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

Question

Aicharya · Accepted Answer

For disc, $\begin{aligned} \mathrm{I}_{\mathrm{D}} & =\frac{\mathrm{I}_z}{2} \\ & =\frac{\frac{\mathrm{MR}^2}{2}}{2}=\frac{\mathrm{MR}^2}{4} \end{aligned}$ For ring,

by using perpendicular and parallel axis theorem, $\begin{array}{ll} & \mathrm{I}_{\mathrm{R}}=\mathrm{I}_y+M \mathrm{x}^2=\frac{\mathrm{MR}^2}{2}+M R^2=\frac{3}{2} \mathrm{MR}^2 \\ \therefore \quad & \frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{R}}}=\frac{\mathrm{MR}^2}{4} \times \frac{2}{3 \mathrm{MR}^2} \\ \therefore \quad & \frac{I_D}{I_R}=\frac{1}{6} \end{array}$

A disc and a ring both have same mass and radius. The ratio of moment of inertia of the disc about its…

Solution