UPSC 2024 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 heavy particle hanging from a fixed point by a light inextensible string of length $l$ starts to move with initial velocity $u$ in a vertical circle. Show that the sum of tensions at the ends of any diameter is constant.

Technique

Energy conservation for $v^2(\theta)$ ; Newton’s second law radially for $T(\theta)$ ; substitute and observe the $\cos\theta$ cancellation at diametrically opposite points.

Solution

$Free-body diagram of the particle in the vertical circle of radius l, at angle \theta from the downward vertical (initial position, with speed u, at the bottom). Two forces act: the string tension T directed inward toward the fixed point O, and the weight mg vertically down. The radial component of Newton's law, T-mg\cos\theta=mv^2/l, together with energy conservation, gives T(\theta); the \cos\theta term reverses sign at the diametrically opposite point, which is why the sum of tensions there is constant.$

Setup. Particle of mass $m$ , string of length $l$ . Let $\theta$ be the angle from the downward vertical. Height above the lowest point: $l(1-\cos\theta)$ .

Step 1 — Energy conservation.

v^2=u^2-2gl(1-\cos\theta).

Step 2 — Tension via radial Newton’s law.

T-mg\cos\theta=\frac{mv^2}{l}\;\Rightarrow\;T(\theta)=mg\cos\theta+\frac{mu^2}{l}-2mg+2mg\cos\theta=\frac{mu^2}{l}+mg(3\cos\theta-2).

Step 3 — Sum at diametrically opposite points.

Diametrically opposite: $\theta$ and $\theta+\pi$ . Since $\cos(\theta+\pi)=-\cos\theta$ :

T(\theta)+T(\theta+\pi)=\Bigl[\frac{mu^2}{l}+mg(3\cos\theta-2)\Bigr]+\Bigl[\frac{mu^2}{l}+mg(-3\cos\theta-2)\Bigr]=\frac{2mu^2}{l}-4mg.

This is independent of $\theta$ , as required.

Note. For a complete revolution the tension must be non-negative at the top ( $\theta=\pi$ ), requiring $u^2\ge 5gl$ .

Answer

\boxed{\;T(\theta)+T(\theta+\pi)=\frac{2mu^2}{l}-4mg=\frac{2m(u^2-2gl)}{l}=\text{constant}.\;}