Comparison Of Moments Of Inertia

Question

A thin ring and a flat disc are made to have exactly equal masses M and equal radii R. What is the ratio of the moment of inertia of the ring to that of the disc, each taken about its own central axis?

RankGuru · Accepted Answer

Moment of inertia depends on how far the mass sits from the axis, so two objects of identical mass and radius can still differ markedly. For a thin ring about its central axis all the mass lies at radius R, giving $I_{ring} = MR^2$. For a uniform disc the mass is spread continuously from the centre to the rim, giving the smaller value $I_{disc} = \frac{1}{2}MR^2$. The required ratio is therefore $\frac{I_{ring}}{I_{disc}} = \frac{MR^2}{\frac{1}{2}MR^2} = 2$, that is $2 : 1$. The ratio 1 : 2 simply inverts the correct answer. The ratio 1 : 1 wrongly assumes that equal mass and equal radius force equal inertia, ignoring how mass is distributed. The ratio 4 : 1 squares the factor incorrectly. These are the standard NCERT results for a ring and a disc. As a conceptual check, since all the ring's mass sits at the maximum radius while the disc has much mass close to the axis, the ring must have the larger inertia, consistent with the 2 : 1 ratio.

Comparison Of Moments Of Inertia

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