$A$ thin uniform rod of mass $m$ and length $l$ is suspended from one end,which can oscillate in a vertical plane about the point of suspension. It is pulled to one side and then released. It passes through the equilibrium position with angular speed $\omega$. The kinetic energy while passing through the mean position is

  • A
    $m l^2 \omega^2$
  • B
    $\frac{m l^2 \omega^2}{4}$
  • C
    $\frac{m l^2 \omega^2}{6}$
  • D
    $\frac{m l^2 \omega^2}{12}$

Explore More

Similar Questions

The rotational kinetic energy and translational kinetic energy of a rolling body are the same. The body is:

If the moment of inertia of an object is $I$ and its angular velocity is $\omega$,then its rotational kinetic energy will be:

An energy of $484\,J$ is spent in increasing the speed of a flywheel from $60\,rpm$ to $360\,rpm$. The moment of inertia of the flywheel is $.............\,kg\cdot m^2$.

Two bodies have their moments of inertia $I$ and $2I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal,their angular momentum will be in the ratio

If the angular momentum of a rotating body is increased by $200\ \%$,then its kinetic energy of rotation will be increased by .......... $\%$

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