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A uniform rod AB of mass .m. length .2a. is allowed to fall under gravity with AB in horizontal. When the speed of the rod is .v., suddenly the end .A. is fixed. Find the angular velocity with which it begins to rotat. |
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Answer» Solution :`L_(1)=L_(2)` `mva = I omega` `mva = (m(2A)^(2))/(3)omega rArr omega = (3V)/(4a)` 
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