of differentiation of constraints to derive expressions for the angular accelerations of bars AB and BC as a P and R . Finally, Let D .0:3 rad=s2 /t 2 and plot the angular function of , , accelerations of AB and BC for 0 t 1 s.
Solution to 6.107 From the geometry of the problem, we have the following position vectors: rEC =O D R.cos {O C sin |O/; rEB=C D L.cos ˇ {O
sin ˇ |O;
rEB=A D H. cos {O C sin |O/; and the velocity vector vEC D vC .sin {O
cos |O/:
O Points B and C are both on bar BC so from rigid body We also define !EBC D !BC kO and !EAB D !AB k. kinematics we have vEB D vEC C !BC kO rEB=C D vC .sin {O
cos |O/ C !BC L.sin ˇ {O C cos ˇ |O/
D .vC sin C !BC L sin ˇ/ {O C .!BC L cos ˇ
vC cos / |O:
(1)
Also, because points A and B are both on bar AB, we have vEB D vEA C !EAB rEB=A D
!AB H.sin {O C cos |O/:
(2)
Equating Eq. (1) with Eq. (2) component by component we have vC sin C !BC L sin ˇ D !BC L cos ˇ
vC cos D
!AB H sin
(3)
!AB H cos :
(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 vC sin C !BC L sin ˇ D tan !BC L cos ˇ vC cos
)
vC sin C !BC L sin ˇ D tan .!BC L cos ˇ )
!BC D vC
vC cos /
sin C tan cos : L.tan cos ˇ sin ˇ/ August 10, 2009
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Solutions Manual
Substituting in given values vC D 7 m=s, D 37ı , D 78:71ı , ˇ D 25:07ı , and L D 1 m, we have !BC D 22:09 rad=s: Solving Eq. (3) for !AB and substituting in known values vC D 7 m=s, !BC D 22:09 rad=s, D 37ı ,
D 78:71ı , ˇ D 25:07ı , H D 1:25 m, and L D 1 m, we have !AB D
vC sin C !BC L sin ˇ D 13:36 rad=s: H sin
We can now begin to look at the acceleration analysis. For the slider we have aEC D aEO C ˛EOC rEC =O
2 !OC rEC =O
2 vC D ˛OC kO R.cos {O C sin |O/ R.cos {O C sin |O/ 2 ! R ! 2 2 vC vC D R˛OC sin cos {O C R˛OC cos sin |O: R R
(5)
Taking the magnitude of aEC and solving for ˛OC we have 2 aC D
R˛OC sin
2 vC cos R
!2 C R˛OC cos
2 vC sin R
s
!2 )
˛OC D ˙
2 aC R2
4 vC R4
(6)
Because the speed is increasing, the angular acceleration must be clockwise, and we choose the negative solution for ˛OC . Substituting in known values R D 0:6 m; L D 1 m; H D 1:25 m; D 37ı ; ˇ D 25:07ı ; and D 78:71ı we have ˛OC D
74:15 rad=s2 :
From rigid body kinematics we have aEB D aEC C ˛EBC rEB=C D
2 L!BC cos ˇ
2 !BC rEB=C 2 vC cos C L˛BC sin ˇ R
! R˛OC sin {O
2 C L˛BC cos ˇ C R˛OC cos C L!BC sin ˇ
2 vC sin R
! |O
(7)
and aEB D aEA C ˛EAB rEB=A 2 D .H !AB cos
2 !AB rEB=A
H˛AB sin / {O C . H˛AB cos
2 H !AB sin / |O:
(8)
Setting Eqs. (7) and (8) equal and solving for ˛AB and ˛BC we have 2 csc .ˇ / ˚ 2 2 HR!AB cos .ˇ / C vC cos .ˇ C / C R L!BC C R˛OC sin .ˇ C / ; HR csc .ˇ / ˚ 2 2 2 LR!BC cos .ˇ / C vC cos . C / C R H !AB C R˛OC sin . C / : ˛BC D LR Substituting in known values as designated above, rounding to three significant figures, and putting in vector form, we have ˛AB D
˛EAB D
1090 kO rad=s2
and ˛EBC D
1190 kO rad=s2 : August 10, 2009
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Dynamics 1e
Solution to 6.108 To solve the problem we first assume that ˇ and are known functions of (and of the geometric parameters R, L, and H ). Then we proceed to determine ˇ and as functions of . Let ˛EAB D ˛AB kO and ˛EBC D ˛BC kO denote the angular accelerations of bars AB and BC , respectively, and therefore we have ˛AB D
R
and
˛BC D
R ˇ:
(9)
Hence, the answer to the problem is found by providing expressions for ˇR and R as a function of the angle and its time derivatives. To do so, we observe that, using trigonometry, the motion of the mechanism is subject to the following two constraint equations: L sin ˇ C H sin D H C R sin ; L cos ˇ C H cos D R C L
(10)
R cos :
(11)
Differentiating the above equations with respect to time, we have LˇP cos ˇ C H P cos D RP cos ; LˇP sin ˇ C H P sin D RP sin :
(12) (13)
Equations (12) and (13) can be viewed as a system of 2 equations in the 2 unknowns ˇP and P whose solution is RP sin. C / RP sin.ˇ C / ˇP D and P D ; (14) L sin.ˇ / H sin.ˇ / where we have used the following trigonometric identities: sin.x C y/ D sin x cos y C cos x sin y and sin.x y/ D sin x cos y cos x sin y. Differentiating the first of Eqs. (14) with respect to time we obtain R sin. C / RP . P C P / cos. C / sin.ˇ / .ˇP P / sin. C / cos.ˇ / R ˇR D L sin.ˇ / L sin2 .ˇ / RR sin. C / RP P cos. C / sin.ˇ / C P sin.ˇ C / ˇP sin. C / cos.ˇ / D ; (15) L sin.ˇ / L sin2 .ˇ / where we have used the trigonometric identity sin.x C y/ D sin x cos y C cos x sin y. Next, substituting R we have Eqs. (14) into Eq. (15), simplifying, and recalling that ˛BC D ˇ, ˛BC D
RR sin. C / L sin.ˇ / RP 2 cos. C / sin2 .ˇ C
/ C
R H
sin2 .ˇ C / C
L sin3 .ˇ
R L
sin2 . C / cos.ˇ
/
:
/
(16)
Repeating similar steps, i.e., differentiating with respect to time the second of Eqs. (14) to obtain an expression P , R ˇP and , for R in terms of , , P recalling that ˛AB D , R and finally using Eqs. (14) again to replace the P expressions for ˇ and , P after simplification, we have ˛AB D
RR sin.ˇ C / H sin.ˇ / RP 2 cos.ˇ C / sin2 .ˇ
/ C
R L
sin2 . C / C
H sin3 .ˇ
/
R H
sin2 .ˇ C / cos.ˇ
/
:
(17)
August 10, 2009
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Solutions Manual
We now need to provide ˇ and as functions of . This can be done in (at least) two ways. The time interval of the needed plots is 0 t 1 s, which corresponds to 0 0:3 rad. Hence, we can use numerical software to obtain ˇ and for 0 0:3 rad and then plot the results. A second but challenging approach, is to find analytical expressions of ˇ and as functions of . We will explore both approaches. Determination of ˇ and via a numerical approach. We begin by going back to Eqs. (10) and (11), which we now differentiate with respect to to obtain Lˇ 0 cos ˇ C H 0 cos D R cos ; 0
0
Lˇ sin ˇ C H sin D
(18)
R sin ;
(19)
where ˇ 0 D dˇ=d and 0 D d =d . The above equations can be solved for ˇ 0 and 0 to obtain ˇ0 D
R sin. C / L sin.ˇ /
and
0 D
R sin.ˇ C / : H sin.ˇ /
(20)
Now observe that for D 0, ˇ D 0 and D =2. Hence, we can obtain ˇ and by numerically solving the system of first order differential equations in Eqs. (20) subject to the initial conditions ˇ.0/ D 0 and
.0/ D =2. Once this is done, recalling that H D 1:25 m, L D 1 m, and R D 0:6 m, D .0:3 rad=s2 /t 2 , P D .0:6 rad=s2 /t , and R D 0:6 rad=s2 , we can then plot the expressions for ˛AB and ˛BC found earlier. The plots in question are shown below.
Angular Acceleration of AB ΑBC !rad#s2 "
ΑAB !rad#s2 "
0.4 0.3 0.2 0.1 0.0 0.0 0.2 0.4 0.6 0.8 1.0 t !s"
Angular Acceleration of BC !0.348 !0.350 !0.352 !0.354 !0.356 !0.358 !0.360 0.0 0.2 0.4 0.6 0.8 1.0 t !s"
These plots and the operations needed to generated them, can be obtained with appropriate mathematical software. The plots given below were obtained using Mathematica using the code below. Parameters ! !R " 0.6, L " 1., H " 1.25, ddΘdtdt " 0.6, dΘdt " 0.6 t, Θ " 0.3 t2 "; ΑAB ! %
ΑBC !
R ddΘdtdt Sin#Β ' Θ$ H Sin#Β % Γ$
R ddΘdtdt Sin#Γ ' Θ$ L Sin#Β % Γ$
R dΘdt 2
%
H Sin#Β % Γ$3 R dΘdt 2
'
ΒΓDifEqandICs ! %Β'#Θ$ !!
L Sin#Β % Γ$3 %R
Cos#Β ' Θ$ Sin#Β % Γ$2 '
Cos#Γ ' Θ$ Sin#Β % Γ$2 '
Sin#Γ#Θ$ ' Θ$
L Sin#Β#Θ$ % Γ#Θ$$
, Γ'#Θ$ )
R
R H
R L
Sin#Γ ' Θ$2 '
Sin#Β ' Θ$2 '
Sin#Β#Θ$ ' Θ$
H Sin#Β#Θ$ % Γ#Θ$$
R L
R H
Sin#Β ' Θ$2 Cos#Β % Γ$ ;
Sin#Γ ' Θ$2 Cos#Β % Γ$ ;
, Β#0$ ) 0, Γ#0$ )
(R " 0.6, L " 1., H " 1.25); ΒΓSolution ! NDSolve#ΒΓDifEqandICs, (Β, Γ), (Θ, 0, 0.3)$; Plot*ΑAB '. Β " +Β#Θ$ '. ΒΓSolution, '. Γ " +Γ#Θ$ '. ΒΓSolution, '. Parameters, (t, 0, 1),
Π 2
& '.
