Question

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocityω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω^‘, which is equal to?

(a) mω/(m+M)

(b) mω/(m+2M)

(c) ((m+2m)ω)/m

(d) ((m-2M)ω)/((m+2M))

This question was posed to me in an interview.

My question comes from Significance of Moment of Inertia in portion Laws of Motion of Engineering Physics I

Answer 1

The correct answer is (b) mω/(m+2M)

Explanation: By conservation of angular momentum,

(m+2M)R^2 ω^‘=mR^2 ω

ω^‘=mω/(m+2M).