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The moment of inertia of a body moving with angular velocity `omega` decreases from `l` to `(l)/(3)` without any external torque. Calculate the new angular velocity of the body. |
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Answer» `I_(1)omega_(1)=I_(2)omega_(2)` `omega_(2)=(I_(1)omega_(1))/(I_(2))=(Iomega)/((I)/(3)))=3omgea` Hint : `I_(1)omega_(1)=I_(2)omega_(2)` |
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