$A$ disc of mass $M$ and radius $R$ is moving with an angular velocity $\omega$ as shown in the figure. Find the angular momentum of the disc about the reference point $O$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The center of mass of the disc moves with a linear velocity $v = R\omega$ and rotates about its center of mass with angular velocity $\omega$.
The total angular momentum $L$ about point $O$ is the sum of the angular momentum due to the motion of the center of mass and the angular momentum due to rotation about the center of mass.
$L = L_{cm} + L_{rot}$
$L = (Mv)R + I_{cm}\omega$
Since $v = R\omega$ and $I_{cm} = \frac{1}{2}MR^2$,we have:
$L = M(R\omega)R + (\frac{1}{2}MR^2)\omega$
$L = MR^2\omega + \frac{1}{2}MR^2\omega$
$L = \frac{3}{2}MR^2\omega$

Explore More

Similar Questions

$A$ mass $m$ moves in a circle on a smooth horizontal plane with velocity $v_0$ at a radius $R_0$. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally $m$ moves in a circle of radius $\frac{R_0}{2}$. The final value of the kinetic energy is

Due to global warming, if the ice in the polar region melts and some of this water flows to the equatorial region, then

$A$ uniform circular disc of mass $50 \ kg$ and radius $0.4 \ m$ is rotating with an angular velocity of $10 \ rad \ s^{-1}$ about its own axis,which is vertical. Two uniform circular rings,each of mass $6.25 \ kg$ and radius $0.2 \ m$,are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in $rad \ s^{-1}$) of the system is:

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are gently placed at opposite ends of a diameter of the ring. The new angular velocity of the ring will be

Difficult
View Solution

$A$ thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane with angular velocity $\omega$ about an axis passing through its center and perpendicular to the plane. If another ring of the same size but mass $\frac{M}{4}$ is placed coaxially on the first ring,then the new angular velocity of the system is:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo