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  4. The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum ( textin kg m2s-1) is approximately:

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Q. The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $ (\text{in kg}\,{{m}^{2}}{{s}^{-1}}) $ is approximately:

EAMCETEAMCET 2006

Solution:

Kinetic energy o rotating body $ K=\frac{1}{2}I{{\omega }^{2}}=62.8\,J $ Angular momentum, $ L=I\omega $ $ =\left( \frac{1}{2}I{{\omega }^{2}} \right)\times \frac{2}{\omega } $ $ =\frac{62.8\times 2}{(2\times \pi \times 300/60)} $ $ =4\,kg\,{{m}^{2}}{{s}^{-1}} $