← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q7a — Solution

15 marks · Section B

Question

A solid sphere rests inside a fixed rough and hemispherical bowl of twice its radius. If a large amount of weight, whatsoever, is attached to the highest point of the sphere, then show that the equilibrium is stable.

Technique

Use the energy criterion: the equilibrium is stable iff the total potential energy is a minimum (the centre of gravity of the system is at minimum height). The rolling constraint with bowl radius $=2\times$ sphere radius makes the loaded point trace a horizontal line — the classical property of a circle rolling inside one of twice its radius.

Solution

Set-up. Let the sphere have radius $r$ and the hemispherical bowl radius $R = 2r$, centre $O$. Because the sphere rolls (rough bowl) without slipping, its centre $C$ stays at distance $R-r = r$ from $O$, moving on a vertical circle of radius $r$ about $O$. Let $\theta$ be the angular displacement of $OC$ from the downward vertical; $\theta=0$ is the equilibrium with the sphere at the lowest point.

Take $O$ as origin and measure heights $y$ upward. Then $y_C = -(R-r)\cos\theta = -r\cos\theta.$

Rolling constraint → rotation of the sphere. No slipping at the contact point: the velocity of $C$ is $(R-r)\dot\theta$ and must equal $r\,\omega$ (sphere spin $\omega$ about its own centre). Hence the absolute rotation angle of the sphere is $\psi = \frac{R-r}{r}\,\theta = \frac{r}{r}\,\theta = \theta,$ in the opposite sense to $\theta$ (the contact constraint fixes the sense; magnitude $=\theta$ since $R=2r$).

Position of the attached weight. The weight $W$ is fixed at the highest point of the sphere, which at $\theta=0$ is the point $P_0 = C + (0, r)$ — directly above $C$ at distance $r$, i.e. at height $y = -r + r = 0$ (the level of $O$). As the sphere rolls, this material point rotates by $\psi=\theta$ (opposite sense) about $C$. Its height is $y_P = y_C + r\cos\psi = -r\cos\theta + r\cos\theta = 0.$

This is the key fact: for $R = 2r$, the point of the sphere that starts at the top always stays at the height of $O$ — it moves on the horizontal diameter of the bowl. (A point on a circle of radius $r$ rolling inside a circle of radius $2r$ describes a straight diameter.) Consequently the attached weight $W$ neither rises nor falls during any displacement.

Potential energy. Let the sphere’s own weight be $w_s$ acting at its centre $C$. The total potential energy (taking $g=1$ scale) is $V(\theta) = w_s\,y_C + W\,y_P = -w_s\,r\cos\theta + W\cdot 0 = -w_s\,r\cos\theta.$

The added weight $W$ contributes a constant ($=0$) to $V$, regardless of how large $W$ is. Therefore $\frac{dV}{d\theta} = w_s\,r\sin\theta = 0 \text{ at } \theta=0,\qquad \frac{d^2V}{d\theta^2}\bigg|_{\theta=0} = w_s\,r\cos\theta\big|_{0} = w_s\,r > 0.$

Since $V''(0) > 0$, the equilibrium $\theta=0$ is a strict minimum of the potential energy. The equilibrium is stable, irrespective of the magnitude of $W$. $\blacksquare$

Answer

$\boxed{\;\text{Because }R=2r,\text{ the loaded top point stays at constant height, so }V(\theta)=-w_s r\cos\theta\text{ has a minimum at }\theta=0;\ V''(0)=w_s r>0\ \Rightarrow\ \text{stable for any }W.\;}$