A disc of mass M and radius R is rolling with angular speed ω 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)

Question

A disc of mass M and radius R is rolling with angular speed ω 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) (3/2) M R2 ω2 (B) M R2 ω (C) M R v (D) (3/2) M R v

Tardigrade Admin · Accepted Answer

Since the disc is rolling, so it has both translational
and rotational motions.

∴

The magnitude of angular momentum of the disc about the origin

O

is

L = L_{t} + L_{r} = MR v + I_{CM} ω = MR v + \frac{1}{2} M R^{2} ω

= MR v + \frac{1}{2} MR (R ω) = MR v + \frac{1}{2} MR v = \frac{3}{2} MR v (

as

v = R ω)

Q. A disc of mass $M$ and radius $R$ is rolling with angular speed $ω$ 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)

$\frac{3}{2} M R^{2} ω^{2}$

$M R^{2} ω$

$MR v$

$\frac{3}{2} MR v$

Q. A disc of mass M and radius R is rolling with angular speed ω 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)

23​MR2ω2

MR2ω

MRv

23​MRv

Since the disc is rolling, so it has both translational and rotational motions. ∴ The magnitude of angular momentum of the disc about the origin O is L=Lt​+Lr​=MRv+ICM​ω=MRv+21​MR2ω =MRv+21​MR(Rω)=MRv+21​MRv=23​MRv( as v=Rω)

Q. A disc of mass $M$ and radius $R$ is rolling with angular speed $ω$ 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)

$\frac{3}{2} M R^{2} ω^{2}$

$M R^{2} ω$

$MR v$

$\frac{3}{2} MR v$