A composite rod of mass `2m` and length `2l` comprises two indentica rods joined end to end at `P`. The composite rod hinged at one of its ends is kept horizontal as shown in the figure. If it is realeased from rest.

a. find its angular speed when it becomes vertical.
b. If the lower rod gets detached with the upper rod due to centrifugal effect at their joint `P`, at the vertical position of the composite rod, find their linear and angular velocities just after their separation.

Explanation

a. Let the composite rod acquire an angular speed `omega`at the vertical position. It centre of mass `C` moves from `C_(1)` to `C_(2)`.
Therefore the potential energy (gr) of the composite rod decreases by `2mgh` where `h=l.` Applying conservation of mechanical energy at horizontal and vertical position.
Using `/_\K+/_\K=(2m)gl`

where `l=MI` of the composite rod about `O=(2m)(2l)^(2)/3=8ml^(2)/3`
`implies omega=sqrt(3/2 g/l)`
b. Referring to figure. we can see that just at the vertical position., during the impact, the weights of the component rods `1` and `2`, the raction force `R` at the pivot and the reaction forces `N` at the joint of the rods, pass through the pivot `O`.

Therefore these forces cannot produce anny moment about `O`, that means the rods fo nt experience any horizontal force during breaking, at vertical position before the angular momentum of the system about `O` remains constant just after the breaking we can also argue that the angular momentum of each rod remains constant just before and after breaking. because all these radial forces cannot produce any moment about the centre of mass of the rods `1` and `2` vertical position. Therefore, the linear velocities of the `CM` of the rods remains constant.
`implies omega_(1)=omega`
`v_(1)^(')=v_(1)=omegal/2{"for rod"1}`
`omega_(2)=omega`
`=v_(2)^(')=v_(2)=(3/2)omegal{"for rod" 2}`
`implies omegaO_(1)=omega_(2)=sqrt((3g)/(2l)) ` and `v_(2)=3/2 sqrt(3/2)gl`