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UPSC 2014 Maths Optional Paper 1 Q7b — Step-by-Step Solution

15 marks · Section B

Constrained motion · Dynamics & Statics · asked 7× in 13 yrs · Read the full method →

Question

A particle of mass $m$, hanging vertically from a fixed point by a light inextensible cord of length $l$, is struck by a horizontal blow which imparts to it a velocity $2\sqrt{gl}$. Find the velocity and height of the particle from the level of its initial position when the cord becomes slack.

Technique

Energy conservation + centripetal force balance + slack condition $T=0$.

Solution

Setup. Particle hangs at the lowest point of a cord (length $l$, fixed top). Horizontal blow gives initial speed $u=2\sqrt{gl}$. Let $\theta$ denote angle of cord from downward vertical. Particle rises along a circle of radius $l$ until the cord goes slack.

Step 1 — Energy conservation

Height above initial position: $h(\theta)=l(1-\cos\theta)$.

With $u^{2}=4gl$:

Step 2 — Centripetal equation; tension

At angle $\theta$, the centripetal direction is toward the fixed point. Tension $T$ pulls along the cord in this direction. Gravity component in the centripetal direction: $-mg\cos\theta$ (negative for $\theta<\pi/2$, i.e., gravity pulls away from centre; positive for $\theta>\pi/2$, gravity pulls toward centre when particle is above horizontal).

Newton’s second law (centripetal):

Step 3 — Cord becomes slack: $T=0$

For $v^{2}>0$, need $\cos\theta<0$, i.e., $\theta>\pi/2$ — particle is above the horizontal through the anchor when the cord goes slack.

Step 4 — Solve for $\theta$

Equating the two expressions for $v^{2}$:

Step 5 — Velocity and height at slack

Velocity: $v^{2}=-gl(-2/3)=\dfrac{2gl}{3}$, so

Height above initial position: $h=l(1-\cos\theta)=l(1+\tfrac{2}{3})=\dfrac{5l}{3}$.

Answer