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) Suppose that the composite rod acquires an angular speed `omega` when it reaches its vertical position. Its centre of mass G moves from `G_(1)"to" G_(2)`. Therefore the potential energy of the composite rod decreases by 2mgh where h = l.
Applying conservation of energy, `(1)/(2) I omega^(2) = (2m) gl`
where I = M.I. of the composite rod about O
`= (2m) (2l)^(2)//3 = 8m l^(2)//3`
`rArr omega = sqrt((3)/(2)(g)/(l))`

(b) Referring to fig. we can see that, just at the vertical position, during the break, the weights of the component rods 1 & 2, the reaction force R at the pivot & the reaction forces N at the joint of the rods, pass through the pivot O. Therefore these forces can not produce any moment about O, that means the rods do not experience any horizontal force and torque during breaking. At vertical position, the angular momentum of the system about O remains same as just after the breaking.
We can also argue that, the angular momentum of each rod remains constant just before & after breaking. The radial forces cannot produce any moment about the centre of mass of the rods 1 & 2 in the vertical position. The linear velocity of the c.m. of the rods remains constant.
`rArr" "omega_(1) = omega`
`v._(1) = v_(1) = omega l//2" "{"for rod 1"}`
`omega_(2) = omega, v._(2) = v_(2) = (3//2) omega l" "{"for rod 2"}`
`rArr" "omega_(1) = omega_(2) = sqrt((3g)/(2l)), v_(1) = (1)/(2) sqrt((3gl)/(2)) gl and v_(2) = (3)/(2) sqrt((3gl)/(2))`