← 2023 Paper 2

UPSC 2023 Maths Optional Paper 2 Q7b — Step-by-Step Solution

15 marks · Section B

Motion of rigid bodies in two dimensions · Mechanics & Fluid Dynamics · asked 4× in 13 yrs · Read the full method →

Question

A perfectly rough ball is at rest within a hollow cylindrical roller. The roller is drawn along a level path with uniform velocity $V$. Let $a$ and $b$ be the radii of the ball and the roller respectively. If $V^2>\tfrac{27}{7}g(b-a)$, then show that the ball will roll completely round the inside of the roller.

Technique

Frame change to the inertial frame of the uniformly-moving roller; rolling kinetic energy $\tfrac{7}{10}mv^{2}$; centripetal force balance at the top.

Solution

Strategy. Move to the roller’s frame (inertial since the roller has uniform velocity $V$). In that frame the ball has initial speed $V$ at the bottom; use energy conservation for a rolling sphere and the contact criterion at the top of the loop.

Step 1 — Choose the inertial frame of the moving roller

The roller is “drawn along” with uniform velocity $V$, so the frame attached to the roller’s centre is inertial. In this frame:

• the roller is at rest;
• the ball, initially at rest in the ground frame, has horizontal velocity $-V$ at the bottom of the roller.

The ball rolls without slipping on the inner surface of the (now stationary) roller — “perfectly rough” guarantees this. The ball’s centre moves on a circle of radius $b-a$ about the roller’s centre.

Step 2 — Kinetic energy of a rolling solid sphere

For a solid sphere of mass $m$, radius $a$, moment of inertia $I=\tfrac{2}{5}ma^{2}$. Rolling without slipping ($\omega=v/a$ where $v$ is the speed of the centre) gives total kinetic energy

Step 3 — Energy conservation: bottom to top

Take the bottom of the ball’s circular trajectory as the zero of potential energy. The top is at height $2(b-a)$. Let $v_b$ and $v_t$ be the speeds at the bottom and top respectively.

Cancel $m$ and rearrange:

Step 4 — Contact condition at the top of the loop

At the top of the inside of the roller, gravity points outward (radially away from the roller’s centre) and the normal force from the roller also points outward (the ball pushes against the inner roller, the roller pushes back radially inward toward the ball’s centre, i.e. downward; but on the inside surface at the top, the normal on the ball is downward — toward the roller’s centre, i.e. toward the ball’s centre of curvature).

Centripetal equation at the top (taking inward — i.e. downward — as positive):

For the ball to maintain contact at the top, $N\ge 0$, giving the minimum-speed condition:

Step 5 — Combine ($\ast$) and ($\dagger$)

Substituting $v_t^{2}=g(b-a)$ (the critical case) into ($\ast$):

The ball completes the loop iff $v_b^{2}>v_{b,\text{crit}}^{2}$, i.e.

since $v_b=V$ (the ball’s speed at the bottom in the roller’s frame).

Answer