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

Options

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)`

Answer

C

Explanation

`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`