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Moment of interia of a thin uniform rod rotating about the perpendicular axis passing through its center is I. If the same rod is bent into a ring and its moment of inertia about its diameter is |
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Answer» `8pi^(2)//3` when rod is BENT into a ring then RADIUS of ring is r `therefore 2pir=limpliesr=(l)/(2pi)` Moment of inertia of ring about its diameter is `I.=(1)/(2)mr^(2)=(1)/(2)m(l^(2))/(4pi^(2))` `therefore (I)/(I.)=(1)/(12)ml^(2)//(ml^(2))/(8pi^(2))=(2pi^(2))/(3)` |
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