Question 20.18: In a party game a flat board in the shape of a space ship is...
In a party game a flat board in the shape of a space ship is made to spin round a fixed vertical axis when hit by a small ‘meteorite’ of mass m. The board is always stationary and facing the player when a meteorite is thrown. Meteorites, which can stick to the board, are not removed between throws. The moment of inertia of the board about its axis is I.
i) Assuming that a meteorite is thrown at the empty board and sticks after hitting it at right angles with speed u at a distance d from the axis, find the initial angular velocity of the board.
ii) The board slows down under the action of a constant frictional couple C and the winner in the game is the person who makes it spin for the longest time. Is this a fair game?
iii) Someone suggests that it would be easier to determine the winner by counting the number of revolutions made by the board before coming to rest. Would the game be fair in this case?
i) During the short period of time taken by the impact, the impulse of the frictional couple is negligible, so the change in total angular momentum about the axis is negligible. In other words, total angular momentum is conserved. You can compare this with conservation of linear momentum when two objects collide.
The angular momentum of the meteorite about the axis just before it hits the board is mu × d. The new moment of inertia of the meteorite and board about the axis after the collision is I + md², so the angular momentum after the collision is (I + md²)ω,
where ω is the angular speed. Hence
(I + md²)ω = mud ⇒ ω = \frac{mud}{(I + md^{2})}. ①
ii) The impulse of the constant frictional couple in time t is Ct and this is equal to the loss of angular momentum of the board and meteorite during that time. When t is the time for the board to stop rotating:
Ct = mud ⇒ t = \frac{mud}{C}.
This depends only on the speed of the meteorite and the position where it hits the board, and is independent of the moment of inertia of the board when it is hit. So it does not matter how many other meteorites are there already. Provided there is sufficient room on the board, the game is fair.
iii) Suppose that In is the moment of inertia of the board about its axis when there are n meteorites attached to it. After another hit this will become I_{n} + md^{2}. The initial angular speed after this additional meteorite hits the space ship is
ω = \frac{mud}{(I_{n} + md^{2})} (see ① ).
When the board has turned through an angle θ, the work done against the couple is Cθ and this is equal to the loss in kinetic energy. Hence
Cθ = \frac{1}{2}(I_{n} + md^{2})ω² ⇒ θ = \frac{(mud)^{2}}{2C(I_{n} + md^{2})}.
This is dependent on the value of I_{n} so it will be more difficult to make the board turn through a given angle as the game progresses. In this case, the game is not fair.