Question

A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal surface as shown in figure. The magnitude of angular momentum of the disc about the origin $O$ is (here $v$ is the linear velocity of the disc)

• A. $\frac{3}{2}M{R}^2{\omega }^2$
• B. $M{R}^2\omega$
• C. $MRv$
• D. $\frac{3}{2}MRv$

Answer 1

Answer: (D)

The angular momentum about the origin
$=L_{\text{translational}}+L_{\text{Rotational}}$
where, $v=$ linear velocity
$I=\frac{1}{2}m{R}^2$
$|L|=MvR+\frac{1}{2}M{R}^2\omega$
$=M(\omega R)\cdot R+\frac{1}{2}M{R}^2\omega$
$=M{R}^2\omega +\frac{1}{2}M{R}^2\omega$
$=\frac{3}{2}M{R}^2\omega$
$=\frac{3}{2}M{R}^2\frac{v}{R}$
$=\frac{3}{2}MRv$