NOTE: Do NOT work Part (b) of Homework H6.B. The material on which that question is based will not be covered until next class.

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Discussion and hints

The four-step plan:

1. FBD: It is recommended that you draw a free body diagram of the plate and cart combined.
2. Kinetics: Your FBD should show that the summation of forces in the direction of motion for the cart (call that the x-direction) is zero; therefore, linear momentum in the x-direction is conserved for ALL time. Also note that energy is conserved up to the point of impact. During impact, energy is not conserved; however, you are given the coefficient of restitution (COR) for the impact.
3. Kinematics: Write down the rigid body velocity equation relating the motion of points O and the plate's center of mass G. Here you will use the fact that the cart moves only in the x-direction.
4. Solve

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Discussion and hints

The four-step plan:

1. FBD: It is recommended that you draw a free body diagram of the bar and particle P combined.
2. Impulse/momentum: Your FBD should show that the summation of moments about point O is zero. From this, you know that angular momentum of the system about point O is zero. In addition to that equation, you can use the coefficient of restitution (COR) equation to relate the "n" components of velocity of P and end A before and after impact.
3. Kinematics: Either write down the rigid body kinematics equation relating points O and A, OR use instant centers to determine the components of velocity of A after impact. (Note that A has both "n" and "t" components of velocity after impact.)
4. Solve

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Discussion and hints

The animation above shows the motion for this system found from numerical integration of the equations of motion for the system. Since we are interested in the velocity of A immediately after its first impact with B, we need to break up our analysis into two parts:

• 1 to 2: From release up to immediately prior to the impact of A and B. For this time period, both energy and angular momentum are conserved.
• 2 to 3: During the time of impact. For this time period, energy is NOT conserved, however, angular momentum is conserved. You are also given information on the coefficient of restitution.

Please note that you are asked to find the VELOCITY of A after the first impact. This velocity includes both radial and transverse components.

This problem should look familiar to you. It is very similar to Homework H4.S from Chapter 4 on angular momentum of particles. The only difference between these two problems is that here you need to take into account the bar for both the energy and angular momentum calculations, whereas for H4.S you treated the bar as being of negligible mass. It might be useful for you to review the solution video for H4.6 before starting this solution.

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Discussion and hints

The four-step plan:

1. FBD
2. Impulse/momentum: The summation of forces in the horizontal direction is used to relate the change in linear momentum to the impulse of the friction force. The summation of moments about the CM point A is used to related the change in angular momentum about A to the moment created by friction. This gives you two equations in terms of three unknowns: omega_2, Delta_t and v_A2.
3. Kinematics: At the instant when slipping ceases, the contact point of the disk with ground (call that point C) becomes the instant center for the disk. This allows you to relate omega_2 and v_A2. You now have three equations and three unknowns.
4. Solve

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Discussion and hints

Here we have two rigid bodies joined together with a pin joint at O. It is recommended that you make your system BIG, as is the usual suggestion for the work/energy equation. What does this mean? Draw a free body diagram of the system of disk + bar. And, when writing down the kinetic and potential energy for the system include both bodies.

The four-step plan:

1. FBD: Draw a free body diagram of the entire system of disk + body.
2. Work/energy equation: Which force(s) do non-conservative work on the system? Recall that the friction force at a no-slip contact point is workless. Write down the kinetic energy for each body and add together. For the disk, it is recommended that you use the no-slip contact point C for your reference point to simplify. For the bar it is recommended that you use the center of mass (call that point G) for your reference point. Also, write down the potential energy expressions for each and add together. Note that the disk rolls on a horizontal surface.
3. Kinematics: Put some thought into the motion of the system at position 2 when the bar is horizontal. Watch the slow-motion animation of the motion of the system shown in the above GIF - where is the instant center of the bar at the horizontal position? What does this imply about the velocity of point O and the rotation rate of the disk? And, what does it say about the relationship between the velocity of G and the angular velocity of the bar?
4. Solve.

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Discussion and hints

Here we have two bodies joined together with a cable connection. It is recommended that you make your system BIG, as is the usual suggestion for the work/energy equation. What does this mean? Draw a free body diagram of the system of drum + block + cable. And, when writing down the kinetic and potential energy for the system include both bodies.

The four-step plan:

1. FBD: Draw a free body diagram of the entire system of drum + block + cable.
2. Work/energy equation: Which force(s) do non-conservative work on the system? Recall that the friction force at a no-slip contact point (call that point C) is workless. Write down the kinetic energy for each body and add together. For the drum, it is recommended that you use the no-slip contact point C for your reference point to simplify. The block moves in pure translation, so treat it as a particle in terms of kinetic energy.  Also, write down the potential energy expressions for each and add together along with the potential energy of the spring. Note that the drum rolls on a horizontal surface.
3. Kinematics: Note that the speed of block A is the same as the speed of the top point on the inner circumference of the drum (call that point B) where the cable comes off the drum. Also, since C is the instant center of the drum, use that to: i) relate the speed of A to the rotation rate of the drum, and ii) relate the distance traveled by A to the amount of stretch in the spring.
4. Solve.

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Discussion and hints

Note that the disk is WELDED to the bar. Because of this, the rotational motion of the disk is the same as the rotational motion of the bar; that is, the angular velocities for the two rigid bodies are the same.

The four-step plan:

1. FBD: Make your system BIG. Include both the bar and the disk in your free body diagram.
2. Work/energy equation: If you included both the bar and the disk in the same FBD above, then the kinetic energy T should include T from both the bar and the disk. For Part (a), include both the translational and rotational component of T for the disk: 0.5*M*v_B^2 + 0.5*I_B*omega_disk^2. For Part (b) include only the translational component of T for the disk: 0.5*M*v_B^2. Same as for the potential energy V: include that for both the bar and the disk. For the work term, note that the applied force F is perpendicular to the bar for all motion; i.e., F is tangent to the circular path of point A at which F acts.
3. Kinematics.
4. Solve.

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Discussion and hints

As the drum rolls up the incline, the no-slip contact point (call it C) is the instant center for the drum. This plays a critical role in solving this problem.

The four-step plan:

1. FBD: Draw a free body diagram of the drum: normal force, friction force and applied force F.
2. Work/energy equation: Since the friction force and normal force act at the no-slip point C, they do no work since the no-slip point is stationary when these forces act on it. Only the applied force F does work. The work will be the force F times the distance that the end of the rope moves up the incline. You might consider writing the kinetic energy using C as the reference point: T = 0.5*I_C*omega^2. The change in potential energy is just m*g*h, where h is the height change for the CM point O.
3. Kinematics:  You task here is to determine the distance that the end of the rope moves up the incline. Here, consider that point C is the instant center of rotation for the drum.
4. Solve.

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