UPSC 2019 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

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

Question

A circular cylinder of radius $a$ and radius of gyration $k$ rolls without slipping inside a fixed hollow cylinder of radius $b$ . Show that the plane through axes moves in a circular pendulum of length $(b-a)\left(1+\dfrac{k^2}{a^2}\right)$ .

Technique

Rolling constraint $a\dot\phi=(b-a)\dot\theta$ ; energy method (conservation of $T+V$ ) or Lagrangian; compare the resulting $\ddot\theta+\frac gL\sin\theta=0$ with the circular pendulum.

Solution

Setup. The inner solid cylinder (radius $a$ , mass $M$ , radius of gyration $k$ about its own axis, so $I=Mk^2$ ) rolls without slipping inside a fixed hollow cylinder of radius $b>a$ . Let $\theta$ be the angle the line of centres (from the fixed axis $O$ to the moving axis $C$ ) makes with the downward vertical. The centre $C$ moves on a circle of radius $(b-a)$ .

Step 1 — Kinematics and the rolling constraint

The centre $C$ is at distance $(b-a)$ from $O$ , so

v_C=(b-a)\dot\theta.

Let $\phi$ be the rotation angle of the rolling cylinder about its own axis. Rolling without slipping inside the fixed cylinder means the contact arcs are equal. The arc swept on the outer (fixed) cylinder as the centre advances by $\theta$ is $b\,\theta$ ; this equals the arc rolled on the inner cylinder, $a\,\psi$ , where $\psi$ is the rotation of the inner cylinder relative to the line of centres. The absolute spin is $\dot\phi$ with

a\,\dot\phi=(b-a)\dot\theta\quad\Longrightarrow\quad \dot\phi=\frac{(b-a)}{a}\dot\theta.

(The contact point of the inner cylinder is instantaneously at rest: $v_C=a\dot\phi$ gives the same relation $a\dot\phi=(b-a)\dot\theta$ .)

Step 2 — Kinetic energy

T=\tfrac12 M v_C^2+\tfrac12 I\dot\phi^2 =\tfrac12 M(b-a)^2\dot\theta^2+\tfrac12 M k^2\Big(\frac{(b-a)}{a}\dot\theta\Big)^2.

T=\tfrac12 M(b-a)^2\dot\theta^2\Big(1+\frac{k^2}{a^2}\Big).

Step 3 — Potential energy

Taking $O$ as reference, the height of $C$ below $O$ is $(b-a)\cos\theta$ , so

V=-Mg(b-a)\cos\theta.

Step 4 — Energy equation and equation of motion

Total energy $E=T+V$ is conserved:

\tfrac12 M(b-a)^2\Big(1+\frac{k^2}{a^2}\Big)\dot\theta^2-Mg(b-a)\cos\theta=E.

Differentiate w.r.t. $t$ and cancel $\dot\theta$ (and $M(b-a)$ ):

(b-a)\Big(1+\frac{k^2}{a^2}\Big)\ddot\theta+g\sin\theta=0,

\boxed{\;\ddot\theta+\frac{g}{(b-a)\big(1+\frac{k^2}{a^2}\big)}\sin\theta=0.\;}

Step 5 — Identify the equivalent simple pendulum

Compare with the exact circular-pendulum equation $\ddot\theta+\dfrac{g}{L}\sin\theta=0$ . The line of centres (the “plane through the axes”) therefore swings exactly like a circular pendulum of length

Answer

\boxed{\;L=(b-a)\Big(1+\frac{k^2}{a^2}\Big).\;}