System of Particles and Rotational Motion - Result Question 70
77. 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) $\frac{1}{2} m r \omega^{2}$
Topic 5: Rolling Motion
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Answer:
Correct Answer: 77. (a)
Solution:
- (a) Kinetic energy $=\frac{1}{2} I \omega^{2}$
and for ring $I=m r^{2}$
Hence, $K E=\frac{1}{2} m r^{2} \omega^{2}$
