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} MR^{2}\omega^{2} $
• B. $ MR^{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
$I = \frac{1}{2} mR^2$
$|L| = MvR + \frac{1}{2} MR^2\omega$
$ = M ( \omega R) \cdot R + \frac{1}{2} MR^2 \omega$
$ =MR^2 \omega + \frac{1}{2} MR^2 \omega$
$ = \frac{3}{2} MR^2 \omega $
$ = \frac{3}{2} MR^2 \frac{v}{R} $
$= \frac{3}{2} MRv$