JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions

The moment of inertia of the rod about O is

1

3

ml2. The kinetic energy of the rod at position A=

1

2

Iω2 where I is the moment of inertia of the rod about O. When the rod is in position B, its angular velocity is zero. In this case, the energy of the rod is mgh where h is the maximum height to which the centre of mass (C.M) rises
Gain in potential energy = Loss in kinetic energy
∴mgh=

1

2

Iω2=

1

2

(

1

3

ml2)ω2
⇒h=

l2ω2

6g