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)

Options

A. `(3)/(2) MR^(2) omega^(2)`
B. `MR^(2) omega`
C. `MRv`
D. `(3)/(2) MRv`

Answer

D

Explanation

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) = MRv + I_(CM) omega = MRv + (1)/(2) MR^(2) omega`
`= MRv + (1)/(2) MR (R omega) = MRv + (1)/(2) MRv = (3)/(2) MRv (as v = R omega)`