A rod of length l is held vertically stationary with its lower end located at a point P , on the horizontal plane. When the rod is released to topple about P , the velocity of the upper end of the rod with which it hits the ground is

Step-by-step solution

In this process potential energy of the metre stick will be converted into rotational kinetic energy. PE of metre stick = m g l 2 Because its centre of gravity lies at the middle of the rod. Rotational kinetic energy E= 1 2 I ^2I= moment of inertia of metre stick about point A= m l^2 3 . By the law of conservation of energy aligned m g ( l 2 ) & = 1 2 I ^2 & = 1 2 m l^2 3 ( v_B l )^2 aligned By solving, we get v_B= 3 g l