A solid sphere of mass M and a disc of mass 2 M have the same radius. The ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent

Question

A solid sphere of mass M and a disc of mass 2 M ​ have the same radius. The ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be

TestHub · Accepted Answer

Let the radius of the solid sphere and the disc be $R$ .

Mass of the solid sphere, $m_{s} = M$

Mass of the disc, $m_{d} = \frac{M}{2}$

Moment of inertia of the disc about its diameter is $\frac{1}{4} m_{d} R^{2}$ . Using the parallel axis theorem, the moment of inertia of the disc about a tangent in its plane is:

$I_{d i sc} = \frac{1}{4} m_{d} R^{2} + m_{d} R^{2} = \frac{5}{4} m_{d} R^{2}$

Substituting $m_{d} = \frac{M}{2}$ :

$I_{d i sc} = \frac{5}{4} (\frac{M}{2}) R^{2} = \frac{5}{8} M R^{2}$

Moment of inertia of the solid sphere about its diameter is $\frac{2}{5} m_{s} R^{2}$ . Using the parallel axis theorem, the moment of inertia of the solid sphere about its tangent is:

$I_{s p h ere} = \frac{2}{5} m_{s} R^{2} + m_{s} R^{2} = \frac{7}{5} m_{s} R^{2}$

Substituting $m_{s} = M$ :

$I_{s p h ere} = \frac{7}{5} M R^{2}$

The required ratio is:

$\frac{I _{d i sc}}{I _{s p h ere}} = \frac{\frac{5}{8} M R ^{2}}{\frac{7}{5} M R ^{2}} = \frac{5}{8} \times \frac{5}{7} = \frac{25}{56}$

Answer: $25 : 56$

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