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 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
Updated On: Jun 11, 2026
- $\frac{1}{2}mr^2\omega^2$
- mr$\omega^2$
- $mr^2\omega^2$
- $\frac{1}{2}mr\omega^2$
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The Correct Option is A
Solution and Explanation
Kinetic energy $= \frac{1}{2}I\omega^2$, and for ring $ I=mr^2$
Hence $KE=\frac{1}{2}mr^2\omega^2$
Hence $KE=\frac{1}{2}mr^2\omega^2$
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