Frame " True, GridLines " Automatic, AspectRatio " 1, FrameLabel " !"t +s,", "ΑAB +rad's2 ,"", PlotLabel " "Angular Acceleration of AB" -
Plot*ΑBC '. Β " +Β#Θ$ '. ΒΓSolution, '. Γ " +Γ#Θ$ '. ΒΓSolution, '. Parameters, (t, 0, 1),
Frame " True, GridLines " Automatic, AspectRatio " 1, FrameLabel " !"t +s,", "ΑBC +rad's2 ,"", PlotLabel " "Angular Acceleration of BC"-
August 10, 2009
887
Dynamics 1e
Determination of ˇ and via a purely analytical approach. To determine ˇ and in terms of analytically, for convenience, we begin by dividing Eqs. (10) and (11) by L and then rewriting them as follows: sin ˇ C sin D A cos ˇ C cos D B
)
sin ˇ D A
sin
and
sin D A
sin ˇ;
(21)
) cos ˇ D B
cos
and
cos D B
cos ˇ;
(22)
where the terms , A, and B are D
H ; L
AD
H R C sin ; L L
and
BD
R .1 L
cos / C 1:
(23)
Squaring the second of Eqs. (21) and (22) and summing, we have sin2 ˇCcos2 ˇ D .A sin /2 C.B cos /2
1 D A2 CB 2 C 2 2A sin 2B cos ; (24)
)
where we have used the identity sin2 x C cos2 x D 1. The last of Eqs. (24) can then be rewritten as A sin C B cos D
A2 C B 2 C 2 2
1
:
(25)
We now introduce the following definition: tan
D
B A
)
sin
B Dp 2 A C B2
and
cos
Dp
A A2
C B2
;
(26)
where we observe that, since A and B pare known functions of , the angle is also a known function of . Then, dividing Eqs. (25) through by A2 C B 2 , and using the last two of Eqs. (26), we have sin cos
C cos sin
D
A2 C B 2 C 2 1 p 2 A2 C B 2
)
sin. C
/D
A2 C B 2 C 2 1 ; p 2 A2 C B 2
(27)
where we have used the following trigonometric identity: sin.x C y/ D sin x cos y C cos x sin y. Finally, solving Eq. (27) for and recalling that transcendental equations admit infinite solutions, we have 8 2 C 2 1 ˆ 1 A2 CB ˆ p ˙ 2n1 tan 1 B ˙ n2 ; or
D (28) 2 C 2 1 ˆ B 1 A2 CB 1 ˆ p ˙ 2n1 tan : sin 2 2 A ˙ n2 ; 2
A CB
where n1 D 0; 1; 2; : : :, n2 D 0; 1; 2; : : : and where it is understood that ranges of the inverse sine and inverse tangent functions are Œ 2 ; 2 and . 2 ; 2 /, respectively. To select the solution that pertains to our problem, we observe that for D 0, we must have D 2 rad. In addition, we observe that, for D 0, we have 2 ˇ ˇ 2 2 ˇ 1 ˇˇ 1 A CB C 1 B ˇ D 0:8961 rad and tan D 0:6630 rad; (29) sin p ˇ A ˇD0 2 A2 C B 2 D0 where we have used the definitions in Eq. (23) and the following numerical data: H D 1:25 m, L D 1 m, and R D 0:6 m. Therefore, in order for us to obtain D 2 rad D 1:571 rad using the results in Eqs. (29) we must choose the solution for given by the lower expression in Eq. (28) and with n1 D n2 D 0. That is, the correct expression for in our problem is 2 2 2 1 1 A CB C 1 B
D sin tan : (30) p A 2 A2 C B 2 August 10, 2009
888
Solutions Manual
Proceeding in a similar manner, we can obtain an expression for the angle ˇ. That is, squaring the last of Eqs. (21) and (22) and summing them we have 2 D A2 C B 2 C 1 ) sin ˇ cos
2A sin ˇ C cos ˇ sin
2B cos ˇ D
)
A sin ˇ C B cos ˇ D
A2 C B 2 C 1 2 ) sin.ˇ C p 2 A2 C B 2
A2 C B 2 C 1 2 2 A2 C B 2 C 1 2 /D ; (31) p 2 A2 C B 2
which can be solved for ˇ to obtain 8 ! ˆ 2 2 2 ˆ 1 A CB ˆ p C1 ˙ 2n3 tan 1 B ˆ
(32)
2 A CB
Then, in order for us to obtain ˇ D 0 when D 0, we must choose the upper formula in Eq. (32) with n3 D n4 D 0. In summary, recalling the definitions in Eqs. (26), we have ! 2 C R.L C R/ R cos H R.L C R/ cos C HR sin 1 LCR 1 tan ;
D sin p H C R sin H .L C R R cos /2 C .H C R sin /2 ! 2 C R.L C R/ L R.L C R/ cos C HR sin R cos 1 1 LCR ˇ D sin tan : p H C R sin L .L C R R cos /2 C .H C R sin /2 Combining the above equations with the expressions for ˛AB and ˛BC found earlier, and recalling that H D 1:25 m, L D 1 m, and R D 0:6 m, D .0:3 rad=s2 /t 2 , P D .0:6 rad=s2 /t, and R D 0:6 rad=s2 , using mathematical software we obtain precisely the same plots given earlier. However, this time we have used the Mathematica code given below. Parameters ! !R " 0.6, L " 1., H " 1.25, ddΘdtdt " 0.6, dΘdt " 0.6 t, Θ " 0.3 t2 "; ΑAB ! %
ΑBC !
R ddΘdtdt Sin#Β ' Θ$ H Sin#Β % Γ$
R ddΘdtdt Sin#Γ ' Θ$ L Sin#Β % Γ$
Β ! ArcSin%
%
R dΘdt 2 H Sin#Β % Γ$3
'
R dΘdt 2 L Sin#Β % Γ$3
Cos#Β ' Θ$ Sin#Β % Γ$2 '
Cos#Γ ' Θ$ Sin#Β % Γ$2 '
L2 ' L R ' R 2 % R &L ' R' Cos#Θ$ ' H R Sin#Θ$ L
Γ ! Π % ArcSin%
&L ' R % R Cos#Θ$'2 ' &H ' R Sin#Θ$'2
( % ArcTan%
H2 ' R &L ' R' % R &L ' R' Cos#Θ$ ' H R Sin#Θ$ H
&L ' R % R
Cos#Θ$'2
' &H ' R
Sin#Θ$'2
R L
R H
Sin#Γ ' Θ$2 '
Sin#Β ' Θ$2 '
L ' R % R Cos#Θ$ H ' R Sin#Θ$
( % ArcTan%
R H
R L
Sin#Β ' Θ$2 Cos#Β % Γ$ ;
Sin#Γ ' Θ$2 Cos#Β % Γ$ ;
(;
L ' R % R Cos#Θ$ H ' R Sin#Θ$
(;
Plot)ΑAB *. Parameters, +t, 0, 1,, Frame " True, GridLines " Automatic, AspectRatio " 1, FrameLabel " !"t &s'", "ΑAB &rad*s2 '"", PlotLabel " "Angular Acceleration of AB" -
Plot)ΑBC *. Parameters, +t, 0, 1,, Frame " True, GridLines " Automatic, AspectRatio " 1, FrameLabel " !"t &s'", "ΑBC &rad*s2 '"", PlotLabel " "Angular Acceleration of BC"-
August 10, 2009
889
Dynamics 1e
Problem 6.109 Complete the acceleration analysis of the slider-crank mechanism using differentiation of constraints that was outlined beginning on p. 498. That is, determine the accleration of the piston C and the angular acceleration of the connecting rod as a function of the given quantities , !AB , R, and L. Assume that !AB is constant, and use the component system shown for your answers.
Solution We have that the vertical distance from A to C , yC is yC D R sin C L cos : We also know that L sin D R cos : Hence, from the trigonometry of , yC can be rewritten as s
yC D R sin C L 1
(1)
R cos L
Because C is constrained to move vertically, we know that 2 6 vEC D yPC |O D RP cos 41 C
2 :
3 R sin 7 q 5 |O 2 cos2 L 1 R L 2
and 2 0
1
R2 cos2 P 2 R4 cos2 sin2 P 2 C q A 2 cos 2 2 cos2 3=2 L2 1 R L L4 1 R L 2 2 1 3 0 2 2 2 2 R cos sin R C 7 B R sin P C sin P 2 C cos R 5 |O: q ACR @ q 2 2 2 2 cos cos L2 1 R L L2 1 R L 2 2
6 B aEC D 4L @
Furthermore, taking two time derivatives of Eq. (1), we have L P cos D R D
RP sin
)
P D
RP sin L cos
R.P 2 cos C R sin / C P 2 tan : L cos
(2) (3) August 10, 2009
890
Solutions Manual
Substituting Eq. (2) for P in Eq. (3), we have R D
R.P 2 cos C R sin / C L cos
RP sin L cos
!2 tan
LR cos .P 2 cos C R sin / C R2 P 2 sin2 tan : L2 cos2 cos , Eq. (4) becomes Realizing that P D !AB , R D 0, and from Eq. (1), D sin 1 R L D
R D
h 2 cos sin LR!AB
1
R L
h 2 sin2 tan sin cos C R2 !AB i h cos L2 cos2 sin 1 R L
cos
i
1
R L
cos
(4)
i :
R in vector form we have Realizing that ˛BC D ,
˛EBC D
h 2 LR!AB cos sin
1
R L
h 2 cos C R2 !AB sin2 tan sin h i cos L2 cos2 sin 1 R L
cos
i
1
R L
cos
i O k:
August 10, 2009
891
Dynamics 1e
Problem 6.110 Obtain the acceleration of point P using the information from the mini-example on p. 522; that is, obtain Eq. (6.71), using polar coordinates.
Solution We let xy be a rotating reference frame with origin at O that rotates with the disc. For the acceleration of P , we have E vEP rel C ˝ EP rEP =O C ˝ E .˝ E rEP =O / aEP D aEO C aEP rel C 2˝ (1) where E aEO D 0;
rEP =O D s {O;
vEP rel D sP {O;
aEP rel D sR {O;
O E D !0 k; ˝
and
O EP D ˛0 k: ˝
(2)
Substituting Eqs. (2) into Eq. (1), we have aEP D sR {O C 2!0 kO sP {O C ˛0 kO s {O C !0 kO !0 kO s {O D sR {O C 2!0 sP |O C ˛0 s |O C !0 kO .s!0 / |O D sR {O C 2!0 sP |O C ˛0 s |O aEP D sR
s!02 {O
s!02 {O C .2!0 sP C ˛0 s/ |O:
August 10, 2009
892
Solutions Manual
Problem 6.111 The reciprocating rectilinear motion mechanism consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and a bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between C and B. For the position shown, determine (a) The angular velocity of the slotted arm CD and the velocity of the bar (b) The angular acceleration of the slotted arm CD and the acceleration of the bar Evaluate your results for !AB D 120 rpm, R= h D 0:5, and d D 0:12 m.
