A uniform rod AB of mass 3m and length 4l, which is free to turn in a vartical plane about a smooth horizontal axis through A, is released from rest when horizontal When the rod first impulse exerted by the peg on the rod it.
(a). The rod is brought to rest by the peg.
(b). The rod rebounds and next comes to instantaneous rest inclined to the downward vertical at an angle `(pi)/(3)` radian.

Explanation

(a). From conservation of mechanical energy
`(3m)(g)(2l)=(1)/(2)Iomega^(2)`
`=(1)/(2)[((3m)(4l)^(2))/(3)]omega^(2)=8ml^(2)omega^(2)`
`thereforeomega=((1)/(2)sqrt((3g)/(l)))`
`thereforeJ=(8)/(3)mlsqrt((3g)/(l))`
or `J=(8)/(3)msqrt(3gl)`
(b). let `omega'` be the angular speed in opposite direction again applying conservation mechanical energy
`(3m)(g)(l)=(1)/(2)I(omega')^(2)=8ml^(2)(omega')^(2)`
`thereforeomega'=(1)/(2sqrt(2))sqrt((3g)/(l))`
Now, applying angular impulse `=` change in angular momentum
`thereforeJ(3l)=l(omega+omega')=(16ml^(2))(1)/(2)sqrt((3g)/(l))(1+(1)/(sqrt(2))`
`thereforeJ=(4)/(3)msqrt(6gl)(sqrt(2)+1)`