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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
guaranteed 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 |
| 3.99 | Use ads=vdv, separate and integrate. Limits of integration on v go from 880 to 0. | s = 8817 ft |
| 3.5 | Note the angle is 25 degrees. |   |
| 3.6 | Recall for projectile motion (using conservation of linear momentum) ax=0, ay = -g. The time to impact is 0.32 s. - be sure to SOLVE for this value in your homework! |
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| Problem Set P-02 | ||
| problem | hint | answer |
| 3.11 | 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 |
| 3.12 | The direction of aB/A is along the incline. |  aB = 5.94 m/s2 at -75.7° |
| 3.96 | Using constant velocity in the x-direction, solve for time of flight in terms of initial velocity. Sub this expression for time into the y-displacement equation and solve for t. Now back out initial v. | a) t=2.73, v0=51.8 ft/s b) t=1.836, and v0=26.4 ft/s |
| 3.38 | Be careful with units. | 303.2 ft/s |
| 3.24 | Resolve v into radial and transverse components. Equate to the polar v and solve for |v| in terms of theta_dot. Repeat for accel, equating transverse components to solve for |a| in terms of theta and its derivatives. | |
| 3.26 | Solve for rddot and use the chain rule to separate variables and integrate. There is an example in the book like this. | N = 2.30 lbf vr = 4.94 ft/s, vq = 7.5 ft/s |
| Problem Set P-03 | ||
| problem | hint | answer |
| 3.29 | 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 |
| 3.31 | 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 |
| 3.32 | vA' = 2.74 m/s down from left at 42º vB' = 2.74 m/s up from right at 42º |
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| 3.34 | 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 |
| 3.35 | No hint (see example in book) | v0=13.9 m/s, |
| Meriam 2/121 | Key is to relate r-theta components of velocity and acceleration to path coordinates. Note that the path is a circle centered at C. R-theta coordinates are centered at O. | |a|=(b2w2)*(b2-e2)-1/2 Where w=d_theta/dt |
| Problem Set P-04 | ||
| problem | hint | answer |
| 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 |
| 4.5 | 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. | a) T = 4.90 lbf b) F = 1.37 lbf to the left |
| TOP | ||
| Problem Set P-05 | ||
| problem | hint | answer |
| 4.1 | No hint | a) 165 rev b) 2200 rev |
| 4.2 | If you first solve for theta as a function of time you should get theta = 0.8t-0.9t2 | vA = 1.64 m/s (up) xA = 3.7 m (up) vB = 0.738 m/s (down) xB = 1.665 m (down) |
| 4.9 | 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. A copy of your solution to this problem needs to be attached to the lab from week 3 and turned in separately on Friday. | Lcg = 0.106 m |
| 4.10 | The spring is stretched at both positions. | omega = 11.1 rad/s CCW |
| 4.11 | 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 |
| 4.16 | 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) |
| TOP | ||
| Problem Set P-06 | ||
| problem | hint | answer |
| 4.18 | 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 |
| P1 | To find the velocity of the point A you may use the IC or vector algebra. Both are about the same amount of work. | a)omegaAD = 4.27 rad/s CW b) 1.33 m/s down c) 1.557 m/s up at 34.7 degrees |
| P2 | 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. | |
| 4.25 | Use instant center of the link to connect all omegas. Place gravity datum at the axle. Notice the geometry. | vA = -7.72j m/s |
| 4.22 | Be careful when finding the work done by the force P. What distance does it travel? | vB = 16.3 ft/s |
| Problem Set P-07 | ||
| problem | hint | answer |
| 4.23 | Have fun. The easiest way to do the kinematics is to use the IC of velocity. | wgear=9.19 rad/s |
| 4.13 | no hint | vGgear = 0.829 m/s (to the left) |
| 4.24 | no hint | vGgear = 0.829 m/s (to the left) |
| 4.26 | Final energy include KE for all bodies (2 terms each for rods), and PE for rods. Initial includes spring energy only. | |vC| = 1.703 m/s |
| 4.29 | Assume the cue applies a horizontal force to the ball. | 2/5 r |
| 4.31 | Don't forget the impact | link |
| TOP | ||
| Problem Set P-08 | ||
| problem | hint | answer |
| 4.32 | 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 |
| 4.33 | 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 |
| 4.35 | 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°) |
| 4.36 | Relate the acceleration of the mass center to both ends of the rod. | |
| 4.37 | 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) |
| 4.38 | no hint | alphaBD = 9.99 rad/s2 (CCW) aEx = 3.75 m/s2 (left) aEy = 1.05 m/s2 (up) |
| TOP | ||
| Problem Set P-09 | ||
| problem | hint | answer |
| 4.41 | Only the magnitudes are given to the right. | aA = 18.4 ft/s^2, aB = 9.2 ft/s^2 |
| 4.54 | Determine alpha as a function of theta and integrate for theta = 0 to 90 degrees. | omega = 2.22 rad/s |
| 4.43 | 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) |
| 4.44 | no hint | a) alpha = 2.81 rad/s2 (CCW) aG = 0.337 m/s2 (left) b) mus = 0.22 |
| 4.47 | no hint | aA = 2P/7m aB = 22P/7m |
| Problem Set P-10 | ||
| problem | hint | answer |
| 4.55 | 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 |
| 4.56 | 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 |
| 4.57 | 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 |
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Prof. Brad Burchett Last modified: Mon Nov 20 at 11:38:25 US Eastern Standard Time 2006 |
