These answers are provided for you to check your work. If you get one of these answers and it does not follow from your work the graders have been told to give you a ZERO for the problem. The answers provided are not guarenteed to be correct. If you get a different answer be sure to talk to your professor and email bradley.burchett@rose-hulman.edu with any corrections. In the answers provided sometimes only the magnitude is given and not the direction. For your homework be sure to include the magnitude and direction when appropriate.

Problem Set P-01
problem	hint	answer
1.1	You will need to integrate twice to find an equation for s as a function of time.	s = 6.98 miles
1.2	You don’t need the initial velocity of the boat because it should cancel from the equations. You may start with the projectile motion equations rather than deriving them from LM rate if you’d like.	d = 0.979 m vD/P=0.765 m/s i – 12.53 m/s j
1.3	To find the answer to b) solve for the magnitude of vB_D in terms of vB and then take the derivative with respect to vB and set it equal to zero. Again, you may start with the projectile motion equations since we have derived them several times	a) 1.175 m/s b) 0.1675 m/s
2.1	Be sure the positive direction you use for LM rate is the same as the positive direction you use for the dependent motion part of the problem. Also, since there are two cables you will need two constraint equations.	aA= 3.098 m/s2 down, aB= 6.196m/s2 down aC= 12.39m/s2 left, T1 = 13.42 N T2 = 24.78 N
2.2	Since A is moving you do not know the direction of the velocity of B, but you do know the direction of the velocity of B with respect to A.	aB= 3.394 m/s2 at 28.6 degrees down from left vB/A= 2.38 m/s at 20 degrees down from left

Problem Set P-02
problem	hint	answer
3.1	No Hint	a) 3.74 m/s c) 3.84 m/s
Text 3.7	Remember the normal force is zero when the block leaves the incline.	theta=27.5º, x = 3.81 ft
3.3	Be careful with units.	a) 9.83 ft/s b) 12.69 ft/s c) 37.5 lbf
4.1	No hint	ans given on handout
Text 3.11	No hint	vr=-4.94 ft/s, vth = 7.5 ft/s
5.1	Use projectile motion from immediately after impact until the ball strikes the ground. In figure 1 the plate has zero velocity right after impact. In figure 2 the plate has non-zero velocity right after impact.	v0=14.3 ft/s, e=0.324
5.2	The cords constrain the large sphere from moving downward after impact, however, they do not constrain the small sphere from moving upwards after impact. Be sure to apply coefficient of restitution to the normal components only.	v'A=1.741m/s, v'B=3.08m/s

Problem Set P-03
problem	hint	answer
6.1	no hint	V0 = 13.88 m/s VB = 15.1 m/s V2 = 13.9 m/s V3 = 5.86 m/s Fimp = 2.75 N-s
Text 3.16	Assume that v0 is known.	theta = 62.7 deg Elost = 0.14 v02 impulse = 0.36v0 N-s
7.2	You will need to use relative motion since you don't know the direction of the motion for AC, but you do know the direction of the relative motion.  For part c) you can either use Energy and LM finite or LM rate and integrate aA/C.	F = 4.85 lbf vB = 1.14 ft/s vAx = 1.63 ft/s vAy = 1.01 ft/s
7.1	no hint	TB = 39 lbf TA = 13 lbf alpha = 8.05 rad/s2 TB = 57.1 lbf TA = 19 lbf
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Problem Set P-04
problem	hint	answer
10.1	No hint	aB = 15.7 rad/s2 aC = 5.24 rad/s2 aE = 3.18 m/s2 (on B) aE = 1.29 m/22 37.4 (on C)
4.3	Rather than finding the center of gravity for the composite shape it is often easier just to put the weights (for the FBD) and the maGs (for the KD) at the c.g. for each individual bars	a = 20.7 ft/s2 TCF = -4.05 lbf TBE = -14.3 lbf
4.4	The disks will stop slipping when the velocity of the point of contact is the same for both disks. Be sure to write the velocities as functions of time and then solve for the time.	a) alphaA = 12.5 rad/s2 CCW alphaB = 33.3 rad/s2 CCW b) omegaA = 240 rpm CW omegaB = 320 rpm CCW
11.2	To maximize w in the vertical, find w2 as a function of Lcg, differentiate with respect to Lcg, set equal to zero and solve for Lcg.	Lcg = 0.1057 m

