Question

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.

Answer 1

The velocity of centre of mass is  |Vc| = √ 3gl ( 1 - sinθ) cos^2θ / ( 1 + 3cos^2θ)

Explanation:

• Applying conservation of mechanical energy.
• Decrease in gravitational potential energy of the rod = increase in rotational kinetic energy about IC

m.g.h =  1  / 2 I ICω^2

or m g (l) /  2

 (1 - sinθ) =  1  / 2  ( ml^2 /  12  +  ml^2  4  cos^2θ)ω^2

Solving this equation we get

ω = √ 12g (1 - sinθ) / l (1 + 3 cos^2θ)

Now |Vc| = ( 1/2 cos θ) ω

|Vc| = √ 3gl ( 1 - sinθ) cos^2θ / ( 1 + 3cos^2θ)

Hence the velocity of centre of mass is  |Vc| = √ 3gl ( 1 - sinθ) cos^2θ / ( 1 + 3cos^2θ)