A uniform thin rod of mass `m` and length `l` is standing on a smooth horizontal suface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of centre of mass of the rod at the instant when it makes an angle `theta` with horizontal.

Explanation

The table is frictionless. The centre of mass of the scale moves vertically downward as the scale falls. Initial kinetic energy K is zero and the potential energy U is `mg l/2`. Total energy of the scale `E = K + U`
`E =0 + mg l/2`
The position of the scale after certain time is shown in Fig. 7.2.54 using the dotted line. Let `theta` be the angle through which the scale has rotated. Let the angular velocity be m and the linear velocity of the centre of mass be v.

`K =1/2 Iomega^(2) + 1/2mv^(2)`
`U = mg(1/2-s)`
Total energy is conserved as there is no friction
`1/2Iomega^(2)+ 1/2mv^(2) + mg(l/2-s) = mg l/2`
`1/2 I omega^(2) + 1/2mv^(2) - mgs =0`........(i)
From Fig. `s=l/2 -l/2 cos theta =l/2(1-cos theta)`
Differentiating `(ds)/(dt) =-l/2 xx sin theta xx (dtheta)/(dt)`
`v=l/2 sin thetaomega, omega = (2v)/(l sin theta)`
Also `I = (ml^(2))/12`
Substituting in eqns (i)
`1/2 xx (ml^(2))/12 xx (4v^(2))/(l^(2) sin^(2) theta) + 1/2 mv^(2) -mgs =0`
`v^(2)/2 [1/(3 sin^(2)theta)+1] =gs, v^(2)[(1+3 sin^(2)theta)/(3 sin^(2)theta)]=2gs`
`v=[(6gs sin^(2)theta)/(1+3 sin^(2)theta)]^(1/2)`