Solution Part (a). Given a fixed reference frame X Y and a rotating reference frame xy attached to the bar CD with origin at point C , we have the relationships IO D cos {O C sin |O
and
JO D
sin {O C cos |O:
(1)
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO R IO D R!AB JO D R!AB . sin {O C cos |O/:
(2)
Relating point B to point C , we must also have E rEB=C D rPB=C |O C !CD kO rB=C |O D vEB D vEC C vEBrel C ˝
rB=C !CD {O C rPB=C |O;
(3)
E Equating Eqs. (2) and (3) component by component, we have where we have used the fact that vEC D 0. !CD D R!AB
sin sin R2 !AB ı 2 !AB D R!AB 2 D D rB=C h C R2 h2 C R 2 1 C ı2
(4)
and rPB=C D R!AB cos ; where we have let
R h
(5)
D ı. From the geometry of the problem at this instant, we have that tan D
R ; h
sin D p
R h2
C
R2
;
and
cos D p
h h2
C R2
:
(6)
Substituting Eqs. (6) into Eq. (5) we have Rh!AB rPB=C D p : h2 C R 2 August 10, 2009
893
Dynamics 1e In vector form, for !E CD , we have
ı 2 !AB O k: 1 C ı2 Because there is no slip between arm CD and the bar, vbar D !CD d , and so we have !E CD D
vEbar D
(7)
ı 2 !AB O d I: 1 C ı2
(8)
Substituting the given values !AB D 120 rpm D 12:57 rad=s, ı D 0:5, and d D 0:12 m into Eqs. (7) and (8), we have !E CD D 2:51 kO rad=s and vEbar D 0:302 IO m=s: Part (b). Obtaining the acceleration of point B by relating it to point A, we must have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A D
2 R!AB IO D
2 R!AB .cos {O
sin |O/;
(9)
where we have used the fact that !AB is constant, so ˛AB D 0. Relating the acceleration of B to C using the rotating reference frame, for aEB we must also have 2 aEB D aEC C aEBrel C 2!E CD vEBrel C ˛ECD rEB=C !CD rEB=C 2 rB=C |O D rRB=C |O C 2!CD kO rPB=C |O C ˛CD kO rB=C |O !CD
ı 2 !AB O Rh!AB kp |O 2 1Cı h2 C R2 2 2 p ı !AB p 2 2 2 O C ˛CD k h C R |O h C R2 |O 1 C ı2 2 p ı !AB Rh!AB 2 2 2 C ˛CD h C R {O p 1 C ı2 h2 C R2 # " 2 2 ı !AB p 2 C rRB=C h C R2 |O; 1 C ı2
D rRB=C |O C 2
D
(10)
where we have substituted in for !CD from Eq. (4). Equating Eqs. (9) and (10), solving for ˛CD , and substituting the relationships from Eqs. (6), we have 2 p Rh!AB ı !AB 2 2 C R2 R!AB cos D 2 C ˛ h p CD 1 C ı2 h2 C R2 2 1 ı !AB Rh!AB 2 ˛CD D p R!AB cos 2 p 1 C ı2 h2 C R2 h2 C R 2 2 2 2 Rh.h R /!AB D 2 .h C R2 /2 R2 R 2 1 !AB h h2 D 2 2 1C R h2 2 2 ı 1 ı !AB O ) ˛ECD D (11) 2 k: 1 C ı2 August 10, 2009
894 Because there is no slip between arm CD and the bar, abar D d˛CD , and we can write 2 ı 1 ı 2 !AB O aEbar D d 2 k: 1 C ı2
Solutions Manual
(12)
Substituting known values !AB D 120 rpm D 12:57 rad=s, ı D 0:5, and d D 0:12 m into Eqs. (11) and (12), we have ˛ECD D 37:9 kO rad=s2 and aEbar D 4:55 IO m=s2 :
August 10, 2009
895
Dynamics 1e
Problem 6.112 The reciprocating rectilinear motion mechanism consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and a bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between C and B. For the position shown, determine (a) The angular velocity of the slotted arm CD and the velocity of the bar (b) The angular acceleration of the slotted arm CD and the acceleration of the bar Evaluate your results for !AB D 90 rpm, R= h D 0:5, and d D 0:12 m.
Solution Part (a). Given a fixed coordinate system X Y and a rotating coordinate system xy attached to the bar CD at point C , we have the relationships IO D {O
and JO D |O:
(1)
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO R JO D
R!AB IO D
R!AB {O (2)
Relating point B to point C , we must also have E rEB=C vEB D vEC C vEBrel C ˝ D rPB=C |O C !CD kO rB=C |O D
rB=C !CD {O C rPB=C |O;
(3)
E Equating Eqs. (2) and (3) component by component, we have where we have used the fact that vEC D 0. !CD D where we have left ı D
R . h
R!AB R ı D !AB D !AB rB=C hCR 1Cı
and
rPB=C D 0;
(4)
In vector form, for !E CD , we have !E CD D
ı O !AB k: 1Cı
(5)
Because there is no slip between arm CD and the bar, we have vbar D !CD d , and so vEbar D
ı !AB d IO: 1Cı
(6)
Substituting the given values !AB D 90 rpm D 9:425 rad=s, ı D 0:5, and d D 0:12 m into Eqs. (5) and (6), we have !E CD D 7:07 kO rad=s and vEbar D 0:848 IO m=s: August 10, 2009
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Solutions Manual
Part (b). Obtaining the acceleration of point B by relating it to point A, we must have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A D
2 R!AB JO D
2 R!AB |O:
(7)
Relating the acceleration of B to C using rotating reference frames, for aEB we must also have 2 aEB D aEC C aEBrel C 2!E CD vEBrel C ˛ECD rEB=C !CD rEB=C 2 D rRB=C |O C 2!CD kO rPB=C |O C ˛CD kO rB=C |O !CD rB=C |O 2 ı !AB .h C R/ |O: D rRB=C |O C ˛CD kO .h C R/ |O 1Cı
(8)
Setting Eqs. (7) and (8) equal, and solving for ˛CD , we have 2 R!AB |O D ˛CD .h C R/ {O C
"
ı !AB 1Cı
2
# .h C R/ |O
)
˛CD .h C R/ D 0 )
˛ECD D 0E rad=s2 :
Because there is no slip between arm CD and the bar, and the bar is confined to move in the horizontal direction, aEbar D d˛CD IO, so we must have aEbar D 0E m=s2 :
August 10, 2009
897
Dynamics 1e
Problem 6.113 The reciprocating rectilinear motion mechanism consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and a bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between C and B. For the position shown, determine (a) The angular velocity of the slotted arm CD and the velocity of the bar (b) The angular acceleration of the slotted arm CD and the acceleration of the bar Evaluate your results for !AB D 60 rpm, R= h D 0:5, and d D 0:12 m.
Solution Part (a). Given a fixed coordinate system X Y and a rotating coordinate system xy attached to the bar CD at point C , we have the relationships IO D {O
and JO D |O:
(1)
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO . R/ JO D R!AB IO D R!AB {O (2) Relating point B to point A, we must also have E rEB=C vEB D vEC C vEBrel C ˝ D rPB=C |O C !CD kO rB=C |O D
rB=C !CD {O C rPB=C |O:
(3)
Equating Eqs. (2) and (3) component by component, we have !CD D
R ı R!AB D !AB D !AB rB=C h R 1 ı
and rPB=C D 0:
(4)
In vector form, for !E CD , we have
ı O !AB k: (5) 1 ı From the concept of instantaneous center of rotation, vbar D !CD d . Observing that if !CD > 0 then vbar > 0, we have ı vEbar D !AB d IO: (6) 1 ı Substituting known values !AB D 60 rpm D 6:283 rad=s, ı D 0:5, and d D 0:12 m into Eqs. (5) and (6), we have !E CD D 6:28 kO rad=s and vEbar D 0:754 IO m=s: !E CD D
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Solutions Manual
Part (b). Obtaining the acceleration of point B by relating it to point A, we must have aEB D aEA C ˛EAB rEB=A
2 2 2 !AB rEB=A D R!AB JO D R!AB |O:
(7)
Using the idea of moving reference frames, for aEB we must also have 2 aEB D aEC C aEBrel C 2!E CD vEBrel C ˛ECD rEB=C !CD rEB=C 2 D rRB=C |O C 2!CD kO rPB=C |O C ˛CD kO rB=C |O !CD rB=C |O 2 ı !AB .h R/ |O: D rRB=C |O C ˛CD kO .h R/ |O 1Cı
(8)
Setting Eqs. (7) and (8) equal, and solving for ˛CD , we have " 2 R!AB |O D
˛CD .h
R/ {O C rRB=C
ı !AB 1Cı
#
2 .h
R/ |O
)
˛CD .h )
R/ D 0
˛ECD D 0E rad=s2 :
Because there is no slip between the arm CD and the bar, and the bar is confined the move in the horizontal direction, abar D d˛CD , so we must have aEbar D 0E m=s2 :
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Dynamics 1e
Problem 6.114 The reciprocating rectilinear motion mechanism consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and a bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between C and B. For the arbitrary position shown, determine (a) The angular velocity of the slotted arm CD and the velocity of the bar (b) The angular acceleration of the slotted arm CD and the acceleration of the bar as functions of , ı D R= h, d , and !AB
Solution Part (a). Given a fixed coordinate system X Y and a rotating coordinate system xy attached to the bar CD at point C , we have the relationships IO D cos {O C sin |O
and
JO D
sin {O C cos |O:
(1)
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO R.cos IO C sin JO / D
R!AB sin IO C R!AB cos JO
D
R!AB sin .cos {O C sin |O/ C R!AB cos . sin {O C cos |O/
D
.R!AB sin cos C R!AB cos sin / {O C .R!AB cos cos R!AB sin sin / |O (2)
Relating point B to point C , we must also have E rEB=C D rPB=C |O C !CD kO rB=C |O D vEB D vEC C vEBrel C ˝
rB=C !CD {O C rPB=C |O:
(3)
Equating Eqs. (2) and (3) component by component, from the {O component we have rB=C !CD D !CD
.R!AB sin cos C R!AB cos sin / sin cos C cos sin D R!AB rB=C sin cos C cos sin D R!AB p h2 C R2 C 2hR sin
(4)
where we have replaced rB=C by its equivalent expression from the Law of Cosines. Also from the geometry of the problem we have tan D
R cos ; and we also have the trig identities h C R sin August 10, 2009
900
Solutions Manual sin D p
tan 1C
and
tan2
1 cos D p : 1 C tan2
(5)
Substituting Eqs. (5) into Eq. (4) we have R!AB .R C h sin / ı!AB .ı C sin / !CD D p D 1 C ı 2 C 2ı sin h2 C R2 C 2hR sin
(6)
rPB=C D R!AB . sin sin C cos cos /:
(7)
and
From the geometry of the problem, we have that Substituting Eqs. (5) into Eq. (7) we have rPB=C D p
Rh!AB cos h2 C R2 C 2Rh sin
:
In vector form, for !E CD , we have !E CD D
ı!AB .ı C sin / O k: 1 C ı 2 C 2ı sin
Because the arm CD doesn’t slip relative to the bar, we have vbar D !CD d , which gives us vEbar D
ı!AB .ı C sin / O d I: 1 C ı 2 C 2ı sin
Part (b). Obtaining the acceleration of point B by relating it to point A, we must have 2 aEB D aEA C ˛EAB rEB=A !AB rEB=A D R! 2 .cos IO C sin JO / AB
D
2 R!AB .cos cos
sin sin / {O
R!AB .cos sin C sin cos / |O:
(8)
Using the idea of moving reference frames, for aEB we must also have 2 aEB D aEC C aEBrel C 2!E CD vEBrel C ˛ECD rEB=C !CD rEB=C 2 D rRB=C |O C 2!CD kO rPB=C |O C ˛CD kO rB=C |O !CD rB=C |O Rh!AB cos ı!AB .ı C sin / O kp |O D rRB=C |O C 2 2 1 C ı C 2ı sin h2 C R2 C 2Rh sin 2 p p ı! .ı C sin / AB C ˛CD kO h2 C R2 |O h2 C R2 |O 1 C ı 2 C 2ı sin ı!AB .ı C sin / Rh!AB cos D 2 {O p 1 C ı 2 C 2ı sin h2 C R2 C 2Rh sin " # p ı!AB .ı C sin / 2 p 2 2 2 2 C ˛CD h C R C rRB=C h C R |O: 1 C ı 2 C 2ı sin
(9)
Setting Eqs. (8) and (9) equal, and solving for ˛CD , and substituting the relationships from Eqs. (5), we have ı!AB .ı C sin / Rh!AB cos 2 R!AB .cos cos sin sin / D 2 p 1 C ı 2 C 2ı sin h2 C R2 C 2Rh sin August 10, 2009
901
Dynamics 1e " p 2 R!AB .cos sin C sin cos / D ˛CD h2 C R2 C rRB=C
˛CD
Rh.h2
2 R2 /!AB
R h
1
cos D 2 D 2 .h C R C 2Rh sin 2 1C
R2 h2 R2 h2
ı!AB .ı C sin / 1 C ı 2 C 2ı sin
2 cos !AB 2 C 2R sin h
)
˛ECD
2 p
# h2 C R2
2 cos ı 2 !AB O D 2 k: 1 C ı 2 C 2ı sin ı 1
Because the arm CD doesn’t slip relative to the bar, we have abar D d˛CD , and so we have aEbar
2 cos ı 2 !AB O Dd 2 k: 1 C ı 2 C 2ı sin ı 1
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Solutions Manual
Problem 6.115 The reciprocating rectilinear motion mechanism consists of a disk pinned at its center at A that rotates with a constant angular velocity !AB , a slotted arm CD that is pinned at C , and a bar that can oscillate within the guides at E and F . As the disk rotates, the peg at B moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between C and B. Determine the angular velocity and angular acceleration of the slotted arm CD as functions of , ı D R= h, d , and !AB . After doing so, plot the velocity and acceleration of the bar as a function of the disk angle for one full cycle of the disk’s motion and for !AB D 90 rpm, d D 0:12 m, and (a) ı D R= h D 0:1 (b) ı D R= h D 0:3 (c) ı D R= h D 0:6 (d) ı D R= h D 0:9 Explain why this mechanism is often referred to as a quick return mechanism.
