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UPSC 2019 Maths Optional Paper 1 Q6a — Step-by-Step Solution

15 marks · Section B

Stability of equilibrium (energy criterion) · Dynamics & Statics · asked 5× in 13 yrs · Read the full method →

Question

A body consists of a cone and underlying hemisphere. The base of the cone and the top of the hemisphere have same radius $a$. The whole body rests on a rough horizontal table with hemisphere in contact with the table. Show that the greatest height of the cone, so that the equilibrium may be stable, is $\sqrt 3\,a$.

Technique

Stability of a spherical-based body $\Leftrightarrow$ centre of gravity at/below the centre $O$ of the spherical surface; compute the composite centroid by volume-weighting the hemisphere ($\bar y=-3a/8$) and cone ($\bar y=+h/4$).

Solution

Setup. A solid hemisphere of radius $a$ sits with its curved surface on the table; a solid cone of base radius $a$ and height $h$ is glued on top, sharing the flat circular face. Both are of the same uniform density $\rho$. Let $O$ be the centre of the flat face (which is the centre of the sphere of which the hemisphere is a part). Measure heights along the common axis, taking up as positive from $O$.

Step 1 — Why the centre $O$ of the spherical surface is the key point

When a body whose base is part of a sphere of radius $a$ rocks on a horizontal plane, the point of contact moves but the geometric centre of the spherical surface, $O$, stays at the same height (the surface rolls on the table at distance $a$ from $O$). The standard stability criterion is:

The equilibrium is stable iff the common centre of gravity $G$ lies below the centre $O$ of the spherical surface (and unstable if above).

So we require the height of $G$ above $O$ to be $\le 0$.

Step 2 — Centroids of the two parts (heights measured from $O$)

• Hemisphere (curved part below the flat face): its centroid is at distance $\tfrac{3a}{8}$ from the flat face, on the side of the curved surface, i.e. below $O$:

• Cone (above the flat face): centroid at $\tfrac{h}{4}$ above the base:

Step 3 — Volumes (masses $\propto$ volumes, equal density)

Step 4 — Height of the combined centre of gravity above $O$

The numerator $=\pi a^2\!\left(-\dfrac{a^2}{4}+\dfrac{h^2}{12}\right)$ and the denominator $=\dfrac{\pi a^2}{3}(2a+h)$, so

Step 5 — Stability condition

Stable equilibrium requires $\bar y\le 0$. Since $2a+h>0$,

The greatest height for which the equilibrium is (just) stable is therefore

Answer