Question

Moment of inertia of a solid sphere of mass M and radius R is ______.

Answer 1

Step 1. Given data:

Mass of solid sphere = $M$

Radius of solid sphere = $R$

Step 2. Calculations:

The moment of inertia of a solid sphere is defined as the quantity expressed by the body resisting angular acceleration which is equal to the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

Let us consider a sphere of radius $R$ and mass $M$. A thin spherical shell of radius $x$, mass $dm$ , and thickness $dx$ is taken as a mass element. Volume density $\left(\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$V$}\right.\right)$ remains constant as the solid sphere is uniform throughout.

Volume $\left(V\right)$ of solid sphere =$\frac{4}{3}{\mathrm{\pi R}}^3$

The volume of thin spherical shell element $\left(dv\right)$ = $4{\mathrm{\pi x}}^2.\mathrm{dx}$

Now, the Volume density of the solid sphere = Volume density of the thin spherical shell element.

$\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$V$}\right.=\raisebox{1ex}{$dm$}\!\left/ \!\raisebox{-1ex}{$dv$}\right.$

$\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)$}\right.$ = $\raisebox{1ex}{$dm$}\!\left/ \!\raisebox{-1ex}{$\left(4{\mathrm{\pi x}}^2\mathrm{dx}\right)$}\right.$

$$\begin{gathered} dm=\left[\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$\left(\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times {\mathrm{\pi R}}^3\right)$}\right.\right]\times 4{\mathrm{\pi x}}^2\mathrm{dx} \\ =\left[3\mathrm{M}/{\mathrm{R}}^3\right].{\mathrm{x}}^2.\mathrm{dx} \end{gathered}$$

Step 3: Calculate the moment of inertia

Now, we can calculate the moment of inertia by putting the value in the following equation:

$$\begin{gathered} I=\int di=(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.)\times \int dm.x^2 \\ =\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\int \left(\raisebox{1ex}{$3M$}\!\left/ \!\raisebox{-1ex}{$R^3$}\right.\right).x^4.dx \\ =\left(\raisebox{1ex}{$2M$}\!\left/ \!\raisebox{-1ex}{$R^3$}\right.\right)\int x^4.dx \end{gathered}$$

Limits: As x increases from $0$ to $R$, the elemental shell covers all the spherical sphere.

Therefore, the moment of inertia of a uniform solid sphere $\left(I\right)=\raisebox{1ex}{$2M{R}^2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$.