Solution Given a fixed coordinate system X Y and a rotating coordinate system xy attached to the bar CD at point C , we have the relationships IO D cos {O C sin |O;
JO D
sin {O C cos |O;
and
O (1) kO D K:
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO R.cos IO C sin JO / D
R!AB sin IO C R!AB cos JO
D
R!AB sin .cos {O C sin |O/ C R!AB cos . sin {O C cos |O/
D
.R!AB sin cos C R!AB cos sin / {O C .R!AB cos cos
R!AB sin sin / |O:
(2)
Relating point B to point C , we must also have E rEB=C D rPB=C |O C !CD kO rB=C |O D vEB D vEC C vEBrel C ˝
rB=C !CD {O C rPB=C |O;
(3)
E Equating the x components of Eqs. (2) and (3), we have where we have used the fact that vEC D 0. rB=C !CD D !CD
.R!AB sin cos C R!AB cos sin / sin cos C cos sin D R!AB rB=C August 10, 2009
903
Dynamics 1e sin cos C cos sin D R!AB p : h2 C R2 C 2hR sin
(4)
where we have replaced rB=C by its equivalent expression from the law of cosines. Also from the geometry of the problem we have tan D
R cos ; h C R sin
tan ; sin D p 1 C tan2
and
1 cos D p : 1 C tan2
(5)
Substituting Eqs. (5) into Eq. (4) we have R!AB .R C h sin / ı!AB .ı C sin / !CD D p D 2 2 1 C ı 2 C 2ı sin h C R C 2hR sin
(6)
rPB=C D R!AB . sin sin C cos cos /:
(7)
and
Substituting Eqs. (5) into Eq. (7) we have rPB=C D p
Rh!AB cos h2 C R2 C 2Rh sin
:
In vector form, for !E CD , we have !E CD D
ı!AB .ı C sin / O K 1 C ı 2 C 2ı sin
:
Because the arm CD does not slip relative to the bar, we have vbar D !CD d , which gives us vEbar D
ı!AB .ı C sin / O dI 1 C ı 2 C 2ı sin
:
Obtaining the acceleration of point B by relating it to point A, we must have 2 aEB D aEA C ˛EAB rEB=A !AB rEB=A 2 D R! .cos IO C sin JO /
D
AB 2 R!AB .cos
cos
sin sin / {O
2 R!AB .cos sin C sin cos / |O:
(8)
Using the idea of moving reference frames, for aEB we must also have 2 aEB D aEC C aEBrel C 2!E CD vEBrel C ˛ECD rEB=C !CD rEB=C 2 D rRB=C |O C 2!CD kO rPB=C |O C ˛CD kO rB=C |O !CD rB=C |O Rh!AB cos ı!AB .ı C sin / O kp |O D rRB=C |O C 2 2 1 C ı C 2ı sin h2 C R2 C 2Rh sin p ı!AB .ı C sin / 2 p 2 2 2 O C ˛CD k h C R |O h C R2 |O 1 C ı 2 C 2ı sin ı!AB .ı C sin / Rh!AB cos D 2 {O p 1 C ı 2 C 2ı sin h2 C R2 C 2Rh sin " # 2 p p ı! .ı C sin / AB C ˛CD h2 C R2 C rRB=C h2 C R2 |O: 1 C ı 2 C 2ı sin
(9) August 10, 2009
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Solutions Manual
Setting Eqs. (8) and (9) equal, and solving for ˛CD , and substituting the relationships from Eqs. (5), we have ı!AB .ı C sin / Rh!AB cos 2 R!AB .cos cos sin sin / D 2 p 1 C ı 2 C 2ı sin h2 C R2 C 2Rh sin # " 2 p p ı! .ı C sin / AB 2 h2 C R 2 R!AB .cos sin C sin cos / D ˛CD h2 C R2 C rRB=C 1 C ı 2 C 2ı sin 2 R2 R 2 2 2 2 cos 1 !AB cos ı 1 ı 2 !AB Rh.h R /!AB cos h h2 O ) ˛ECD D ˛CD D 2 D K: 2 2 2 2 C 2ı sin 2 .h C R C 2Rh sin R2 R 1 C ı 1 C h2 C 2 h sin Because the arm CD does not slip relative to the bar, we have abar D d˛CD , and so we have aEbar
2 cos ı 2 !AB Dd 2 IO 2 1 C ı C 2ı sin ı 1
:
The expressions for vEbar and aEbar just derived is a function of when !AB , d , and ı are specified. For each of the several values of ı, the function can be plotted with software such as Mathematica or M ATLAB. Using Mathematica, the following code was used, with the value for ı changed for each part: Parameters ! ! ∆ # 0.1, d # 0.12, ΩAB # 9.425"; vbar !
∆ ΩAB #∆ % Sin$Θ%&
1 % ∆2 % 2 ∆ Sin$Θ%
abar ! d
'. Parameters;
∆ (1 ' ∆2 ) ΩAB 2 Cos$Θ%
(1 % ∆2 % 2 ∆ Sin$Θ%)
2
'. Parameters;
Plot$vbar , !Θ, 0, 2 Π", PlotLabel # "vbar ", Frame # True,
FrameLabel # !"Θ#rad&", "vbar #m's&"", ImageSize # Medium%
Plot*abar , !Θ, 0, 2 Π", PlotLabel # "abar ", Frame # True,
FrameLabel # +"Θ#rad&", "abar #m's2 &",, ImageSize # Medium-
August 10, 2009
905
Dynamics 1e Part (a).
We let ı = 0.1. Using the above code we obtain the following plots:
vbar
abar 1.0 abar !m#s2 "
vbar !m#s"
0.5 0.0 !0.5
0.5 0.0 !0.5 !1.0
!1.0 0
1
2
3
4
5
0
6
1
2
3
4
5
6
4
5
6
4
5
6
Θ!rad"
Θ!rad"
Part (b). We let ı = 0.3. Using the above code we obtain the following plots:
vbar
abar
2
4 abar !m#s2 "
vbar !m#s"
1 0 !1 !2 !3
2 0 !2 !4
!4 0
1
2
3
4
5
6
0
1
2
3 Θ!rad"
Θ!rad"
Part (c). We let ı = 0.6. Using the above code we obtain the following plots:
vbar
abar 20 abar !m#s2 "
vbar !m#s"
0 !5 !10
10 0 !10 !20
0
1
2
3 Θ!rad"
4
5
6
0
1
2
3 Θ!rad"
Part (d). We let ı = 0.9. Using the above code we obtain the following plots:
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Solutions Manual
abar
4 3 2 1 0 !1 !2 !3
abar !m#s2 "
vbar !m#s"
vbar
0
1
2
3 Θ!rad"
4
5
6
6 4 2 0 !2 !4 !6 0
1
2
3
4
5
6
Θ!rad"
August 10, 2009
907
Dynamics 1e
Problem 6.116 The reciprocating rectilinear motion mechanism of Probs. 6.111–6.115 is often referred to as a quick return mechanism since it can move much more quickly in one direction than the other. To see this, we will determine the velocity of the bar in each of the two positions shown (position 1 when B is farthest from C and position 2 when B is closest to C ) under the assumption that the disk, whose center is at A, rotates with a constant angular velocity !AB D 120 rpm. Find the velocity of the bar in positions 1 and 2 for d D 0:12 m and for (a) R= h D 0:3 (b) R= h D 0:8 Comment on which of the two R= h values would provide a better quick return and why.
