A disc and a ring both have same mass and radius. The ratio of moment of inertia of the disc about its diameter to that of a ring about a tangent in its plane is

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TestHub · Accepted Answer

Let the mass of the disc and the ring be $M$ and their radius be $R$ .

The moment of inertia of a disc about its diameter is given by:
$I_{1} = \frac{1}{4} M R^{2}$

The moment of inertia of a ring about its diameter is:
$I_{d iam e t er} = \frac{1}{2} M R^{2}$

Using the parallel axis theorem, the moment of inertia of the ring about a tangent in its plane is:
$I_{2} = I_{d iam e t er} + M R^{2} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$

The required ratio is:
$\frac{I _{1}}{I _{2}} = \frac{\frac{1}{4} M R ^{2}}{\frac{3}{2} M R ^{2}} = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6}$

Answer: $1 : 6$

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