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| 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 |
| P1 | To do the integral (if you use separation of variables) you may use Maple, but be sure to attach a copy of the Maple Worksheet to the problem. You may also use Dsolve on this problem if you prefer. | a) 0.044 v2 - 32.2 (up) b) 27.1 ft/s (down) |
| P2 | You will need to use two relative motion
equations - one for each direction. The magnitude of the velocity of
the wind with respect to the boat will change in the two cases, so you
should have 4 equations and 4 unknowns.
The answers for the wind should be: vwindx = 0.423v0 vwindy = v0 |
|
| P3 | Recall for projectile motion (using conservation of linear momentum) ax=0, ay
= -g) |
 |
| Problem Set P-02 | ||
| problem | hint | answer |
| P4 | Be careful with the signs. What you define to be positive for your dependent motion also needs to be the direction you put the acceleration in you kinetic diagrams. | a) aA = 3.10 m/s2 (down), aB =
6.20 m/s2 (down), aC = 12.39 m/s2 (to the
left) b) T1 = 13.42 N, T2 = 24.78 N c) 6.20 m |
| P5 | The acceleration of B is not directed down the incline! | aB = 3.39 m/s2
down from horizontal left at 28.6 degrees vB/A = 2.38 m/s down from horizontal left at 20 degrees. |
| P6 | The two critical locations are at the bottom and at the top of the loop. In the case of the second loop in order for it to make it to the top the normal force must always be positive. | a) 3.74 m/s b) c) 3.84 m/s |
| P7 | Remember the spring is initially stretched 0.1 m in position 1. | a) 5.51 m/s b) 85.8 N up |
| P8 | Assume the motion takes place in the horizontal plane. Use kinematics to determine the acceleration components in the radial and transverse directions. From these you should be able to find the normal and tangential components of the accelerations. Remember that the tangential direction is in the same direction as the velocity. One method is to use a dot product to determine the tangential component of acceleration. Alternatively, you could just use geometry. Finally use LM rate in the normal and tangential directions. | P = tangential force = 12.1 N, Normal force = 34.5 N Note: These are the magnitudes. Be sure to also include the directions in your answers. |
| 3.10 | Remember you know the direction the plane is traveling. |  |
| Problem Set P-03 | ||
| problem | hint | answer |
| 3.13 | The spring force is not impulsive. The initial energy in the spring is not zero! You need to determine the initial compression of the spring that mass B is sitting on. | h = 0.072 ft k = 72.2 lbf/ft |
| 3.14 | Use projectile motion or conservation of energy to find the velocity of the ball as it strikes the ground. In the first impact A will not move, but in the second impact C will have a velocity downwards. The x-component of velocity will not change in this problem since there are no forces in the x-direction. | e = 0.923 h1 = 1.28 m |
| 3.16 | Remember that this is a constrained impact problem. | theta = 62.7 deg Elost = 0.14 v02 impulse = 0.36 v0 |
| 3.25 | No hint | v0=13.9 m/s, |
| Problem Set P-04 | ||
| problem | hint | answer |
| P9 | There is really no need to find the center of gravity of the cat and the chair. Just use a kinetic diagram with two CGs. Note: Since this is a new problem the answer shown may be wrong or have a typo. | a) T = 4.90 lbf b) F = 1.37 lbf to the left |
| 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 bar. You know the direction of the acceleration of each CG using normal and tangential coordinates. For a translating object, every point has the save velocity and the same acceleration. | a = 20.7 ft/s2 TCF = -4.05 lbf TBE = -14.3 lbf |
| 4.1 | No hint | a) 165 rev b) 2200 rev |
| 4.37 | No hint. Be sure to include your directions for the accelerations of E on B and E on C. | aB = 15.7 rad/s2 aC = 5.24 rad/s2 aE = 3.18 m/s2 (on B) aE = 1.29 m/s2 (on C) |
| TOP | ||
| Problem Set P-05 | ||
| problem | hint | answer |
| 4.5 | Be sure your answers include the directions for Cx and Cy. You do not need to find the center of gravity of the object - just treat it as two different CGs. | a) alpha = 43.6 rad/s2 CW b) Cx = 21 N, Cy = 54.6 N |
| 4.38 | 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. You can do this in Maple or by hand. | Lcg = 0.106 m |
| P10 | When you draw your impulse-momentum diagrams and your FBD/KD make sure they are large enough to include everything. Treat the problem as if you have three centers of gravity (the two bars and the glob). There are two parts to the problem 1) the impact, 2) after the impact. | omega = 1.68 rad/s CW Ax = 3.13 N (left) Ay = 17.4 N (up) |
| 4.40 | The spring is stretched at both positions. | omega = 11.1 rad/s CCW |
| 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 |
| TOP | ||
| Problem Set P-06 | ||
| problem | hint | answer |
| P11 | 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. | |
| 4.13 | no hint | vGgear = 0.829 m/s (to the left) |
| 4.12 | Be careful when finding the work done by the force P. What distance does it travel? | vB = 16.3 ft/s |
| 4.41 | Have fun. The easiest way to do the kinematics is to use the IC of velocity. | wgear=9.19 rad/s |
| 4.14 | Assume the cue applies a horizontal force to the ball. | 2/5 r |
| 4.45 | Which way will the chair swing if it is moving too slow when the cable engages? Minimum velocity at 2 is 5.56 m/s. | Ramp length = 6.09 m |
| TOP | ||
| 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 |
| TOP | ||
| Problem Set P-08 | ||
| problem | hint | answer |
| 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. | |
| 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) |
| 4.47 | no hint | aCD = 0.0468 rad/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 |
| TOP | ||
| Problem Set P-09 | ||
| problem | hint | answer |
| 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 |
| 4.27 | no hint | aA = 2P/7m aB = 22P/7m |
| 4.30 | no hint | Some answers and a plot is given in the notes. |
| 4.48 | See the problem assignment for Maple hints. I had to assign the acceleration of the top to be a function of time, that is b:=t-> ... | Lwcg~22 cm |
| 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 vrel = 1.344 m/s |
| 4.32 | Use frames with origins at A, and both oriented horizontal-vertical. | vB = (-0.25i + 2j) ft/s aB = (4.8i -2.4j) ft/s^2 |
| 4.33 | Don't forget the relative acceleration will have a component to the left since it is traveling in a curved path. | -20 rad/s2 CW |
| TOP | ||
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Prof. Brad Burchett Last modified: Mon Nov 20 at 11:38:25 US Eastern Standard Time 2006 |