Question

A solid sphere of mass $M$, radius $R$ and having moment of inertia about an axis passing through the center of mass as $I$, 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 $I$. Then, radius of the disc will be:
A. $\dfrac{2R}{\sqrt{15}}$
B. $R\sqrt{\dfrac{2}{15}}$
C. $\dfrac{4R}{\sqrt{15}}$
D. $\dfrac{R}{4}$

Answer 1

Hint: Since, the disc is made by recasting the solid sphere, the same amount of material must have been used and hence the masses of both the disc and the solid sphere will be the same. We can find out the radius of the disc in terms of the radius of the sphere by equating their moment of inertia about the respective axes as given in the question.

Formula used:
For a solid sphere of mass $M$ and radius $R$, the moment of inertia $I$ about an axis passing through its center is given by
$I=\dfrac{2}{5}M{{R}^{2}}$
For a solid disc of mass $M$ and radius $R$, the moment of inertia $I$ about an axis passing through its edge and perpendicular to its plane is given by
$I=\dfrac{3}{2}M{{R}^{2}}$

Complete step-by-step answer:
As explained in the hint, the mass of the solid sphere and the disc will be the same since they are made of the same amount of single material. Let the radius of the disc be ${{R}_{disc}}$ Therefore, the mass of the disc is $M$.
The radius of the solid sphere is $R$. The mass of the sphere is $M$.
For a solid sphere of mass $M$ and radius $R$, the moment of inertia ${{I}_{sphere}}$ about an axis passing through its center is given by ${{I}_{sphere}}=\dfrac{2}{5}M{{R}^{2}}$ --(1)
For a solid disc of mass $M$ and radius ${{R}_{disc}}$, the moment of inertia ${{I}_{disc}}$ about an axis passing through its edge and perpendicular to its plane is given by ${{I}_{disc}}=\dfrac{3}{2}M{{R}_{disc}}^{2}$ --(2)
Now, according to the question, ${{I}_{sphere}}={{I}_{disc}}$
Therefore, equating (1) and (2), we get,
$\dfrac{2}{5}M{{R}^{2}}=\dfrac{3}{2}M{{R}_{disc}}^{2}$ $\therefore {{R}_{disc}}^{2}=\dfrac{2}{5}{{R}^{2}}\times \dfrac{2}{3}=\dfrac{4}{15}{{R}^{2}}$
Square rooting both sides, we get, $\sqrt{{{R}_{disc}}^{2}}=\sqrt{\dfrac{4}{15}{{R}^{2}}}$ $\therefore {{R}_{disc}}=R\sqrt{\dfrac{4}{15}}=\dfrac{2R}{\sqrt{15}}$
Hence, the required radius of the disc is $\dfrac{2R}{\sqrt{15}}$. Therefore, the correct option is A) $\dfrac{2R}{\sqrt{15}}$.

Note: Students must take care that since the two objects were recast from each other, they had the same mass. If the two objects were of different materials, they would have different densities and hence different masses, even if they had the same volume. However, upon seeing this question, students should realize that the masses of the two objects will be the same and not waste time in equating the volumes of the objects. Students must know the moment of inertia of differently shaped-objects about at least one axis. By using perpendicular and parallel axis theorems, the moments of inertia about other axes could have been found out. For example, for a disc, even if the student does not know the moment of inertia of the disc about an axis passing through its edge and perpendicular to the plane, if they know of the moment of inertia about the axis passing through its center and perpendicular to the plane they can find out the required moment of inertia using the parallel axis theorem.