A uniform thin rod of mass m and length l is standing on a smooth horizontal surface. 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

As the floor is smooth, mechanical energy of the rod will remain conserved. Further, no horizontal force acts on the rod, hence the centre of mass moves vertically downwards in a straight line. Thus velocities of COM and the lower end B are in the direction shown in figure. The location of IC at this instant can be found by drawing perpendiculars to
`vec (v)_(C) and vec(v)_(B)` at respective points. Now, the rod may be assumed to be in pure rotational motion about IAOR passing through IC with angular speed ` omega`.
Applying conservation of mechanical energy. Decrease in gravitational potential energy of the rod = increase in rotational kinetic energy about
IAOR
`mgh=(1)/(2) I_(I"AOR")omega^(2) `or
`mg(l)/(2) (1 - sin theta) = (1)/(2) ((ml^(2))/(12) + (ml^(2))/(4) cos^(2) theta) omega^(2)`
Solving this equation, we get
`omega = sqrt((12 g (1 - sin theta))/(l(1 + 3 cos ^(2) theta )))`
Now, `|vec(v)_(c)| = ((l)/(2) cos theta) omega `
`= sqrt(( 3g l (1 - sin theta ) cos^(2) theta)/((1 + 3 cos^(2) theta)))`