Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A rigid body of mass $m$ rotates with angular velocity $\omega$ about an axis at a distance $a$ from the centre of mass $C$. The radius of gyration about $C$ is $K$. Then, kinetic energy of rotation of the body about new parallel axis is

Bihar CECEBihar CECE 2012System of Particles and Rotational Motion

Solution:

MI of body about centre of mass
$I_{ cm }=m k^{2}$
MI of body about new parallel axis
$I_{\text {new }} =I_{ cm}+m a^{2}$
$=m k^{2}+m a^{2}$
$=m\left(K^{2}+a^{2}\right)$
$\therefore $ Kinetic energy, $K=\frac{1}{2} I_{\text {new }} \omega^{2}$
$=\frac{1}{2} m\left(K^{2}+a^{2}\right) \omega^{2}$