Problem Set P-05
problem	hint	answer
12.1	The spring is stretched at both positions.	omega = 11.1 rad/s CCW
12.2	The answers shown neglect the mass of the bullet (after the impact) since we don't know exactly where it stops in the plate and its mass is much less than the mass of the plate.	a) vG = 2.57 m/s (to the right) b) Ax = 3.79 kN (to the right) Ay = 4.5 kN c)Ax = 491 N (to the left) Ay =  35.3 N (up)
4.8	For part c) be sure to consider that the center of gravity is moving and it is rolling.	a)omega = 3 rad/s CW b) vA = 9 in/s to the right c) 15 in of cord unwound per second
4.9	To find the velocity of the midpoint I think it is easier to use the vector algebra approach rather than instantaneous center of velocity.	a)omegaBD = 12 rad/s CCW b) 3.9 m/s at 67.4
4.11	The hardest part in this problem is to relate all the angular velocities and velocities of the centers of gravities in the kinetic energy expression.  You will also need to use the velocity of point B. Also, don't forget the work done by the moment.	vGrod = 0.48 m/s
P6	Write the position vector between A and B in terms of the general angle, q, and the position vector from B to D in terms of b.  You will then need to relate b and q using geometry.	pdf
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Problem Set P-06
problem	hint	answer
4.11	The hardest part in this problem is to relate all the angular velocities and velocities of the centers of gravities in the kinetic energy expression.  You will also need to use the velocity of point B. Also, don't forget the work done by the moment.	vGrod = 0.48 m/s
4.12	Be careful when finding the work done by the force P. What distance does it travel?	vB = 16.3 ft/s
4.15	no hint	8.8 ft/s
P7, P8	Have fun	pdf
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Problem Set P-07
problem	hint	answer
4.16	You may assume the mass moments of inertia of the rods about their own CGs are zero when they are vertical since you are not given the radius of the rods.	32.0 rev/min
4.17	The velocity of the point of contact between the sphere and the cart is not zero. Therefore, the point of contact is not the instantaneous center of velocity. Therefore, I would suggest using the vector algebra approach to relate vcart to vGsphere and omegacyl. You will need to use conservation of energy plus another conservation principle in order to get enough equations to solve the problem. Since you are not given the radius of the sphere, r, just leave it in your equations (note: it is not one of your unknowns). When solving the equations if you get a RootOf just use evalf(allvalues(solve( ...)))	vcart = 2.10 ft/s
4.18	no hint	a) x = 0.6m from A b) x = 0.2 m from A
4.19	The translational acceleration of the mass center is zero.	a) aB = 5407 ft/s2 (down) b) aC = 5407 ft/s2 (up) c) aD = 5407 ft/s2 (down at 60°)
4.20	Relate the acceleration of the mass center to both ends of the rod.
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Problem Set P-08
problem	hint	answer
4.21	The first step is to do the geometry.  The angle between BD and vertical turns out to be 16.779° (be sure you calculate this value and don't just use this hint).	aD = 296 m/s2 (up)
4.22a	no hint	aC = {-7.79i -33.5j} m/s2
4.23	Only the magnitudes are given to the right.	aA = 18.4 ft/s^2, aB = 9.2 ft/s^2
4.24	You will need to use kinematics to relate the acceleration of the string to the acceleration of the center of gravity.	a) T = 0.197 lbf b) alpha = 65.0 rad/s2
4.25	It will start to rotate without slipping when vG = omega r	a)  3.22 ft/s2 b)  24.15 rad/s2 c)  1.60 s d)  vG = 9.86 ft/s e)  xG = 19.9 ft f)
4.26	no hint	a) alpha = 2.81 rad/s2 (CCW) aG = 0.337 m/s2 (left) b) mus = 0.22
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Problem Set P-09
problem	hint	answer
4.27	no hint	aA = 2P/7m aB = 22P/7m
4.28	no hint	theta = 54.7 deg
4.29	Determine alpha as a function of theta and integrate for theta = 0 to 90 degrees.	omega = 2.22 rad/s
4.30	The easiest way to write the position vector for bar BD is in terms of an angle with respect to the horizontal, beta, where beta = sin-1(b sin(theta)/L). Be sure you know where this equation comes from.	Answer for when theta = 180°: By = 0, Bx = 805 N (left) Dy = 0, Dx = 426 N (right). A plot of Bx was provided with the problem statement.
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Problem Set P-10
problem	hint	answer
4.31	Attach the body (rotating) frame to arm AD at A. Attach the fixed frame to bar BP at B. Rotate the body frame velocity to line up with the fixed frame (horizontal-vertical).	omega_BP = -5.17 rad/s k nu_B = 1.344 m/s
4.32	Use frames with origins at A, and both oriented horizontal-vertical.	v_B = (-0.25i + 2j) ft/s a_B = (4.8i -2.4j) ft/s^2
4.33	Don't forget the normal component of acceleration in the rotating frame (the slot is curved). Use fixed frame centered at C and oriented i =up j = left. Rotating frame centered at D with same orientation.	Alpha_AD = 20 rad/s^2 CW
4.34	Use velocity kinematics from 28.1. Use normal-tangential coordinates (RAFA) on your FBD=KD. Take moments about P.  The problem statement has an error - it should read the mass of bar BP is 5 kg.	nu_dot_B = 24.48 m/s^2 alpha_BP = -64.41 rad/s^2 k in normal tangential:   Bt = 44.73 ut   Bn = -45.5 un in x-y:   Bx = -27.47 i   By =57.6 j
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Prof. Brad Burchett Last modified: Mon Jan 04 at 11:38:25 US Eastern Standard Time 2006