Question

A disc of mass $4m$ is resting on a horizontal frictionless surface and pivoted at point $\text{O}$ as shown in the figure, perpendicular to its plane. A point object of mass $m$, moving horizontally with velocity $v$, hits the disc and sticks to it. Find the angular velocity of the system after the hit.

• A. $\frac{3v}{22R}$
• B. $\frac{v}{11R}$
• C. $\frac{2v}{R}$
• D. $\frac{v}{6R}$

Answer 1

The correct option is D $\frac{v}{6R}$

Given, mass of disc $=M=4m$
Mass of particle $=m$
Speed of particle $=v$
Initially (before hit) angular momentum
$L_i=mv{R}_{\perp }$
where $R_{\perp }=(R+R\sin {30}^{\circ })$ from point $\text{O}$
Hence, $L_i=mv(R+\frac{R}{2})=\frac{3mvR}{2}$ ... (1)
After hit, angular momentum will be
$L_f=I\omega$
where $I=\frac{M{R}^2}{2}+M{R}^2+m{r}_{\perp }^2$
(using parallel axis theorem)
Here, $r_{\perp }=\sqrt{(R+R\sin {30}^{\circ }{)}^2+(R\cos {30}^{\circ }{)}^2}$
$r_{\perp }=\sqrt{3}R$
Hence, $I=\frac{3}{2}M{R}^2+3m{R}^2$ $(\because M=4m)$
$I=6m{R}^2+3m{R}^2=9m{R}^2$

From angular momentum conservation:
$L_i=L_f$
$\Rightarrow \frac{3mvR}{2}=\left(9m{R}^2\right)\omega$
$\Rightarrow \omega =\frac{v}{6R}$