Question

A rigid body of mass m rotates with angular velocity w about a distance a from the centre a mass G. The radius of gyration about a parallel axis through G is k. The Kinetic energy of rotation of the body is

Answer 1

Consider a right body consisting of n particles rotating about xy with uniform angular velocity.

Let $r_1,r_2$ be the distances of particles from axis of rotation consider $i^{th}$ particle of mass $m_i$ at distance $r_i$ from axis of rotation

$\therefore$ K.E of particle due to rotation about xy axis is

$k_1=\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_i{r}_i^2{\omega }^2$

Total kinetic energy of rotation of the rigid body about xy is

$k=\sum k_i$

$=\sum \frac{1}{2}{m}_i{r}_i^2{\omega }^2$

$=\frac{1}{2}{\omega }_i^2\sum m_i{r}_i^2$

$=\frac{1}{2}2{\omega }^2$

$\therefore K.E=\frac{1}{2}I{\omega }^2$.