UPSC 2015 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

Moment of inertia · Mechanics & Fluid Dynamics · asked 7× in 13 yrs · Read the full method →

Question

Calculate the moment of inertia of a solid uniform hemisphere $x^2+y^2+z^2=a^2$ , $z\ge 0$ with mass $m$ about the $OZ$ -axis.

Technique

Standard spherical-coordinate triple integral; $\sin^3\phi$ integral via $u=\cos\phi$ ; factor independently.

Solution

Setup. Hemisphere of radius $a$ , mass $m$ , uniform density $\rho=m/V$ where $V=\tfrac{2}{3}\pi a^3$ .

So $\rho=\dfrac{3m}{2\pi a^3}$ .

Moment of inertia about $z$ -axis:

I_z=\iiint_H(x^2+y^2)\rho\,dV.

Step 1 — Use spherical coordinates

$x=r\sin\phi\cos\theta$ , $y=r\sin\phi\sin\theta$ , $z=r\cos\phi$ , $dV=r^2\sin\phi\,dr\,d\phi\,d\theta$ .

$x^2+y^2=r^2\sin^2\phi$ .

Domain: $0\le r\le a$ , $0\le\phi\le\pi/2$ (upper hemisphere), $0\le\theta\le 2\pi$ .

Step 2 — Set up integral

I_z=\rho\int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^a r^2\sin^2\phi\cdot r^2\sin\phi\,dr\,d\phi\,d\theta=\rho\int_0^{2\pi}d\theta\int_0^{\pi/2}\sin^3\phi\,d\phi\int_0^a r^4\,dr.

Step 3 — Evaluate each factor

$\int_0^{2\pi}d\theta=2\pi$ .

$\int_0^a r^4\,dr=a^5/5$ .

$\int_0^{\pi/2}\sin^3\phi\,d\phi=\int_0^{\pi/2}\sin\phi(1-\cos^2\phi)\,d\phi$ . Let $u=\cos\phi$ , $du=-\sin\phi\,d\phi$ : $=\int_1^0(1-u^2)(-du)=\int_0^1(1-u^2)du=1-1/3=2/3$ .

Step 4 — Combine

I_z=\rho\cdot 2\pi\cdot\dfrac{2}{3}\cdot\dfrac{a^5}{5}=\dfrac{4\pi\rho a^5}{15}.

Substitute $\rho=3m/(2\pi a^3)$ :

I_z=\dfrac{4\pi a^5}{15}\cdot\dfrac{3m}{2\pi a^3}=\dfrac{4\cdot 3\,m\,a^2}{15\cdot 2}=\dfrac{12 ma^2}{30}=\dfrac{2ma^2}{5}.

Answer

\boxed{\;I_z=\dfrac{2}{5}ma^2.\;}