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6.System of Particles and Rotational Motion
medium
A ring of radius $0.5\, m$ and mass $10 \,kg$ is rotating about its diameter with an angular velocity of $20 \,rad/s.$ Its kinetic energy is .......... $J$
A
$10$
B
$100$
C
$500$
D
$250$
Solution
Rotational kinetic energy $\frac{1}{2}I{\omega ^2}$
$= \frac{1}{2}\left( {\frac{1}{2}M{R^2}} \right)\,{\omega ^2}$
$ = \frac{1}{2}\left( {\frac{1}{2} \times 10 \times {{(0.5)}^2}} \right)\,{\left( {20} \right)^2}$
$= 250\,J$
Standard 11
Physics
