UPSC 2013 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

Four solid spheres $A,B,C$ and $D$ , each of mass $m$ and radius $a$ , are placed with their centres on the four corners of a square of side $b$ . Calculate the moment of inertia of the system about a diagonal of the square.

Technique

Parallel-axis theorem applied to each sphere; sum contributions.

Solution

Strategy. Use the parallel-axis theorem for each sphere; the MI contribution depends on the sphere’s perpendicular distance from the diagonal.

Step 1 — Geometry

Place the square in a plane with corners at $A=(b/2,b/2)$ , $B=(-b/2,b/2)$ , $C=(-b/2,-b/2)$ , $D=(b/2,-b/2)$ . The diagonal $AC$ (from $A$ to $C$ ) lies along the line $y=x$ .

Perpendicular distance from each corner to this diagonal:

$A=(b/2,b/2)$ : on the diagonal, $d_A=0$ .
$C=(-b/2,-b/2)$ : on the diagonal, $d_C=0$ .
$B=(-b/2,b/2)$ : distance to line $y=x$ is $|b/2-(-b/2)|/\sqrt 2=b/\sqrt 2$ .
$D=(b/2,-b/2)$ : distance $|-b/2-b/2|/\sqrt 2=b/\sqrt 2$ .

Step 2 — MI of a single solid sphere

About a diameter (axis through centre): $I_{\text{CM}}=\dfrac{2}{5}ma^{2}$ .

By the parallel-axis theorem, about an axis parallel to the diameter at perpendicular distance $d$ :

I=I_{\text{CM}}+md^{2}=\frac{2}{5}ma^{2}+md^{2}.

Step 3 — Sum contributions

Sphere	Position	Distance to diagonal	MI contribution
$A$	on diagonal	$0$	$\tfrac{2}{5}ma^{2}$
$B$	off	$b/\sqrt 2$	$\tfrac{2}{5}ma^{2}+m\cdot\tfrac{b^{2}}{2}$
$C$	on diagonal	$0$	$\tfrac{2}{5}ma^{2}$
$D$	off	$b/\sqrt 2$	$\tfrac{2}{5}ma^{2}+m\cdot\tfrac{b^{2}}{2}$

Total:

I=4\cdot\frac{2}{5}ma^{2}+2\cdot m\cdot\frac{b^{2}}{2}=\frac{8}{5}ma^{2}+mb^{2}.

Answer

\boxed{\;I=\dfrac{8ma^{2}}{5}+mb^{2}=m\!\left(\dfrac{8a^{2}}{5}+b^{2}\right).\;}