EQUILIBRIUM OF RIGID BODIES Mech 65 Interpreting center of mass A T-shaped board
ID: 1790875 • Letter: E
Question
EQUILIBRIUM OF RIGID BODIES Mech 65 Interpreting center of mass A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is through the right hole, of mass of the board. The board is then held in place. 1. A. Pivot Board placed which corresponds to the center Predict the motion of the board after it is released from rest. Explain. 1. Check your prediction by observing the demonstration. a. Describe the angular acceleration of the board. Explain how you can tell 2. What does your answer imply about the net torque about the pivot? Explain. Describe the acceleration of the center of mass of the board. Explain how you can tell. b. What does your answer imply about the net force acting on the board? Explain. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest 3. B. Imagine that the board is now hung from the hole to the Pivot Center of mass left of the center of mass. Predict the subsequent motion of the board after it is released. Explain. 1. 2. On the diagram at right, draw and label an extended free-body diagram for the board just after it is released (ie., for each force, indicate explicitly a single point on the object at which the force can be regarded as acting). Extended free-body diagram for board Explain how the diagram can be used to support your prediction for the motion of the board.Explanation / Answer
1. The net force and net force and net torque about center of mass is zero. So the board will remain at rest when released.
2. (a) Since net torque is zero, so no change in angluar momentum.
(b) Acceleration of center of mass will be zero.
The net force acting on the board will be zero in this case.
3. If the object remains at rest when released, so friction will not change the scenerio, still the net torque and net friction will remain zero.
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