# Thread: Pure rolling motion on snooker table?

1. ## Pure rolling motion on snooker table?

Find the ratio of height h of a cushion on a snooker table to the radius r of a ball as shown in the figure, such that when the ball hits the cushion with a pure rolling motion it rebounds with a pure rolling motion. (Assume that the force exerted on the ball by the cushion is horizontal during the impact and that the ball hits the cushion normally)

2. Originally Posted by fardeen_gen
Find the ratio of height h of a cushion on a snooker table to the radius r of a ball as shown in the figure, such that when the ball hits the cushion with a pure rolling motion it rebounds with a pure rolling motion. (Assume that the force exerted on the ball by the cushion is horizontal during the impact and that the ball hits the cushion normally)

However, both snooker & billiard tables have the cushion slightly higher than the radius of the ball -- the cushion rails do NOT compress symetrically. If a player imparts a high spin hitting the rail the ball will bounce from the table.

With the cushion higher than the radius some energy will be lost since the resulting force will effectively drive the ball into the table.

Spoiler:
$\displaystyle \frac{7}{5}$
. This question is more difficult than it seems.

4. Originally Posted by fardeen_gen
Find the ratio of height h of a cushion on a snooker table to the radius r of a ball as shown in the figure, such that when the ball hits the cushion with a pure rolling motion it rebounds with a pure rolling motion. (Assume that the force exerted on the ball by the cushion is horizontal during the impact and that the ball hits the cushion normally)
The cushion exerts a horizontal impulse I at a height h–r above the centre of the ball. This impulse must reverse both the momentum and the angular momentum of the ball if the pure rolling motion is to be preserved.

For the momentum to be reversed, $\displaystyle I = 2mr\omega$ (m is the mass of the ball, $\displaystyle \omega$ its angular velocity). For the angular momentum to be reversed, $\displaystyle I(h-r) = 2\cdot\tfrac25 mr^2\omega$ (where $\displaystyle \tfrac25 mr^2$ is the moment of inertia of a solid sphere). Put those two equations together to deduce that $\displaystyle h=\tfrac75r$.