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UPSC 2017 Maths Optional Paper 1 Q6c — Step-by-Step Solution

17 marks · Section B

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

Question

A uniform solid hemisphere rests on a rough plane inclined to the horizon at an angle $\phi$ with its curved surface touching the plane. Find the greatest admissible value of the inclination $\phi$ for equilibrium. If $\phi$ be less than this value, is the equilibrium stable?

Technique

Sphere reaction passes through $O$; equilibrium $\Leftrightarrow G$ vertically over contact; the realizability of the tilt angle bounds $\phi$.

Solution

Step 1 — Geometry of the hemisphere

For a uniform solid hemisphere of radius $r$, the centre of mass $G$ lies on the axis of symmetry at distance

from the centre $O$ of the plane (flat) face, measured into the solid (towards the curved part).

Because the contact is on the spherical surface, the reaction at the contact point $C$ acts along the line $CO$ (the normal to a sphere passes through its centre). Thus $OC=r$ and $OC\perp$ plane.

Step 2 — Condition for equilibrium

Three forces act: weight $W$ (vertically down through $G$), normal reaction $N$ (along $CO$), and friction $F$ (along the plane). For equilibrium the line of action of the weight must pass through the contact point $C$ (so that the moment about $C$ vanishes), i.e. $G$ is vertically above $C$.

Work in the vertical plane of greatest slope. Let $\hat n$ be the outward normal to the incline (making angle $\phi$ with the vertical) and $\hat u$ the up-slope direction. Place $C$ at the origin, so $O=r\hat n$. Let the symmetry axis make angle $\alpha$ with $\hat n$, so

With $\hat n=(-\sin\phi,\cos\phi)$ and $\hat u=(\cos\phi,\sin\phi)$, the horizontal coordinate of $G$ is

Equilibrium requires $G_x=0$:

Step 3 — Greatest admissible $\phi$

A real tilt angle $\alpha$ exists only if $\big|\tfrac{8}{3}\sin\phi\big|\le1$, i.e.

The greatest admissible inclination is therefore

Answer