System Of Particles And Rotational Motion Ques 75

75. 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]

(a) $\frac{1}{2} m r^{2} \omega^{2}$

(b) $m r \omega^{2}$

(c) $m r^{2} \omega^{2}$

(d) $\frac{1}{2} m r \omega^{2}$

Show Answer

Answer:

Correct Answer: 75.(a)

Solution:

  1. (a) Kinetic energy $=\frac{1}{2} I \omega^{2}$

and for ring $I=m r^{2}$

Hence, $K E=\frac{1}{2} m r^{2} \omega^{2}$

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