Question

A disc is freely rotating with an angular speed $\omega$ on a smooth horizontal plane. If it is hooked at a rigid point $P$(near to its circumference) and rotates without bouncing about $P$. Its angular speed after the impact will be equal to $\frac{\omega }{n},$ value of $n$ is

Answer 1

During the impact, the impact forces pass through point $P$.
Therefore, the torque produced by it about $P$ is equal to zero.
Consequently, the angular momentum of the disc about $P$, just before and after the impact, remains the same
$\Rightarrow L_2=L_1$ ... (i)
Where $L_1=$ angular momentum of the disc about $P$ just before the impact
$I_0\omega =\frac{1}{2}m{r}^2\omega$
$L_2=$angular momentum of the disc about $P$ just after the impact
$I_0\omega =(\frac{1}{2}m{r}^2+m{r}^2){\omega }^{\prime }=\frac{3}{2}m{r}^2{\omega }^{\prime }$
Just before the impact, the disc rotates about $O$. But just after the impact, the disc rotates about $P$.
$\Rightarrow \frac{1}{2}m{r}^2\omega =\frac{3}{2}m{r}^2{\omega }^{\prime }\Rightarrow {\omega }^{\prime }=\frac{1}{3}\omega$