A billiard ball, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance $h$ above the centerline as in the figure below. The ball leaves the cue with a speed $V$ and, because of its "forward English," eventually acquires a final speed of $9V/7$. Show that $h = 4R/5$, where $R$ is the radius of the ball.

I've attempted the problem by using the conservation of angular momentum about the point of contact, but I can't seem to get the right answer.

To be more clear, I'm having trouble figuring out how I can apply the conservation of angular momentum (if applicable).

## Best Answer

Assume the cue acts with a force $F$ on the ball and that the friction is negligible during this short time interval $\Delta t$.

The translational accerelation of the ball is given by $F = ma$. The force $F$ also produces a torque $Fh = I\alpha$ on the point of contact.

Since the ball is initially at rest, we have $$mhv = Fh\Delta t = I\alpha\Delta t = I\omega_0$$ where $\omega_0$ is the angular velocity immediately after the cue hits the ball.

The ball is now rolling while slipping forward (hence the "forward English") until the friction with the ground slows down the rotation so that the ball rolls without slipping, i.e. until the angular velocity is $\frac{9v}{7R}$.

As in this answer we conclude that total angular momentum is still conserved so $$mRv+mhv = mRv+I\omega_0 = mR\left(\frac97v\right)+I\left(\frac{9v}{7R}\right)$$

which after using $I = \frac25mR^2$ yields $h = \frac45R$.