Interactive Lecture Demonstration 14 – Rotating Board

Consider the situation depicted in the diagram above. A board of uniform density, mass M and length L is hinged to a table at one end. The variable r represents the distance from the hinge to points along the board. A small ball sits on a stand at the end of the board. A small cup sits at some distance closer to the hinge. The board is lifted to an arbitrary angle q and then released. The purpose of this problem is to show that the vertical acceleration of the end of the board can exceed g, so that the ball can lose contact with the end of the board. If the cup is properly placed, the ball will fall into the cup.

1. The board undergoes angular acceleration due to the application of a net torque. The force responsible for this torque is the board's own weight. The weight can be taken to act on the center of mass of the board as if all the mass were concentrated at that point. What is the lever arm of this force in terms of L and q?

2. What is the torque acting on the board in terms of L, q, M, and g? Check your expression by seeing whether the torque is maximum at q = 0. (Why should it be?)

3. The angular acceleration of the board is given by a = t_net/I, where I is the rotational inertia of the board. Use your result from #2 and a formula for I that you look up in order to obtain an expression for a in terms of L, q, and g. (Assume that the board can be approximated by a thin rod.) Check your expression for the correct units.

4. For a given q, a is the same for all points on the board. The tangential acceleration, however, will depend on r, the distance from the axis of rotation. Let's focus now on the end of the board. Find a_t for that part of the board in terms of L, g, and q.

5. The tangential acceleration of any part of the board has both a horizontal and a vertical component. Find an expression for the vertical component of the tangential acceleration of the end of the board, again in terms of L, g, and q. You will need to draw a triangle to find the components of the tangential acceleration. Note that every part of the rotating board will also have a centripetal acceleration that will have its own vertical component. The centripetal acceleration will be zero, though, when the board is just released (why ?).

6. Now let's examine your result from #5. If we want to have the ball fall into the cup, the ball is going to have to lose contact with the board. For that to happen, the end of the board where the ball is located must accelerate downward (i.e. have a vertical component of its acceleration) greater than g. This will only be possible for a certain range of q. Use your answer to #5 to find the maximum value of q for which the end of the board will have a vertical component of its tangential acceleration greater than g. Why should your answer be a maximum (not a minimum) ? 

7. One more thing: if the ball is going to fall into the cup, the cup will have to reach the position on the table directly underneath where the ball starts falling, before the ball reaches that point. Assuming that the end of the board has a greater vertical acceleration than the ball does, the end of the board will reach the table before the ball does. Since the board is a rigid object, the cup will therefore also reach the tabletop before the ball does. In order for the ball to go into the cup, though, the cup must be located at the correct position along the board. Find an expression for that position in terms of the length L of the board and the angle q at which the board is released.