Moment Of Inertia Of A Solid Sphere - Practice Questions & MCQ

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 figure

So, $A C=\sqrt{R^2-x^2}$
So, $d V=\pi\left(R^2-x^2\right) d x$

$
d m=\rho d V=\frac{3 M}{4 \pi R^3} * \pi\left(R^2-x^2\right) d x=\frac{3 M}{4 R^3} *\left(R^2-x^2\right) d x
$

The moment of inertia of the elementary disc about the axis

$
d I=\int \frac{(A C)^2 d m}{2} \quad\left(\because I_{d i s c}=\frac{m R^2}{2}\right)
$

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

$
\begin{aligned}
& I=\int d I=\int \frac{(A C)^2 d m}{2} \\
& =\int_{-R}^R\left(R^2-x^2\right) \frac{3 M}{8 R^3}\left(R^2-x^2\right) d x \\
& =\int_{-R}^R \frac{3 M}{8 R^3}\left(R^2-x^2\right)^2 d x \\
& =\frac{3 M}{8 R^3} \int_{-R}^R\left(R^4-2 R^2 x^2+x^4\right) d x \\
& \Rightarrow \mathbf{I}=\frac{2}{5} \mathbf{M R}^2
\end{aligned}
$