Question

A thin uniform metal rod of mass 'm' and length ‘l’ standing vertically is now allowed to fall due to its own weight. The velocity with which its mid-point will strike ground will be : 

(1) √5/2 gl

(2) √5 gl

(3) √3 gl

(4) √3/4 gl

Answer 1

Explain: Loss of gravitational =  P.E mg l/2

Gain in rotational = KE = 1/2 lW²

Now l = 1/3 ml²

So, mg 1/2 = 1/2 x 1/2 ml² x W² ⇒ w = √3g/l

So, the velocity of mid point  ⇒v = w l/2

=√3g/l. l/2

= √3gl/4.