Question

From a uniform circular disc of radius R, a circular disc of radius R/6 and having centre at a distance + R/2 from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc.

(A) [(– R) / 70]

(B) [(+ R) / 70]

(C) [(– R) / 7]

(D) [(+ R) / 7]

Answer 1

Correct option: (A) [(– R) / 70]

Explanation:

Let mass per unit area of disc = m

Mass of disc = M = πR2 ∙ m

Mass of removed disc = M' = π(R / 6)2 ∙ m = [(πR2m) / 36] from figure

M × 0 = M' × (R / 2) + (M – M') x --- here point P is new centre of mass at distance x

O = [(M'R) / 2] + (M – M')x

x(M – M') = [(– M'R) / 2]

x = [(– M'R) / {2(M – M')]

= [{{(– πR2m) / (36)} ∙ R} / {2m[πR2 – {(πR2) / (36)}}]

= [{(– πR2mR) / {{(36 × 2m × {(35πR2) / (36)}}]

= [(– R) / (70)]

hence centre of mass [(– R) / 70]