Kinetic Energy of a Rotating Ring – Rankers Physics

Kinetic Energy of a Rotating Ring

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: (1988)

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

Solution:

For a ring rotating about its central perpendicular axis, the moment of inertia is $I = mr^2$. The rotational kinetic energy is defined as $K = \frac{1}{2} I \omega^2$. Substituting $I$, we get $K = \frac{1}{2} mr^2 \omega^2$.

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