A rod of length L and mass M is hinged at point O. A small bullet of mass m ihits the rod with velocity v as shown in the figure. The bullet gets embedded in the rod. Calculate the angular velocity of the system just after the impact?

Explanation

Before impact, the bullet is moving. The initial impact, angular momentum of the system about O is
`L_(i)=mvxxL=mvL`……..1
After the bullet gets embedded in the rod, suppose the system attains angular velocity `omega`.
The moment of inerita of the bullet rod system about the axis through O is,
`I=` (M.I of bullet + M.I of rod)
`=mL^(2)+1/3ML^(2)`
`I=((M+3m)/3)L^(2)`
Final angular momentum of the system is
`L_(f)=Iomega=((M+3m)/3)L^(2)omega`..2
By conservation of angular momentum,
`L_(f)=L_(i)`
`((M+3m)/3)L^(2)omega=mvL`
`:.` Angular velocity
`omega=(3mv)/((M+3m)L)`