# Rotational Kinetic Energy - Work-Kinetic Theorem

If a rigid body is rotating about a fixed axis with angular velocity $\omega$ and a force is applied on it to increase its angular speed (and thus rotational kinetic energy), then the relation between the work done by the torque of the force and the change in its kinetic energy is predicted by **work-kinetic theorem for rotation**.

## Work-Kinetic Theorem for Rotation

According to

work-kinetic theorem for rotation, the amount of work done by all the torques acting on a rigid body under a fixed axis rotation (pure rotation) equals the change in its rotational kinetic energy:${W_\text{torque}} = \Delta K{E_\text{rotation}}.$

Work done by a torque can be calculated by taking an analogy from work done by force.
Work done by force is calculated as the dot product of force and displacement of point of application of force.
In case of angular motion, force is replaced by torque and linear displacement is replaced by angular displacement.
Thus,

$W = \int_{}^{} {\vec \tau \cdot d\vec \theta }.$

Consider a rigid body, rotating freely about a fixed axis of rotation. Its initial angular speed is ${\omega_i}$. Suppose a force $F$ is now applied (at a distance of $r$ from the axis of rotation) to increase its angular speed. This force will produce a torque about the axis of rotation: $\vec \tau = \vec r \times \vec F.$ From the rotational form of Newton's second law, ${{\vec \tau }_\text{rot}} = {I_\text{rot}}\vec \alpha .$ Here, ${I_\text{rot}}$ is the moment of the body about the axis of rotation and $\alpha$ is the angular acceleration produced in the body.

Also, the work done by torque equals $W = \int_{}^{} {\tau \cdot d\theta } .$ Angular acceleration changes the angular speed of the body and $\alpha = \frac{{d\omega }}{{dt}}.$ Therefore, the work done by toque can be written as follows: $W = \int_{}^{} {{I_\text{rot}}\frac{{d\omega }}{{dt}}\cdot d\theta }.$ Angular velocity is related to the rate of rotation of the body: $\omega = \frac{{d\theta }}{{dt}}.$

On calculating, one finds $W = \int_{{\omega _o}}^\omega {I_\text{rot} \omega \, d\omega } = \frac{1}{2}{I_\text{rot}}{\omega ^2} - \frac{1}{2}{I_\text{rot}}\omega _{^o}^2.$Thus, the work done by the torque equals the change in rotational kinetic energy of the body.

## A ring, a solid sphere and a thin disc of different masses rotate with the same kinetic energy. Equal constant torques are applied to stop them. Which will make the least number of rotations before coming to rest?

Work done by a constant torque is $W = \tau \theta$ According to the work-kinetic theorem for rotation, work done by torque equals change in rotational kinetic energy

$W = \Delta K{E_\text{rot}}$ $\tau \theta = \Delta K{E_\text{rot}}$ $\theta = \frac{{\Delta K{E_\text{rot}}}}{\tau }$ Since, the change in rotational kinetic energy and torque applied are equal, thus the angle rotated by all the objects are same.

**Cite as:**Rotational Kinetic Energy - Work-Kinetic Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/rotational-kinetic-energy-work-kinetic-theorem/