Question

Moment of inertia of a ring of radius R and mass M about an axis parallel to horizontal diameter and situated at distance of R/2 from axis of rotation is:

(a) MR²

(b) \(\frac{1}{2}\)MR²

(c) 2MR²

(d) \(\frac{3}{4}\)MR²

Answer 1

Correct option is : (d) \(\frac{3}{4}\)MR²

From theorem of parallel axes:

I = I_cm + mh² = \(\frac{1}{2}MR^2 + M(\frac{R}{2})^2\)

\( = \frac{1}{2}MR^2 + \frac{MR^2}{4} = \frac{3}{4}MR^2\)