Question

A point object of mass $m$ moving horizontally hits the uniform solid disc and stick to it. The disc is resting on a horizontal frictionless surface and pivoted at its centre as shown in figure. Find out the $MOI$ of the disc if it rotates with angular velocity $\omega =\frac{v}{5R}$ after the strike.

• A. $I=\frac{m{R}^2}{2}$
• B. $I=m{R}^2$
• C. $I=4m{R}^2$
• D. $I=\frac{3}{2}m{R}^2$

Answer 1

The correct option is D $I=\frac{3}{2}m{R}^2$

Given, mass of point object $=m$
Speed of point object $=v$
Angular momentum of object about disc centre before hitting the disc $=mv\left(R\sin {30}^{\circ }\right)=\frac{mvR}{2}$


Hence, after hitting, angular momentum of the system (disc + particle) $=I^{\prime }=(I+m{R}^2)$
From angular momentum conservation,
$\frac{mvR}{2}=(I+m{R}^2)\omega$
$\Rightarrow \frac{mvR}{2}=(I+m{R}^2)\frac{v}{5R}$
$\Rightarrow \frac{5m{R}^2}{2}-m{R}^2=I$
$\therefore I=\frac{3m{R}^2}{2}$