A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, when it is in vertical position is
√3gL
√2gL
√g2L
√gL
Solution:
The loss in gravitational potential energy of the rod's center of mass is equal to the gain in rotational kinetic energy of the rod.HencemgL2=12Iω2=12(mL23)ω2⟹ω=√3gL