Question 594: Centre of Mass and Rotational Motion

Question

A solid sphere of mass $$\mathrm{M}$$, radius $$\mathrm{R}$$ has moment of inertia '$$\mathrm{I}$$' about its diameter. It is recast into a disc of thickness 't' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains 'I'. Radius of the disc will be

Accepted Answer

Moment of inertia of the solid sphere I = $$\frac{2}{5}$$MR$$^2$$
Moment of inertia of the disc about an axis passing through its edge and perpendicular to the plane is given by
$$\begin{aligned} & \mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2} \ & \mathrm{I}^{\prime}=\mathrm{I} \quad 	herefore \frac{3}{2} \mathrm{MR}^{\prime 2}=\frac{2}{5} \mathrm{MR}^2 \ & 	herefore \quad \mathrm{R}^{\prime}=\frac{4}{15} \mathrm{R}^2 \quad 	herefore \mathrm{R}^{\prime}=\frac{2}{\sqrt{15}} \mathrm{R} \ & \end{aligned}$$

Answer

$$\frac{4 R}{\sqrt{11}}$$

Answer

$$\frac{3 R}{4}$$

Answer

$$\frac{2 R}{\sqrt{15}}$$

Answer

$$\frac{2 R}{3}$$

Centre of Mass and Rotational Motion

A solid sphere of mass $\mathrm{M}, radius \mathrm{R} has moment of inertia '\mathrm{I}$' about its diameter. It is recast into a disc of thickness 't' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains 'I'. Radius of the disc will be

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