$A$ disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

  • A
    $\frac{1}{2} M R^2 \omega$
  • B
    $M R^2 \omega$
  • C
    $\frac{3}{2} M R^2 \omega$
  • D
    $2 M R^2 \omega$

Explore More

Similar Questions

The angular momentum of a moving body remains constant, if

$A$ child is standing at the center of a rotating platform with hands folded. The kinetic energy of the system is $K$. The child now stretches his hands,doubling the moment of inertia. The kinetic energy of the system will now become ........

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about a transverse axis passing through its centre with constant angular velocity $\omega$. Two objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. What is the new angular velocity?

$A$ particle of mass $m$ moves with a velocity $v_0$ in a circle of radius $R_0$ on a smooth horizontal plane. The mass is tied to a string passing through a hole in the smooth plane as shown in the figure. The tension in the string is gradually increased,and finally,the mass $m$ moves in a circle of radius $\frac{R_0}{2}$. What is the ratio of the final kinetic energy to the initial kinetic energy?

Difficult
View Solution

$Assertion$ : For a system of particles under a central force field,the total angular momentum is conserved.
$Reason$ : The torque acting on such a system is zero.

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