Write the formula for rotational kinetic energy.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The rotational kinetic energy $(K_{rot})$ of a rigid body rotating about a fixed axis is given by the formula:
$K_{rot} = \frac{1}{2} I \omega^2$
Where:
$I$ is the moment of inertia of the body about the axis of rotation.
$\omega$ is the angular velocity of the body.

Explore More

Similar Questions

Two bodies $A$ and $B$ have their moments of inertia $I_1$ and $I_2$ respectively about their axis of rotation. If their kinetic energies of rotation are equal and their angular momenta $L_1$ and $L_2$ respectively are in the ratio $1 : \sqrt{3}$,then $I_2$ will be

$A$ flywheel rotating about a fixed axis has a kinetic energy of $500 \text{ J}$. If its angular frequency is $5 \text{ Hz}$, calculate the moment of inertia of the wheel about the axis of rotation. (Take $\pi^2 = 10$) (in $\text{ kg m}^2$)

$A$ solid sphere of mass $M$ and radius $R$ is rotating about its diameter. $A$ solid cylinder of the same mass and radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($K_{\text{sphere}}$ to $K_{\text{cylinder}}$) will be:

Rotational kinetic energy of a given body about an axis is proportional to

If $L, M$,and $P$ are the angular momentum,mass,and linear momentum of a particle respectively,which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius $R$?

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