Question

A disc is freely rotating with an angular speed $\omega$ on a smooth horizontal plane. If it is hooked at a rigid peg P and rotates without bouncing about P, its angular speed after the impact will be equal to

• A. $\frac{2\omega }{3}$
• B. $\frac{\omega }{3}$
• C. $\frac{\omega }{2}$
• D. none of these

Answer 1

The correct option is B $\frac{\omega }{3}$

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.
$\Rightarrow \frac{1}{2}m{r}^2\omega =\frac{3}{2}m{r}^2{\omega }^{\prime }\Rightarrow {\omega }^{\prime }=\frac{1}{3}\omega$