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Q.
When a ring of mass 10 kg and diameter 0.4 m is rotated about its axis it makes 2100 rev/min. The angular momentum of the ring will be
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Solution:
The moment of inertia of the ring, $ I=m{{r}^{2}}=10\times {{(0.2)}^{2}}=0.4\,kg-{{m}^{2}} $ Angular frequency ofthe ring, $ \omega =2\pi n=2\pi \times \frac{2100}{60}\,rad\text{/}s $ $ \therefore $ Angular momentum of the ring, $ L=I\omega =\frac{0.4\times 2\pi \times 2100}{60} $ $ =88\,kg\,{{m}^{2}}\text{/}s $