Question

A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60? and released. Its angular velocity when it reaches the equilibrium position will be__ 

(A) √(g / 3R) 

(B) √(2g / 3R) 

(C) √(2g / R) 

(D) 2√(2g / R)

Answer 1

Correct option: (B) √(2g / 3R)

Explanation:

From given information distance between axis rotation & centre of mass

disc = R = i.e. x = R

I = I_CM + Mx² 

I = (MR² / 2) + MR² ------ x = R

I = (3/2)MR² 

Gain in KE when disc reaches equilibrium position = (1/2)Iω² 

= (3/4)MR²ω² 

PE at θ = 60°, = mgh(1 – cos 60)

= mgR(1/2)

PE = KE gives

[(MgR) / 2] = (3/4)MR²ω² 

g = (3/2)Rω² 

ω = √(2g / 3R)