Question

a rod of length l is hinged at point P vertically standing upward due to a slight push rod starts to come down. the angular velocity of rod after falling angle theta is

A. √6g/l sin(theta/2)
B. √6g/l cos(theta/2)
C. √(6gcostheta)/l
D. √3g(1+costheta)/l​

Answer 1

The angular velocity of rod after falling angle theta is √6/gl sinθ/2

Length of the rod = l (Given)

Point of hang = P (Given)

As tge reaction force at the height is idle, the total mechanical energy is thus conserved.

ΔPE = mgl/2(1−cosθ)

ΔK =1/2 I0 ×w²

= 1/2 × ml²/3 × w²

= 1/2 × ml²/3w²=mgl. sin²θ/2

w = √6/gl sinθ/2

Therefore, the angular velocity is √6/gl sinθ/2.