Solution Position 1 Given a fixed coordinate system XY and a rotating coordinate system xy attached to the bar CD at point C , we have the relationships IO D {O;
JO D |O;
and
O KO D k:
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO R JO D
R!AB IO
D
R!AB {O
(1)
Relating point B to point C , we must also have E rEB=C D rPB=C |O C !CD kO rB=C |O D vEB D vEC C vEBrel C ˝
rB=C !CD {O C rPB=C |O;
(2)
E Equating Eqs. (1) and (2) component by component, we have where we have used the fact that vEC D 0. !CD D where we have let ı D
R . h
R ı R!AB D !AB D !AB rB=C hCR 1Cı
and
rPB=C D 0;
(3)
In vector form, for !E CD , we have !E CD D
ı O !AB k: 1Cı
Because there is no slip between arm CD and the bar, we have vbar D !CD d , and so vEbar D
ı !AB d IO 1Cı
: August 10, 2009
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Solutions Manual
Position 2 Given a fixed coordinate system X Y and a rotating coordinate system xy attached to the bar CD at point C , we have the relationships IO D {O;
JO D |O
and
O KO D k:
From rigid body kinematics, we have vEB D vEA C !EAB rEB=A D !AB kO . R/ JO D R!AB IO D R!AB {O
(4)
Relating point B to point A, we must also have E rEB=C D rPB=C |O C !CD kO rB=C |O D vEB D vEC C vEBrel C ˝
rB=C !CD {O C rPB=C |O:
(5)
Equating Eqs. (4) and (5) component by component, we have !CD D
R!AB R ı D !AB D !AB rB=C h R 1 ı
In vector form, for !E CD , we have
and rPB=C D 0:
ı
O !AB k: 1 ı From the concept of instantaneous center of rotation, vbar D !CD d . Observing that if !CD > 0 then vbar > 0, we have !E CD D
vEbar D
Part (a) For
R h
ı 1
ı
!AB d IO
:
D 0:3 and !AB D 12:57 rad=s, we have vbar D 2:90 m=s at position 1 and
vbar D 5:39 m=s at position 2. Part (b) For
R h
D 0:8 and !AB D 12:57 rad=s, we have vbar D 5:59 m=s at position 1 and
vbar D 50:3 m=s at position 2. The bar is moving through the return phase when it is in position 2. Therefore, the R value from Part (b) h would provide a better quick return because it is moving back into position faster than the R value from Part h (a).
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Dynamics 1e
Problem 6.117 The Pioneer 3 spacecraft was a spin-stabilized spacecraft launched on December 6, 1958, by the U.S. Army Ballistic Missile agency in conjunction with NASA. It was designed with a despin mechanism consisting of two equal masses A and B that could be spooled out to the end of two wires of variable length `.t/ when triggered by a hydraulic timer. As a prelude to Probs. 7.64, 7.65, and 8.77, we will find the velocity and acceleration of each the two masses. To do this, assume that masses A and B are initially at positions A0 and B0 , respectively. After the masses are released, they begin to unwind symmetrically, and the length of the cord attaching each mass to the spacecraft of radius R is `.t/. Given that the angular velocity of the spacecraft at each instant is !s .t /, determine (a) The velocity of mass A (b) The acceleration of mass A in components expressed in the rotating reference frame whose origin is at Q, as well as R, `.t /, and !s .t /. Note that the rotating frame is always aligned with the unwinding cord, and Q is the point on the cord that is about to unwind at time t.
Solution We let be the unwind angle of the rope relative to the satellite. As stated, we will use a rotating reference frame attached at Q, as indicated on the diagram. For the velocity of A, we have vEA D vEQ C EPrel C !E OQ EA=Q
(1)
where vEQ D .!S C P /R {O; EPrel D
`P {O; !E OQ D
O EA=Q D .!S C P / k;
` {O; and D
` R
)
`P P D : (2) R
Substituting Eqs (2) into Eq. (1), we have ! `P vEA D R !S C {O R
`P {O
! `P O !S C k . `/ {O R )
! P / `.t vEA D R!S .t / {O C `.t / !S .t / C |O: R
(3)
For the acceleration we have aEA D aEQ C ERrel C 2!E OQ EPrel C ˛EOQ EA=Q
2 !OQ EA=Q ;
(4)
where aEQ D
R.!S C P /2 |O C R.!P S C R / {O;
ERrel D
`R {O;
˛EOQ D
.!P S C R /;
and
`R R D kO R
(5)
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Solutions Manual
Substituting Eqs. (5) into Eq. (4), we have aEA D
R.!S C P /2 |O C R.!P S C R / {O
`R {O
2.!S C P / kO . `P {O/
.!P S C R / kO . ` {O/ .!S C P /2 . ` {O/
" P /2 `.t / R `.t `.t / P 2 . `.t / C R! .t // C R ! P .t / { O C `.t / C aEA D S S R2 R R
# R!S .t /2 C `!P S .t / |O:
August 10, 2009
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Dynamics 1e
Problem 6.118 Let frames A and B be the frames with origins at points A and B, respectively. Point B does not move relative to point A. The velocity and acceleration of point P relative to frame B are vEP rel D . 6:14 {OB C 23:7 |OB / ft=s
and aEP rel D .3:97 {OB C 4:79 |OB / ft=s2 :
Knowing that, at the instant shown, frame B rotates relative to frame A at a constant angular velocity !B D 1:2 rad=s, that the position of point P relative to frame B is rEP =B D .8 {OB C 4:5 |OB / ft, and that frame A is fixed, determine the velocity and acceleration of P at the instant shown and express the results using the frame A component system.
Solution From the kinematic equations for the velocity of a point relative to a rotating reference frame we have E rEP =B D PEPrel C !B kO rEP =B : vEP D vEB C vEP rel C ˝
(1)
P We have known values PErel D . 6:14 {OB C 23:7 |OB / ft=s, !B D 1:2 kOB rad=s, and rEP =B D .8 {OB C 4:5 |OB / ft, but they must be transformed from the B component system to the A component system where {OB D cos 23ı {OA C sin 23ı |OA and |OB D sin 23ı {OA C cos 23ı |OA . Therefore, in the A component system, our known values are P PErel D . 14:91 {OA C 19:42 |OA /;
rEP =B D .5:606 {OA C 7:268 |OA / ft;
and
!EB D 1:2 kOA rad=s:
Substituting these known values into Eq. (1) we have vEP D . 23:6 {OA C 26:1 |OA / ft=s: Similarly, for the acceleration of point P we have E vEP rel C ˝ EP rEP =B C ˝ E .˝ E rEP =B / aEP D aEB C aEP rel C 2˝ R P D PErel C 2!B kO PErel C !P B kO rEP =B C !B kO .!B kO rEB=P /:
(2)
R We must also transform PErel into the A component system. Doing so we have R PErel D .2:103 {OA C 6:447 |OA / ft=s2 :
(3) August 10, 2009
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Solutions Manual
P Substituting this value, as well as the known values PErel D . 14:91 {OA C 19:42 |OA /; 7:268 |OA / ft; and !EB D 1:2 kOA rad=s into Eq. (2), we have aEP D . 52:9 {OA
rEP =B D .5:606 {OA C
40:3 |OA / ft=s2 :
August 10, 2009
913
Dynamics 1e
Problem 6.119 Repeat Prob. 6.118, but express the results using the component system of frame B.
Solution From the kinematic equations for the velocity of a point relative to a rotating reference frame we have E rEP =B vEP D vEB C vEP rel C ˝ P D PErel C !B kO rEP =B :
(1)
P Substituting known values PErel D . 6:14 {OB C23:7 |OB / ft=s, !EB D 1:2 kO rad=s, and rEP =B D .8 {OB C4:5 |OB / ft into Eq. (1) we have vEP D . 11:5 {OB C 33:3 |OB / ft=s: Similarly, for the acceleration of point P we have E vEP rel C ˝ EP rEP =B C ˝ E .˝ E rEP =B / aEP D aEB C aEP rel C 2˝ R P D PErel C 2!B kO PErel C !P B kO rEP =B C !B kO .!B kO rEB=P /
(2)
P R Substituting the known values PErel D . 6:14 {OB C 23:7 |OB / ft=s, PErel D .3:97 {OB C 4:79 |OB / ft=s2 , !EB D 1:2 kO rad=s, and rEP =B D .8 {OB C 4:5 |OB / ft into Eq. (2), we have aEP D . 64:4 {OB
16:4 |OB / ft=s2 :
August 10, 2009
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Solutions Manual
Problem 6.120 A vertical shaft has a base B that is stationary relative to an inertial reference frame with vertical axis Z. Arm OA is attached to the vertical shaft and rotates about the Z axis with an angular velocity !OA D 5 rad=s and an angular acceleration ˛OA D 1:5 rad=s2 . The ´ axis is coincident with the Z axis but is part of a reference frame that rotates with the arm OA. The x axis of the rotating reference frame coincides with the axis of the arm OA. At the instant shown, the y axis is perpendicular to the page and directed into the page. At this instant, the collar C is sliding along OA with a constant speed vC D 5 ft=s and is at a distance d D 1:2 ft from the ´ axis. Compute the inertial acceleration of the collar, and express it relative to the rotating coordinate system.
Solution Using the idea of rotating reference frames, we have O E rC =O D vC {OC!OA kd vEC D vEO CrEPC =O C˝E {O D vC {OCd!OA |O; where we have used the fact that vO D 0. For the acceleration, we then have E OC vEC =O C ˝ EP OC rEC =O C ˝ E OC .˝ E OC rEC =O / aEC D aEO C aEC =O C 2˝ D 2!OA kO vC {O C ˛OA kO d {O C !OA kO .!OA kO d {O/ D 2!OA vC |O C d˛OA |O C !OA kO d!OA |O D
2 d!OA {O C .2!OA vC C d˛OA / |O;
(1)
O and ˝ O Substituting E OC D !OA k, EP OC D ˛OA k. where we have used the fact that aO D 0, vC D constant, ˝ 2 the given values d D 1:2 ft, vC D 5 ft=s, !OA D 5 rad=s, and ˛OA D 1:5 rad=s into Eq. (1), and rounding to three significant figures, we have aEC D . 30:0 {O C 51:8 |O/ ft=s2 :
August 10, 2009
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Dynamics 1e
Problem 6.121 A vertical shaft has a base B that is stationary relative to an inertial reference frame with vertical axis Z. Arm OA is attached to the vertical shaft and rotates about the Z axis with an angular velocity !OA D 5 rad=s and an angular acceleration ˛OA D 1:5 rad=s2 . The ´ axis is coincident with the Z axis but is part of a reference frame that rotates with the arm OA. The x axis of the rotating reference frame coincides with the axis of the arm OA. At the instant shown, the y axis is perpendicular to the page and directed into the page. At this instant, the collar C is sliding along OA with a constant speed vC D 3:32 m=s and is rotating with a constant angular velocity !C D 2:3 rad=s relative to the arm OA. At the instant shown, point D happens to be in the x´ plane and is at a distance ` D 0:05 m from the x axis and at a distance d D 0:75 m from the ´ axis. Compute the inertial acceleration of point D, and express it relative to the rotating reference frame.
