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Question:

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