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. $M R v$
• D. $\frac{3}{2} M R v$

Answer 1

Answer: (D)

Since the disc is rolling, so it has both translational
and rotational motions.
$\therefore $ The magnitude of angular momentum of the disc about the origin $O$ is
$L=L_{t}+L_{r}=M R v+I_{ CM } \omega=M R v+\frac{1}{2} M R^{2} \omega$
$=M R v+\frac{1}{2} M R(R \omega)=M R v+\frac{1}{2} M R v=\frac{3}{2} M R v($ as $v=R \omega)$