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If I_(1) is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass, and I_(2) is the moment of inertia (about central axis) of the ring formed by bending the rod, then |
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Answer» `I_(1):I_(2)=1:1` and `I_(2)=m[(L)/(2pi)]^(2)""(because 2pir=l)` `=(ml^(2))/(4pi^(2))""...(ii) [r=l//2pi]` `therefore (I_(1))/(I_(2))=(ml^(2))/(12)xx(4pi^(2))/(ml^(2))=(pi^(2))/(3)` |
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