← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q5c — Step-by-Step Solution

10 marks · Section B

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

Question

A uniform rod $OA$, of length $2a$, free to turn about its end $O$, revolves with angular velocity $\omega$ about the vertical $OZ$ through $O$, and is inclined at a constant angle $\alpha$ to $OZ$; find the value of $\alpha$.

Technique

Steady precession; take moments about the hinge $O$ to eliminate the reaction. Gravity moment $=Mga\sin\alpha$; integrated centrifugal moment $=\frac{4}{3}Ma^2\omega^2\sin\alpha\cos\alpha$ (uses $I_O=\frac43 Ma^2$).

Solution

Setup. Mass $M$, length $2a$, hinged at $O$, sweeping out a cone of semi-angle $\alpha$ about the vertical $OZ$ with steady angular velocity $\omega$. Take moments about the fixed point $O$ so the (unknown) hinge reaction drops out. The motion is steady precession: each element travels in a horizontal circle, requiring a centripetal force, equivalently each element carries a “centrifugal” reversed-effective force $\omega^2(\text{horizontal distance})$ per unit mass directed outward.

Step 1 — Take moments about $O$ (vertical plane containing the rod)

Two distributed effects produce moments about the horizontal axis through $O$ (perpendicular to the rod’s vertical plane):

(i) Gravity. Acts at the centroid, distance $a$ from $O$. Weight $Mg$ vertical. Its moment about $O$ (tending to decrease $\alpha$, i.e. lower the rod):

(ii) Centrifugal (inertial) reaction. A rod element at distance $r$ from $O$ sits at horizontal distance $r\sin\alpha$ from the axis. Its element of centrifugal force is $dm\,\omega^2(r\sin\alpha)$, horizontal, outward. Its moment arm about the horizontal axis through $O$ is the vertical height $r\cos\alpha$. So the elemental moment (tending to increase $\alpha$, throwing the rod outward) is

Step 2 — Integrate the centrifugal moment

Linear density $\lambda=M/2a$, $dm=\lambda\,dr$, $r:0\to2a$:

(The integral $\int_0^{2a} r^2\lambda\,dr=\frac13 M(2a)^2=\frac43 Ma^2$ is just the moment of inertia of the rod about $O$.)

Step 3 — Balance the moments (steady state)

For a steady cone the net moment about $O$ vanishes: the centrifugal moment (outward) balances the gravity moment (inward):

Cancel $M$, $a$, and (for the non-vertical solution) $\sin\alpha\ne0$:

Answer