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Q.

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

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a

$\frac{1}{2} m r^{2} ω^{2}$

b

$m r ω^{2}$

c

$m r^{2} ω^{2}$

d

$\frac{1}{2} m r ω^{2}$

answer is A.

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Detailed Solution

Kinetic energy $= \frac{1}{2} l ω^{2}$

and for ring $I = {mr}^{2}$

Hence $KE = \frac{1}{2} {mr}^{2} ω^{2}$

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