In the shown figure a mass `m` slides down the frictionless surface from height `h` and collides with the uniform vertical rod of length `L` and mass `M` after collision the mass `m` sticks to the rod. The rod is face to rotate in a vertical plane about fixed axis through `O`. Find the maximum angular deflection of the rod from its initial position.

Explanation

Just before collision, velocity of the mass `m` is along the horizontal and is equal to `v_(0)=sqrt(2gh)`
In the process of collision only angular momentum of the system will be conserved about the point `O`.
If `L_(1)` and `L_(2)` are the angular momentum of the system just before and just after the collision then
`L_(1)=mv_(0)L`
and `L_(2)=Iomega=((ML^(2))/3+mL^(2))omega`
From `COAM`
`(M/3+m)L^(2)omega=mv_(0)L`
`omega=(mv_(0))/((M/3+m)L)`

Let the rod deflects through an angle theta
Initial energy of rod and mass system `=(1/2)Iomega^(2)`
where `I=((ML^(2))/2+mL^(2))`
Gain in potential energy of the system
`=mgL[1-costheta]+MgL/2[1-costheta]`
`=(m+M/2)gL(1-costheta)`
`:'` From conservation of energy
`1/2Iomega^(2)=(m+M/2)gL(1-costheta)`
`1/2((ML^(2))/3+mL^(2))xx(m^(2)v_(0)^(2))/((M/3+m)^(2)L^(2))=(m+M/2)gL(1-costheta)`
`1/2(m^(2)v_(0)^(2))/([M/3+m])=(m=M/2)gL(1-costheta)`
`costheta=1-1/2(m^(2)v_(0)^(2))/([M/3+m][M/2+m]gL)`