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A uniform thin rod AB of mass M and length l attached to a string OA of length `(l)/(2)` is placed on a smooth horizontal plane and rotates with angular velocity omega around a vertical axis through O. A peg P is inserted in the plane in order that on striking it the bar will come exactly to rest A. Location of peg for rod coming to rest is `r=(5l)/(6)`B. Location of peg for rod coming to rest is `r=(3l)/(4)`C. Location of peg for rod coming to rest is `r=(13)/(12)`D. Location of peg for rod coming to rest is `r=(2l)/(3)` |
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Answer» Correct Answer - C `MV_(cm)L^(2)+I_(cm)omega` `MomegaL^(2)+(ML^(2))/(12)omega` `I_(0)=(13)/(12)ML^(2)omega` By impulse momentum theorem `int Ndt =mV` By angular Impulse `rintNdt=mVl+(Ml^(2))/(12)(V)/(L)` `rMV=(13)/(12)MVL` |
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