Question

A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is

• A. $\frac{1}{2}mr^2\omega^2$
• B. mr$\omega^2$
• C. $mr^2\omega^2$
• D. $\frac{1}{2}mr\omega^2$

Answer 1

Answer: (A)

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

Answer 2

Kinetic energy $= \frac{1}{2}I\omega^2$, and for ring $I=mr^2$
Hence $KE=\frac{1}{2}mr^2\omega^2$