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

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 directions shown in figure. The location of IC at this instant can be found by drawing perpendiculars to `V_(C)` and `V_(B)` at respective points. Now the rod may be assumed to be in pure rotational motion about 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 IC
`thereforemgh=(1)/(2)I_(IC)omega^(2)`
or `mg(l)/(2)(1-sintheta)=(1)/(2)((ml^(2))/(12)+(ml^(2))/(4)cos^(2)theta)omega^(2)`
Solving this equation we get
`omega=sqrt((12g(1-sintheta))/(l(1+3cos^(2)theta)))`
Now. `|V_(C)|=((l)/(2)costheta)omega`
`=sqrt((3gl(1-sintheta)cos^(2)theta)/((1+3cos^(2)theta))`