UPSC 2024 Maths Optional Paper 2 Q6c — Step-by-Step Solution

15 marks · Section B

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

Question

Find the moment of inertia of a quadrant of an elliptic disk $x^2/a^2+y^2/b^2=1$ (mass $M$ , density $\propto xy$ ) about the axis through its centre perpendicular to its plane.

Technique

Use scaled-polar coordinates $x=ar\cos\theta$ , $y=br\sin\theta$ (Jacobian $abr$ ); first determine $k$ from the total mass, then compute $I$ .

Solution

Setup. First-quadrant region, density $\rho=k\cdot xy$ , scaled coordinates $x=ar\cos\theta$ , $y=br\sin\theta$ , $r\in[0,1]$ , $\theta\in[0,\pi/2]$ , Jacobian $abr$ .

Step 1 — Determine $k$ from total mass.

M=\int_0^{\pi/2}\!\!\int_0^1 k(ar\cos\theta)(br\sin\theta)\cdot abr\,dr\,d\theta=ka^2 b^2\int_0^1 r^3\,dr\cdot\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=ka^2 b^2\cdot\frac{1}{4}\cdot\frac{1}{2}=\frac{ka^2 b^2}{8}.

So $k=8M/(a^2 b^2)$ .

Step 2 — Moment of inertia.

I=\iint(x^2+y^2)\rho\,dA=k\int_0^{\pi/2}\!\!\int_0^1 r^2(a^2\cos^2\theta+b^2\sin^2\theta)\cdot\frac{a^2 b^2}{2}r^3\sin 2\theta\,dr\,d\theta.

$\int_0^1 r^5\,dr=1/6$ . For the angular integral, substitute $u=\sin^2\theta$ :

\int_0^{\pi/2}(a^2\cos^2\theta+b^2\sin^2\theta)\cdot 2\sin\theta\cos\theta\,d\theta=\int_0^1[a^2(1-u)+b^2 u]\,du=a^2+\frac{b^2-a^2}{2}=\frac{a^2+b^2}{2}.

I=\frac{ka^2 b^2}{2}\cdot\frac{1}{6}\cdot\frac{a^2+b^2}{2}=\frac{ka^2 b^2(a^2+b^2)}{24}=\frac{8M}{a^2 b^2}\cdot\frac{a^2 b^2(a^2+b^2)}{24}=\frac{M(a^2+b^2)}{3}.

Answer

\boxed{\;I=\frac{M(a^2+b^2)}{3}.\;}