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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 jones5@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 | selected answers |
| 3.4 |
You will need to separate variables and integrate. Also, be careful with your units! If you do a unit consistency check, you will see that the mass of the sphere cannot be input as 29 lb. |
Depth:
6.98 miles |
| 3.5 |
VB/A=109.5i+20.29j km/hr rB/A=2.96 km |
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| Problem Set P-02 | ||
| problem | hint | selected answers |
| 3.11 |
Be careful
with the directions you use for dependent motion and the direction of
your acceleration on your kinetic diagram. |
T1 = 13.42 N T2 = 24.78 N SA = 6.2 |
| 3.12 |
The direction of aB/A is along the incline. | answers |
| 3.14 |
When the car loses contact the normal force goes to zero. | answers |
| 3.24 |
|
answers |
| 3.26 |
No hint |
answers |
| Problem Set P-04 | ||
| problem | hint | answer |
| 4.3 | Since the system is released from rest, the velocity immediately after release is negligably small. What does this imply for the normal acceleration? | a = 20.7 ft/s2 FCF = 4.1 lbf FCF = 14.3 lbf |
| 4.5 | Since you are concerned with the forces between the cat and the chair, it would be helpful to consider each body as its own system. | a) T = 4.9 lbf b) The cat does not slip (be sure to quantify why this is the case) |
| Problem Set P-05 | ||
| problem | hint | answer |
| 4.1 |
Uniformly accelerated motion means constant acceleration; this is a pure rotational kinematics problem. |
θspin up=165 revolutions θspin down=2200 revolutions |
| 4.8 |
No hint. |
αB = 15.7
rad/s2 CCW αC = 5.24 rad/s2 CW (on B) aE = 3.18 m/s2 at an angle of 14.3 deg. up from left horiztonal. (on C) aE = 1.29 m/s2 at an angle 37.4 deg. up from horizontal right. |
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| 4.9 |
To maximize omega when the object is in the vertical position, find omega2 as a function of Lcg; differentiate with respect to Lcg; set equal to zero and solve for Lcg. | Lcg = 0.1057 m |
| 4.10 | The spring is stretched at both positions. | omega = 11.1 rad/s CCW alpha = 28.3 rad/s2 CCW |
| 4.11 |
Treat the system has having two centers of gravity | alpha = 43.6 rad/s2 (CW) Cx = 21 N (left) Cy = 54.6 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 |
| 4.19 |
No hint |
a) 12 rad/s CCW b) -1.5i-3.6j m/s |
| 4.20 | No hint | a) omegaFBD = 3.333 rad/s CCW b)vF = 1.11i - 1.67j m/s |
| TOP | ||
| Problem Set P-07 | ||
| problem | hint | answer |
| 4.21 | At state 2 (i.e. after the disk rotates 90 deg), is the bar translating, rotating, or exhibiting general plane motion? Compute the instant center; there may be fewer unknowns than you think! Also, don't forget to account for the external moment; how are you going to "work" that into your equations? | vR = 0.48 m/s → |
| 4.26 |
Final energy includes KE for all bodies (2 terms each for rods), and PE for rods. Initial includes spring energy only. You are going to need a bunch of kinematic equations to relate the speeds at the C.G.'s to the angular velocities of the bars. | vC = 1.70 m/s ← |
| 4.29 |
Ball rolls without slip and the cue imparts a sharp impulsive force on the ball. The cue impulse would therefore be much greater
than, say, the impulse related to the weight or the frictional force. |
h = (2r)/5 |
| 4.31 |
This isn't as simple as it looks! The ball will have to impact the ramp before it start rolling up the slope;
choose wisely the location to sum angular momentum. |
h = R + (R2 ω2)(1+4cosθ+4cos2θ)/(12g) |
| 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 (in Maple) 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 |
aG = L/2 [(αcosθ - ω2sinθ)i - (αsinθ + ω2cosθ)j] |
| 4.37 | Remember, the method of instantaneous centers does not work (in general) for accelerations. You may use it for computing velocities, but do not use it for accelerations! | aD = 296 m/s2 (up) |
| 4.38 | This is a straight-forward question, but you must be diligent in writing down the equations.
It is very easy to write down the incorrect subscript or drop a minus sign; be careful and double check your equations
before you try to solve them. |
αBD = 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 provided in
the answers shown on the right |
aA = 18.4 ft/s2, aB
= 9.2 ft/s2 |
| 4.42 | T = 0.197 Ibf alpha = 65 rad/s^2 CW |
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| 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) answer not given |
| 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. Jones Last modified: 9 March 2014 |