Solution Using the idea of rotating reference frames, we have E rED=O vED D vEO C rEPD=O C ˝ D .vC {O
O d {O `!C |O/ C .!C {O C !OA k/
D vC {O C .d!OA
`!C / |O;
where we have used the fact that vO D 0 and D is making a circular motion about the bar OA, and moving out of the page at the instant shown. For the acceleration, we then have E vED=O C ˝ EP rED=O C ˝ E ˝ E rED=O aED D aEO C aED=O C 2˝ D 2 !C {O C !OA kO .vC {O `!C |O/ C ˛OA kO d {O C ` kO h i O O C !C {O C !OA k !C {O C !OA k d {O 2 O D 2`!C k C 2!OA .vC |O C `!C {O/ C d˛OA |O C !C {O C !OA kO .d!OA |O/ 2 2 O D 2`!C !OA d!OA {O C .d˛OA C 2vC !OA / |O C `!C k (1) where we have used the fact that aO D 0, vC D constant, !C D constant, Dis making a circular E D !C {O C !OA kO , and motion about the bar OA, and moving out of the page at the instant shown, ˝ O Substituting the given values d D 0:75 m, ` D 0:05 m, vC D 3:32 m=s, !C D 2:3 rad=s, EP OC D ˛OA k. ˝ August 10, 2009
916
Solutions Manual
!OA D 5 rad=s, and ˛OA D 1:5 rad=s2 into Eq. (1), and rounding to three significant figures, we have aEC D
17:6 {O C 34:3 |O C 0:264 kO m=s2 :
August 10, 2009
917
Dynamics 1e
Problem 6.122 The wheel D rotates with a constant angular velocity !D D 14 rad=s about the fixed point O, which is assumed to be stationary relative to an inertial frame of reference. The xy frame rotates with the wheel. Collar C slides along the bar AB with a constant velocity vC D 4 ft=s relative to the xy frame. Letting ` D 0:25 ft, determine the inertial velocity and acceleration of C when D 25ı . Express the result with respect to the xy frame.
Solution From the geometry of the problem, we have rEC =O D . ` cot {O C ` |O/. From the concept of rotating reference frames, we have E rEC =O D vC {O C !D kO .` cot {O vEC D vEO C vEC =O C ˝
` |O/ D .vC C `!D / {O C `!D cot |O; (1)
O Substituting known values ` D 0:25 ft, E D !D k. where we have used the fact that vO D 0 and ˝ ı !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (1), and rounding to three significant figures we have vEC D .7:50 {O C 7:51 |O/ ft=s: For the acceleration, we then have E vEC =O C ˝ EP rEC =O C ˝ E ˝ E rEC =O aEC D aEO C aEC =O C 2˝ h i D 2!D kO vC {O C !D kO !D kO . ` cot {O C ` |O/ 2 2 D `!D cot {O 2vC !D C `!D |O;
(2)
O ˝ E D !D k, EP D 0. E Substituting the known values ` D 0:25 ft, where we have used the fact that vO D 0, ˝ ı !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (2), and rounding to three significant figures we have aEC D .105 {O
161 |O/ ft=s2 :
August 10, 2009
918
Solutions Manual
Problem 6.123 The wheel D rotates without slipping over a flat surface. The X Y frame shown is inertial, whereas the xy frame is attached to D at O and rotates with it at a constant angular velocity !D D 14 rad=s. Collar C slides along the bar AB with a constant velocity vC D 4 ft=s relative to the xy frame. Letting ` D 0:25 ft and R D 1 ft, determine the inertial velcity and acceleration of C when D 25ı and the xy frame is parallel to the X Y frame as shown. Express your result in both the xy and X Y frames.
Solution From the geometry of the problem, we have rEC =O D . ` cot {O C ` |O/. From the concept of rotating reference frames, we have E rEC =O vEC D vEO C vEC =O C ˝ D .R!D C vC /O{ C !D kO .` cot {O
` |O/
D .R!D C vC C `!D / {O C `!D cot |O;
(1)
O Substituting known values R D 1 ft, E D !D k. where we have used the fact that vEO D R!D {O and ˝ ı ` D 0:25 ft, !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (1), and rounding to three significant figures we have vEC D .21:5 {O C 7:51 |O/ ft=s: For the acceleration, we then have E vEC =O C ˝ EP rEC =O C ˝ E ˝ E rEC =O aEC D aEO C aEC =O C 2˝ h i D 2!D kO vC {O C !D kO !D kO . ` cot {O C ` |O/ 2 2 D `!D cot {O 2vC !D C `!D |O;
(2)
O ˝ E D !D k, EP D 0. E Substituting the known values ` D 0:25 ft, where we have used the fact that vO D 0, ˝ ı !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (2), and rounding to three significant figures we have aEC D .105 {O
161 |O/ ft=s2 :
August 10, 2009
Dynamics 1e
919
Problem 6.124 The wheel D rotates without slipping over a flat surface. The X Y frame shown is inertial whereas the xy frame is attached to D and rotates with it at a constant angular velocity !D . Collar C slides along the bar AB with a velocity vC relative to the xy frame. Suppose that `, , and R are given and that we want to determine the inertial acceleration of C when the xy frame is parallel to the X Y frame as shown. Would the expression of the inertial acceleration of the collar in the two frames be different or the same?
Solution At the instant shown, the two reference frames are coincidental, i.e. {O D IO and |O D JO . Therefore, a vector will have exactly the same representation with respect to the two frames.
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Solutions Manual
Problem 6.125 At the instant shown, the wheel D rotates without slipping over a flat surface with an angular velocity !D D 14 rad=s and an angular acceleration ˛D D 1:1 rad=s2 . The X Y frame shown is inertial whereas the xy frame is attached to D. At the instant shown, the collar C is sliding along the bar AB with a velocity vC D 4 ft=s and acceleration aC D 7 ft=s2 , both relative to the xy frame. Letting ` D 0:25 ft and R D 1 ft, determine the inertial velocity and acceleration of C when D 25ı and the xy frame is parallel to the XY frame as shown. Express your result in both the xy and the X Y frames.
Solution From the geometry of the problem, we have rEC =O D . ` cot {O C ` |O/. From the concept of rotating reference frames, we have E rEC =O vEC D vEO C vEC =O C ˝ D .R!D C vC /O{ C !D kO .` cot {O
` |O/
D .R!D C vC C `!D / {O C `!D cot |O;
(1)
O Substituting known values R D 1 ft, E D !D k. where we have used the fact that vEO D R!D {O and ˝ ı ` D 0:25 ft, !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (1), and rounding to three significant figures we have vEC D .21:5 {O C 7:51 |O/ ft=s: For the acceleration, we then have E vEC =O C ˝ EP rEC =O C ˝ E ˝ E rEC =O aEC D aEO C aEC =O C 2˝ h i D 3!D kO vC {O ˛D kO . ` cot {O C ` |O/ C !D kO !D kO . ` cot {O C ` |O/ 2 2 D aC C `˛D C R˛D C `!D cot {O C 2vC !D `!D C `˛D cot |O;
(2)
O ˝ E D !D k, EP D R˛D {O. Substituting the known values R D 1 ft, where we have used the fact that vO D 0, ˝ ` D 0:25 ft, !D D 14 rad=s, ˛D D 1:1 rad=s2 , vC D 4 ft=s, D 25ı into Eq. (2), and rounding to three significant figures we have aEC D .113 {O
160 |O/ ft=s2 :
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Dynamics 1e
Problem 6.126 A floodgate is controlled by the motion of the hydraulic cylinder AB. If the gate BC is to be lifted with a constant angular velocity !BC D 0:5 rad=s, determine dPAB and dRAB , where dAB is the distance between points A and B when D 0. Let ` D 10 ft, h D 2:5 ft, and d D 5 ft.
Solution We define a fixed reference frame X Y such that X is horizontal and pointing to the right, and Y is vertical and pointing up, and a rotating reference frame xy attached at A that rotates with the bar. We are asked to find dPAB and dRAB , which correspond to the {O component of rPB=A and rRB=A , respectively. If we write the velocity of B using the rotating reference frame, we have E rEB=A vEB D vEA C vEB=A C ˝ q D dPB=A {O C !AB kO .` d /2 C h2 {O q (1) D dPB=A {O C !AB .` d /2 C h2 |O; p where we used the fact that vA D 0 and rEB=A D .` d /2 C h2 |O. We must also have E rEB=C D vEB D vEC C vEB=C C ˝
!BC kO . d / IO D
d!BC JO D d!BC sin {O C d!BC cos |O; (2)
p where we used the fact that vC D 0, rEB=A D .` d /2 C h2 |O, and JO D sin {O C cos |O. Setting Eqs. (1) and (2) equal component by component, we have dPA=B D d!BC sin ; q !AB .`
d /2 C h2 D d!BC cos
(3) )
!AB D p
Substituting the known values d D 5 ft, !BC D 0:5 rad=s, and D tan have dPAB D 1:12 ft=s:
1
d!BC cos
: (4) d /2 C h2 h D 26:57ı into Eq. (3), we ` d
.`
Similarly for the acceleration we have E vEB=A C ˝ EP rEB=A C ˝ E .˝ E rEB=A / aEB D aEA C aEB=A C 2˝ August 10, 2009
922
Solutions Manual D dRB=A C 2!AB kO dPB=A {O C ˛AB kO dB=A C !AB kO .!AB kO dB=A {O/ q q 2 2 2 2 2 R D dAB !AB .` d / C h {O C ˛AB .` d / C h C 2d!BC !AB sin |O
(5)
O and dAB D !AB . We must also have where we have used the fact that aA D 0, ˝ D !AB , ˝P D ˛AB k, E vEB=C C ˝ EP rEB=C C ˝ E .˝ E rEB=C / aEB D aEC C aEB=C C 2˝ h i D !BC kO !BC . rB=C /IO 2 D !BC rB=C IO 2 D d!BC .sin {O
D
2 d!BC
sin {O
cos |O/ 2 d!BC cos |O;
(6)
where we have used ˝ D !BC and IO D cos {O sin |O. Setting the y components of Eqs. (5) and (6) equal to each other, we have q 2 2 R .` d /2 C h2 (7) dAB D d! sin C ! BC
AB
Substituting the expression for !AB from Eq. (4) into Eq. (7), we have !2 q
.d!BC cos /2 2 d /2 C h2 D d!BC sin C p (8) .` d /2 C h2 .` d /2 C h2 Substituting the known values h D 2:5 ft, d D 5 ft, ` D 10 ft, !BC D 0:5 rad=s, and D tan 1 ` hd D 26:57ı into Eq. (8) and rounding to three significant figures, we have dRAB D
2 d!BC
sin C
d!BC cos
p
.`
dRAB D 2:01 ft=s2 :
August 10, 2009
923
Dynamics 1e
Problem 6.127 The manually operated road barrier shown has a total length l D 15:7 ft and has a counterweight C whose position is identified by the distances d D 2:58 ft and ı D 1:4 ft. The barrier is pinned at O and can move in the vertical plane. Suppose that the barrier is raised and then released so that the end B of the barrier hits the support with a speed vB D 1:5 ft=s. Determine the speed of the counterweight C when B hits A.
Solution The barrier rotates about the fixed point O. Since both points B and C are on the barrier, ie.e, belong to the same rigid body, we must have !barrier D
vB `
d
and !barrier D
vC ı
)
vC D ı
vB `
d
:
(1)
Substituting given values ı D 1:4 ft, ` D 15:7 ft, d D 2:58 ft, and vB D 1:5 ft=s into the last of Eqs. (1), we have vC D 0:160 ft=s:
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Solutions Manual
Problem 6.128 A slender bar AB of length L D 1:45 ft is mounted on two identical disks D and E pinned at A and B, respectively, and of radius r D 1:5 in: The bar is allowed to move within a cylindrical bowl with center at O and diameter d D 2 ft. At the instant shown, the center G of the bar is moving with a speed v D 7 ft=s. Determine the angular velocity of the bar at the instant shown.
