# Question 93dc2

Aug 17, 2017

Consider a solid sphere of radius $R$ and mass $M$ as shown in the figure below

We need to calculate Moment of Inertia of this sphere about its diameter along $x$-axis. The sphere can be considered to be made up of infinitesimal solid cylinders.
Consider a solid cylinder of thickness $\mathrm{dx}$ located at a distance $x$ from the center of sphere. Let $r$ be radius of this cylinder.
We know that moment of inertia for a solid cylinder is given by the expression

$I = \frac{1}{2} M {R}^{2}$

$\therefore$ Moment of inertia of the infinitesimal cylinder is

$\mathrm{dI} = \frac{1}{2} \mathrm{dm} \cdot {r}^{2}$ ......(1)

$\mathrm{dm}$ can be calculated with help of density of the sphere $\rho$ and volume of the infinitesimal cylinder

dm=ρcdotπr^2dx

Substitute this value in equation (1)

dI=1/2ρπr^4dx

Writing $r$ in terms of variable $x \mathmr{and} R$ above expression becomes

dI=1/2ρπ(R^2–x^2)^2dx

Integrating between limits $- R \text{ to } R$ we get

I=1/2ρπint_(-R)^R(R^2–x^2)^2dx
=>I=1/2ρπint_(-R)^R(R^4–2R^2x^2+x^4)dx
=>I=1/2ρπ|R^4x–2R^2x^3/3+x^5/5|_(-R)^R
=>I=1/2ρπ|(R^4xxR–2R^2xxR^3/3+R^5/5)-(R^4(-R)–2R^2(-R)^3/3+(-R)^5/5)|
=>I=1/2ρπR^5(1-2/3+1/5+1-2/3+1/5)
=>I=1/2ρπR^5((15-10+3+15-10+3)/15)
=>I=1/2ρπ16/15R^5

Density of the sphere is given by
ρ=M/V
=>ρ=M/(4/3πR^3)

Substituting in above we get

I=1/2(M/(4/3πR^3))π16/15R^5#
$\implies I = \frac{2}{5} M {R}^{2}$