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UPSC 2020 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

Motion in a Plane (Resolved Components / Polar) · Dynamics & Statics · Read the full method →

Question

A light rigid rod $ABC$ has three particles each of mass $m$ attached to it at $A$, $B$ and $C$. The rod is struck by a blow $P$ at right angles to it at a point distant from $A$ equal to $BC$. Prove that the kinetic energy set up is $\frac{1}{2}\frac{P^2}{m}\frac{a^2-ab+b^2}{a^2+ab+b^2}$, where $AB=a$ and $BC=b$.

Technique

Impulsive motion of a free rigid body: linear impulse $=Mv$, moment of impulse about $G$ $=I_G\omega$; then $T=\tfrac12Mv^2+\tfrac12 I_G\omega^2$. Cross-check via $T=\tfrac12 P\,V_D$.

Solution

Setup

Measure distance $x$ along the rod from $A$. Particle positions:

Total mass $=3m$. The blow $P$ (an impulse) is applied perpendicular to the rod at the point $D$ with $AD=b$.

Step 1 — Centre of mass

The rod is free in the plane, struck by an impulse perpendicular to it. After the blow it has a transverse velocity $v$ of the centre of mass $G$ and an angular velocity $\omega$ about $G$.

Step 2 — Impulse–momentum equations

Linear impulse = change in linear momentum (all transverse):

Angular impulse about $G$ = change in angular momentum about $G$. The blow acts at $D$, whose distance from $G$ (measured along the rod) is

Moment of impulse about $G$ is $P\cdot d$ (taking the transverse direction consistently). Thus

where $I_G$ is the moment of inertia about $G$.

Step 3 — Moment of inertia about $G$

Distances of the particles from $G$:

Expand:

Sum $=6a^2+6ab+6b^2=6(a^2+ab+b^2)$. Hence

Step 4 — Velocities

From (1): $v=\dfrac{P}{3m}$.

From (2): $\omega=\dfrac{P\,d}{I_G}=\dfrac{P\cdot\frac{2(a-b)}{3}}{\frac{2m}{3}(a^2+ab+b^2)}=\dfrac{P(a-b)}{m(a^2+ab+b^2)}$.

Step 5 — Kinetic energy

First term:

Second term:

Add, with common factor $\dfrac{P^2}{6m}$:

Numerator:

Therefore

Answer