Q. A rigid body of mass $m$ rotates with the angular velocity $\omega$ about an axis at a distance $'a'$ from the centre of mass $G$. The radius of gyration about $G$ is $K$. Then kinetic energy of rotation of the body about new parallel axis is
A) $\frac{1}{2}mK^{2}\omega^{2}$
B) $\frac{1}{2}ma^{2}\omega^{2}$
C) $\frac{1}{2}m(a^{2} + K^{2})\omega^{2}$
D) $\frac{1}{2}m(a + K^{2})\omega^{2}$
Explanation
(c) M.I. of body about centre of mass $= I_{cm} = mK^{2}$
M.I. of a body about new parallel axis
$I_{new} = I_{cm} + ma^{2} = mK^{2} + ma^{2}$
$I_{new} = m(K^{2} + a^{2})$
$K_{R} = \frac{1}{2}I_{new}\omega^{2} = \frac{1}{2}m\left( K^{2} + a^{2} \right)\omega^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.