Question

A billiard ball is hit by a cue at height $h$ above the centre. It acquires a linear velocity $V_0$. Mass of the ball is $m$ and radius is $r$. The angular velocity ${\omega }_0$ acquired by the ball is

• A. $\frac{2V_{0}h}{5r^2}$
• B. $\frac{5V_{0}h}{2r^2}$
• C. $\frac{2V_0{r}^2}{5h}$
• D. $\frac{5V_0{r}^2}{2h}$

Answer 1

The correct option is B $\frac{5V_{0}h}{2r^2}$


Let $J$ be the linear impulse imparted to the ball.
Impulse = Change in momentum
$\Rightarrow J=m{v}_0$
And torque $\left(\tau \right)$ is related to angular momentum $\left(L\right)$ by
$\tau =\frac{dL}{dt}\Rightarrow dL=\tau dt$
And torque is related by Linear momentum $\left(J\right)$ by
$\tau =F\times h=\frac{d\left(m{v}_0\right)}{dt}\times h=h\frac{dJ}{dt}$
Therefore,
$\Rightarrow \tau dt=h\times dJ=dL$
Integrating ,we get
$h\times J=h\times \left(m{v}_0\right)=L=I{\omega }_0$,
Now, moment of inertia of solid ball=$\frac{2m{r}^2{\omega }_0}{5}$
Therefore, $h\times \left(m{v}_0\right)=\frac{2m{r}^2{\omega }_0}{5}$
$\Rightarrow {\omega }_0=\frac{5v_{0}h}{2r^2}$