Question

A uniform rod of mass m is hinged at its upper end. It is released from a horizontal position. When it reaches the vertical position, what force does it exert on the hinge ? 

(a) 1.5mg 

(b) 2mg 

(c) 2.5mg 

(d) 3.5mg

Answer 1

Correct Answer is: (c) 2.5mg

Loss in PE = 1/2 mg l/2 = gain in KE = 1/2 Iω²

or mgl = (ml2/3) ω² or ω² = 3g/l.

N - mg = mv²/l/2 = 2m/l (ω l/2)² = 1/2 ml (3g/l)

or = 5/2 mg.