Solution By construction, the bar rotates about the fixed point O. Therefore, its angular speed is given by j!E bar j D hv , where h is the vertical distance between G and O. From the geometry of the problem we have that s 2 d L2 hD r : 2 4 Given the direction of v, G is moving clockwise about O, so we must have v !E bar D r kO d 2
2
r
L2 4
Substituting in the known values v D 7 ft=s, d D 2 ft, r D 1:9 in: D 0:1583 ft, and L D 1:45 ft, we have !E bar D
14:3 kO rad=s
:
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Dynamics 1e
Problem 6.129 Assuming that the rope does not slip relative to any of the pulleys in the system, determine the velocity and acceleration of A and D, knowing that the angular velocity and acceleration of pulley B are !B D 7 rad=s and ˛B D 3 rad=s2 , respectively. The diameters of pulleys B and C are d D 25 cm and ` D 34 cm, respectively.
Solution We let Q be the point of contact between pulley B and the cord to the right of pulley B. This cord is not moving because it is fixed to the wall, so from the no slip condition at Q, we must have vA D !B
d 2
and
d aA D ˛B : 2
Substituting given values d D 25 cm D 0:25 m, !B D 7 rad=s, and ˛B D 3 rad=s2 , we have vA D 0:8750 m=s and aA D 0:3750 m=s2 : (1) Using the directions shown in the problem statement, in vector form and with three significant figures, we have vEA D 0:875 |O m=s
and aEA D 0:375 |O m=s2
:
By pulley kinematics, we have yD C 2yB D constant
)
yPD D
2yPB
and yRD D
2yRB )
yPD D
2vA
and
yRD D
2aB ; (2)
where we have used that fact that A and B must have the same velocity and acceleration. Substituting the values from Eqs. (1) into Eqs. (2), for D we have vED D
1:75 |O m=s
and aED D
0:750 |O m=s2
:
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Solutions Manual
Problem 6.130 At the instant shown, the hammer head H is moving to the right with a speed vH D 45 ft=s and the angle D 20ı . Assuming that the belt does not slip relative to wheels A and B and assuming that wheel A is mounted on the shaft of the motor shown, determine the angular velocity of the motor at the instant shown. The diameter of wheel A is d D 0:25 ft, the radius of wheel B is R D 0:75 ft, and point C is at a distance ` D 0:72 ft from O, which is the center of the wheel A. Finally, let CD have a length L D 2 ft, and assume that, at the instant shown, D 25ı .
Solution From the geometry of the problem, we have the geometrical constraints xD D ` cos C L cos
and
L sin D ` sin C h; (1)
where h is the vertical distance between O and D. Taking the first time derivative of Eqs. (1), we have
xPD D vH x D
`P sin C L P sin
(2)
and P cos D P ` cos L
)
` cos P D P : L cos
(3)
Substituting the expresion for P from Eq. (3) into Eq. (2), we have vH x D
`P .sin cos C sin cos / D cos
`P sin . C / cos
)
P D
vH x cos ` sin . C /
Since the belt does not slip, and because !A is indicated as being clockwise, we have d!A D 2R!B
)
!EA D
2R vH x cos O k; d ` sin . C /
(4)
P Subsituting given values R D 0:75 ft, d D 0:25 ft, ` D 0:72 ft, where we have used the fact that !B D . ı ı vH x D 45 ft=s, D 25 , and D 20 into Eq. (4), we have !EA D
481 kO rad=s:
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Dynamics 1e
Problems 6.131 and 6.132 An overhead fold-up door with height H D 30 ft consists of two identical sections hinged at C . The roller at A moves along a horizontal guide, whereas the rollers at B and D, which are the midpoints of sections AC and CE, move along a vertical guide. The door’s operation is assisted by a counterweight P . Express your answers using the component system shown. If at the instant shown, the angle D 55ı and P is moving upward with a speed vP D 15 ft=s, determine the velocity of point E as well as the angular velocities of sections AC and CE. Problem 6.131
If at the instant shown, D 45ı , and A is moving to the right with a speed vA D 2 ft=s while decelerating at a rate of 1:5 ft=s2 , determine the acceleration of point E. Problem 6.132
Solution to 6.131 From the geometry of the problem, we have the constraints xA D
H cos ; 4
xE D xA ;
and
yE D
H sin :
(1)
Taking a time derivative of xA from the first of Eq. (1) and observing that xPA D vA D vA D
H P sin D 4
Observing that !AC D P , and !EAC D H D 30 ft, and D 55ı gives us P D 2:442 rad=s
)
)
vP
4vP : H sin
(2)
!E CE by symmetry, substituting given values vP D 15 ft=s,
!EAC D 2:44 kO rad=s
For the position of E, we have H cos {O 4 Taking a time derivative of Eq. (3), we have rEE D
vEE D
P D
vP , we have
H P sin {O 4
H
p
and
1
H P cos |O D
!E CE D
cos2
vP {O
2:44 kO rad=s:
(3)
4vP |O: tan
Substituting known values vP D 15 ft=s and D 55ı , we have vEE D . 15:0 {O
42:0 |O/ ft=s:
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Solutions Manual
Solution to 6.132 From the geometry of the problem, we have the constraints xE D xA D
H cos D 5:303 ft 4
and yE D yE D
For the position of E, we have H cos {O 4 Taking two time derivatives of Eq. (4), we have rEE D
vEE D vEE D and aEE
H P sin {O 4 0
B B D aA {O C H B @
H
H P cos |O D
p
1
vP {O
H sin :
cos2
(4)
4vP |O: tan 1
C 16vA2 256xA2 vA2 16xA aA C C q q C |O: 3=2 C 2 2 A 16xA 16xA 2 2 1 2 1 16xA H H 4 H 1 H2 H2 H2
Substituting known values xA D 5:303 ft, vA D 2 ft=s, aA D 1:5 ft=s2 , and D 45ı , we have aEE D . 1:50 {O
0:320 |O/ ft=s2 :
August 10, 2009
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Dynamics 1e
Problem 6.133 At the instant shown, bar AB rotates with a constant angular velocity !AB D 24 rad=s. Letting L D 0:75 m and H D 0:85 m, determine the angular acceleration of bar BC when bars AB and CD are as shown, i.e., parallel and horizontal.
Solution From the concept of instantaneous center of rotation, we know that the center of rotation of bar BC is at infinity, so !BC D 0: Therefore, the velocity of B is the same as that of C , so we have !CD D
!AB
!E CD D 24 kO rad=s:
)
From rigid body kinematics, we know that aEB D aEA C ˛EAB rEB=A
2 2 !AB rEB=A D L!AB {O
L˛AB |O
where we have used the fact that aA D 0. We must also have 2 !BC rEC =B L˛AB |O C ˛BC kO .2L {O
aEC D aEB C ˛EBC rEC =B 2 D L!AB {O
2 D .L!AB C H˛BC / {O C .2L˛BC
H |O/
2 !BC .2L {O
H |O/
L˛AB / |O:
(1)
We must also have aEC D aED C ˛ECD rEC =D
2 !CD rEC =D D
2 L!AB {O
L˛CD |O:
(2)
Eqsuations (1) and (2) are both expressions for aEC , and equating them component by component yields 2 L!AB C H˛BC D
2 L!AB
)
˛BC D
2 2L!AB H
L˛AB D
L˛CD
)
˛CD D
˛AB
2L˛BC
(3) 2 4L!AB : H
Substituting known values L D 0:75 m, !AB D 24 rad=s, and H D 0:85 m, in vector form to three significant figures, we have ˛EBC D
1020 kO rad=s2 :
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Solutions Manual
Problem 6.134 The bucket of a backhoe is the element AB of the four-bar linkage system ABCD. The bucket’s motion is controlled by extending or retracting the hydraulic arm EC . Assume that the points A, D, and E are fixed and that the bucket is made to rotate with a constant angular velocity !AB D 0:25 rad=s. In addition, suppose that, at the instant shown, point B is vertically aligned with point A and point C is horizontally aligned with B. Letting dE C denote the distance between points E and C , determine dPE C and dRE C at the instant shown. Let h D 0:66 ft, e D 0:46 ft, l D 0:9 ft, w D 1:0 ft, d D 4:6 ft, and q D 3:2 ft.
Solution Because point A is stationary, it is the center of rotation of bar AB, and therefore we have vEB D `!AB {O: Then, for point C , we must have vEC D vEB C !EBC rEC =B D `!AB {O C !BC kO . w/ {O D `!AB {O
w!BC |O;
(1)
and also vEC D vED C !E CD rEC =D D !CD kO Œ.h
w/ {O C .` C e/ |O D
.` C e/!CD {O C .h
w/!CD |O: (2)
Equations (1) and (2) are two equations in !BC and !CD , and when solved give !BC D
`.w h/!AB w.e C `/
and !CD D
`!AB : eC`
(3)
Substituting !BC from Eq. (3) into Eq. (1), we have vEC D `!AB {O C
`.w h/!AB |O D .0:2250 {O C 0:05625 |O/ ft=s; .e C `/
(4)
where we have used the given values ` D 0:9 ft, w D 1:0 ft, e D 0:46 ft, h D 0:66 ft, and !AB D 0:25 rad=s. Similarly for the accelerations, we have aEB D aEA C ˛EAB rEB=A
2 !AB rEB=A D
2 !AB ` |O;
(5)
where we have used the fact that aA D ˛AB D 0. Relating point C to point B gives us 2 aEC D aEB C ˛EBC rEC =B !BC rEC =B 2 O w/ {O ! 2 . w/ {O D !AB ` |O C ˛BC k. BC
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Dynamics 1e
`.w h/!AB Dw w.e C `/ 2 `2 .w h/2 !AB {O D w.e C `/2
2 {O
2 .`!AB C w˛BC / |O
2 .`!AB C w˛BC / |O:
(6)
Similarly, relating point C to point D we can have 2 aEC D aED C ˛ECD rEC =D !CD rEC =D D ˛CD kO Œ.h w/ {O C .e C `/ |O
D
e˛CD
`˛CD
2 !CD Œ.h !
