Moment Of Inertia Of A Solid Sphere MCQ - Practice Questions with Answers

Let I=Moment of inertia of a SOLID SPHERE about an axis through its centre

To calculate I

Consider a sphere of mass M, radius R and centre O. Mass per unit volume of the sphere =  $\rho =\frac{M}{\frac{4}{3}\pi R^3}$

Take an elementary disc of mass dm, whose centre is C and which lies between two planes perpendicular to the axis at a distance x and x + dx from its centre

Disc has Radius=AC and thickness =dx.  As shown in the figure

${\mathrm{So}}_{,}AC=\sqrt{R^2-x^2}$
So, $dV=\pi (R^2-x^2)dx$

$$dm=\rho dV=\frac{3M}{4\pi R^3}\ast \pi (R^2-x^2)dx=\frac{3M}{4R^3}\ast (R^2-x^2)dx$$

The moment of inertia of the elementary disc about the axis

$$dI=\int (AC{)}^{2}dm$$

Now integrate this dl between the limits $x=-R$ to $x=+R$

$$\begin{array}{rl}& I=\int dI=\int (AC{)}^{2}dm\\ & ={\int }_{-R}^R(R^2-x^2)\ast \frac{3M}{4R^3}\ast (R^2-x^2)\cdot dx\\ & ={\int }_{-R}^R\frac{3M}{4R^3}\ast {(R^2-x^2)}^2\cdot dx\\ & =\frac{3M}{4R^3}{\int }_{-R}^R(R^4-2R^2{x}^2+x^4)dx\\ & \Rightarrow \mathbf{I}=\frac{2}{5}\ast {\mathbf{M}\mathbf{R}}^2\end{array}$$