Question

A circular disc of radius R is removed from a bigger disc of radius 2R. such that the circumferences of the disc coincide. The centre of mass of the remaining portion is αR from the centre of mass of the bigger disc. The value of α is. 

(A) 1/2 

(B) 1/6 

(C) 1/4 

(D) [(– 1) / 3]

Answer 1

Correct option: (D) [(– 1)  / 3]

Explanation:

Let m be mass per unit area

M = mass of big disc = π(2R)²m

M₁ = mass of removed disc = πR²m

M₂ = M – M₁ = Mass of remaining Portion

M₂ = M – M₁ = Mass of remaining Portion

M × O = M₁(OO') + (M – M₁)(OG)

M₁ ∙ R = – (M – M₁) (∝R)

M₁ = – α(M – M₁)

πR²m = – α(π ∙ 4R²m – πR²m)

R² = – α(3R²)

α = [(– 1)  / 3]