2 `2 .h w/!AB .e C `/2
w/ {O C .e C `/ |O
{O C h˛CD
w˛CD
2 `2 !AB e`
! |O:
(7)
Equations (6) and (7) are two equations in ˛BC and ˛CD , which when solved give ˛BC D
`Πh3 ` C 2h2 w` C e.e C `/2 w .e C `/3 w 2
2 h`w 2 !AB
(8)
and ˛CD D
2 h`2 .w h/!AB : .e C `/w
(9)
Substituting ˛BC from Eq. (8) into Eq. (6), we have 2 `2 .w h/2 !AB aC D {O w.e C `/2
" 2 `!AB
`Πh3 ` C 2h2 w` C e.e C `/2 w .e C `/3 w
2 h`w 2 !AB
# |O
D .0:003164 {O C 0:03876 |O/ ft=s2 ; (10) where we have used the given values ` D 0:9 ft, w D 1:0 ft, e D 0:46 ft, h D 0:66 ft, and !AB D 0:25 rad=s. Now that the motion of both ends of the hydraulic arm EC is known, we can determine its time rate of change in length. We begin by defining the horizontal distance from E to C as xE C , and the vertical distance from E to C as yE C . The distance between E and C is then q 2 2 (11) dE C D xE C C yE C ; where xE C D d C h have
w D 4:260 ft and yE C D q D 3:200 ft: Taking two time derivatives of Eq. (11) we
2xE C xP E C C 2yE C yPE C dPE C D q 2 2 2 xE C C yE C and dRE C D
2 2 2xP E R E C C 2yE C yRE C .2xE C xP E C C 2yE C yPE C /2 C C 2yPE C C 2xE C x C : q 3=2 2 2 2 2 4 xE 2 xE C y C C yE C C EC
(12)
Observing that xP E C D vC x D 0:2250 ft=s, yPE C D vCy D 0:05625 ft=s, xR E C D aC x D 0:003164 ft=s2 , and yRE C D aCy D 0:03876 ft=s2 , we have dPE C D 0:214 ft=s
and dRE C D
0:0192 ft=s2 :
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Solutions Manual
Problem 6.135 Let frames A and B have their origins at points A and B, respectively. Point B does not move relative to point A. The velocity and acceleration of point P relative to frame A, which is fixed, are vEP D . 14:9 {OA C 19:4 |OA / ft=s
and
aEP D .1:78 {OA C 5:96 |OA / ft=s2 :
Knowing that frame B rotates relative to frame A at a constant angular velocity !B D 1:2 rad=s and that the position of point P relative to frame B is rEP =B D .8 {OB C4:5 |OB / ft, determine the velocity and acceleration of P relative to frame B at the instant shown and express the results using the frame A component system.
Solution Given the two reference frames in the problem statement, we have the following relationships between them: {OB D cos 23ı {OA C sin 23ı |OA
and |OB D
sin 23ı {OA C cos 23ı |OA :
(1)
From the concept of rotating reference frames, we have E rEP =B vEP D vEB C vEP rel C ˝ )
vP xA {OA C vP yA |OA D vP xB {OB C vP yB |OB C !B kO .8 {OB C 4:5 |OB /
)
vP xA {OA C vP yA |OA D vP xB .cos 23ı {OA C sin 23ı |OA / C vP yB . sin 23ı {OA C cos 23ı |OA / C !B kO .8.cos 23ı {OA C sin 23ı |OA / C 4:5. sin 23ı {OA C cos 23ı |OA //
)
. 14:9 {OA C 19:4 |OA / D .vP xB cos 23ı
vP yB sin 23ı
8!B sin 23ı
C .vP xB sin 23ı C vP yB cos 23ı 8!B cos 23ı
4:5!B sin 23ı / {OA 4:5!B sin 23ı / |OA ; (2)
where we’ve used the fact that vB D 0. Solving Eq. (2) for vP xB and vP yB using the known value !B D 1:2 rad=s, we have vP xB D
6:178 ft=s
and
vP yB D 12:67 ft=s
)
vEP rel D . 6:18 {OA C 12:7 |OA / ft=s:
Similarly for the acceleration, we have
) )
E vEP rel C ˝ EP rEP =B C ˝ E .˝ E rEP =B / aEP D aEB C aEP rel C 2˝ .1:78 {OA C 5:96 |OA / ft=s2 D aP xB {OB C aP yB {OB C 2!B kO . 6:18 {OA C 12:7 |OA / C !B kO Œ!B kO .8 {OB C 4:5 |OB / .1:78 {OA C 5:96 |OA / ft=s2 D aP xB {OB C aP yB {OB .25:4!B {OA C !B 12:36 |OA / C !B kO . 4:5!B {OB C 8!B |OB /:
(3)
Substituting Eqs. (1) and the given value !B D 1:2 rad=s into Eq. (3), and solving for aP xB and aP yB , we have aP xB D 40:27 ft=s2
and aP yB D 31:25 ft=s2
)
aEP rel D .40:3 {OA C 31:3 |OA /: August 10, 2009
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Dynamics 1e
Problem 6.136 A vertical shaft has a base B that is stationary relative to an inertial reference frame with vertical axis Z. Arm OA is attached to the vertical shaft and rotates about the Z axis with an angular velocity !OA and an angular acceleration ˛OA . The ´ axis is coincident with the Z axis but is part of a reference frame that rotates with the arm OA. The x axis of the rotating reference frame coincides with the axis of the arm OA. At the instant shown, the y axis is perpendicular to the page and directed into the page. At this instant, the collar C is at a distance d from the Z axis, is sliding along OA with a constant speed vC (relative to the arm OA), and is rotating with a constant angular velocity !C (relative to the xy´ frame). Point D is attached to the collar and is at a distance ` from the x axis. At the instant shown, point D happens to be in the plane that is rotated by an angle from the x´ plane. Compute the expression of the inertial velocity and acceleration of point D at the instant shown in terms of the parameters given, and express it relative to the rotating coordinate system.
Solution Using the concept of rotating reference frames, we have E rED=O ; vED D vEO C vEDrel C ˝
(1)
where E vEO D 0;
vEDrel D vC {O
`!C cos |O
O `!C sin k;
O E D !OA k; ˝
(2)
and O rED=O D d {O C `. sin |O C cos k/:
(3)
Substituting Eqs. (3) into Eq. (1), we have vED D vC {O
`!C cos |O
O `!C sin kO C !OA kO Œd {O C `. sin |O C cos k/; (4)
which can be simplified to read vED D .vC C !OA ` sin / {O C .d!OA
`!C cos / |O
O `!C sin k:
Similarly, for the accelerations, we have E vEDrel C ˝ EP rED=O C ˝ E .˝ E rED=O /; aED D aEO C aEDrel C 2˝
(5) August 10, 2009
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Solutions Manual
where E aEO D 0;
vEDrel D vC {O
O `!C sin k;
`!C cos |O
O E D !OA k; ˝
O EP D ˛OA k; ˝
2 aEDrel D !C ` sin |O
and
2 O !C cos k;
O rED=O D d {O C `. sin |O C cos k/: (6)
Substituting Eqs. (6) into Eq. (5), we have 2 2 O aED D !C ` sin |O !C cos kO C 2!OA kO ŒvC {O `!C cos |O `!C sin k O C !OA kO Œ!OA kO .d {O ` sin |O C ` cos k/; O C ˛OA kO Œd {O C `. sin |O C cos k/
which can be simplified to read 2 2 2 aED D . d!OA C 2!OA `!C cos C `˛OA sin / {O C .d˛OA C .!C C !OA /` sin C 2!OA vC / |O O ! 2 cos k: C
August 10, 2009
935
Dynamics 1e
Problem 6.137 The wheel D rotates without slip over a curved cylindrical surface with constant radius of curvature L D 1:6 ft and center at the fixed point E. The X Y frame is attached to the rolling surface, and it is inertial. The xy frame is attached to D at O and rotates with it at a constant angular velocity !D D 14 rad=s. Collar C slides along the bar AB with a constant velocity vC D 4 ft=s relative to the xy frame. At the instant shown, points O and E are vertically aligned. Letting ` D 0:25 ft and R D 1 ft, determine the inertial velocity and acceleration of C at the instant shown when D 25ı and the xy frame is parallel to the XY frame. Express your result in the xy and X Y frames.
Solution From the geometry of the problem, we have rEC =O D frames, we have
` cot {O C ` |O. From the concept of rotating reference
E rEC =O vEC D vEO C vEC =O C ˝ D .R!D C vC /O{ C !D kO .` cot {O
` |O/
D .R!D C vC C `!D / {O C `!D cot |O;
(1)
O Substituting known values R D 1 ft, E D !D k. where we have used the fact that vEO D R!D {O and ˝ ı ` D 0:25 ft, !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (1), and rounding to three significant figures we have vEC D .21:5 {O C 7:51 |O/ ft=s: For the acceleration, we then have E vEC =O C ˝ EP rEC =O C ˝ E ˝ E rEC =O aEC D aEO C aEC =O C 2˝ 2 R2 !D |O L R
h i 2!D kO vC {O C ˛D kO . ` cot {O C ` |O/ C !D kO !D kO . ` cot {O C ` |O/ ! 2!2 R 2 D 2 2vC !D `!D C D `!D cot {O C |O; (2) L R
D
2 2 O ˝ E D !D k, EP D ˛D kO D 0, E and aEO D R !D |O. Substituting where we have used the fact that vO D 0, ˝ L R known values ` D 0:25 ft, !D D 14 rad=s, vC D 4 ft=s, D 25ı into Eq. (2), and rounding to three significant figures we have
aEC D .105 {O
166 |O/ ft=s2 : August 10, 2009
936
Solutions Manual
August 10, 2009
937
Dynamics 1e
Problem 6.138 The wheel D rotates without slip over a curved cylindrical surface with constant radius of curvature L D 1:6 ft and center at the fixed point E. The X Y frame is attached to the rolling surface, and it is inertial. The xy frame is attached to D at O and rotates with it at an angular velocity !D D 14 rad=s and an angular acceleration ˛D D 1:3 rad=s2 . Collar C slides along the bar AB with a constant velocity vC D 4 ft=s relative to the xy frame. Letting ` D 0:25 ft and R D 1 ft, determine the inertial velocity and acceleration of C , at the instant shown, when D 25ı and the xy frame is parallel to the X Y frame. Express your result in the xy and X Y frames.
Solution From the geometry of the problem, we have rEC =O D frames, we have
` cot {O C ` |O. From the concept of rotating reference
E rEC =O vEC D vEO C vEC =O C ˝ D .R!D C vC /O{ C !D kO .` cot {O
` |O/
D .R!D C vC C `!D / {O C `!D cot |O;
(1)
O Substituting known values R D 1 ft, E D !D k. where we have used the fact that vEO D R!D {O and ˝ ı ` D 0:25 ft, !D D 14 rad=s, vC D 4 ft=s, D 25 into Eq. (1), and rounding to three significant figures we have vEC D .21:5 {O C 7:51 |O/ ft=s: For the acceleration, we then have E rEC =O E vEC =O C ˝ EP rEC =O C ˝ E ˝ aEC D aEO C aEC =O C 2˝ 2 R2 !D |O L R
i h 2!D kO vC {O C ˛D kO . ` cot {O C ` |O/ C !D kO !D kO . ` cot {O C ` |O/ ! 2!2 R 2 2 D D `˛D C `!D cot {O C 2vC !D `!D C C `˛D cos |O; (2) L R
D
O and aEO D R !D |O. Substituting O ˝ E D !D k, EP D ˛D k, where we have used the fact that vO D 0, ˝ L R known values ` D 0:25 ft, !D D 14 rad=s, vC D 4 ft=s, D 25ı into Eq. (2), and rounding to three significant figures we have 2
aEC D .107 {O
2
166 |O/ ft=s2 : August 10, 2009
