← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q7a — Solution

20 marks · Section B

Question

A particle of mass $2.5$ kg hangs at the end of a string, $0.9$ m long, the other end of which is attached to a fixed point. The particle is projected horizontally with a velocity $8$ m/sec. Find the velocity of the particle and tension in the string when the string is (i) horizontal (ii) vertically upward.

Technique

Energy conservation for speed; radial Newton’s-second-law for tension (centripetal equation).

Solution

Setup. $m=2.5$ kg, $\ell=0.9$ m, $u=8$ m/s, $g=9.8$ m/s². Let $\theta$ be the angle the string makes with the downward vertical (so $\theta=0$ at start, $\theta=\pi/2$ horizontal, $\theta=\pi$ vertically up).

Step 1 — Energy conservation to find speed at angle $\theta$

The particle rises by $\ell(1-\cos\theta)$ from the starting point. Conservation of energy:

Step 2 — Centripetal equation

At angle $\theta$, the centripetal force toward the pivot is $mv^{2}/\ell$, supplied by tension $T$ and the radial component of gravity. With “outward” defined as away from the pivot:

• Tension $T$ acts radially inward (toward pivot).
• Gravity’s radial component: $mg\cos\theta$ outward when the particle is below the pivot ($\theta<\pi/2$), zero at horizontal, $-mg|\cos\theta|=mg\cos\theta$ inward when above ($\theta>\pi/2$).

A clean unified form (taking inward as positive): $T-mg\cos\theta=mv^{2}/\ell$ for $\theta\le\pi/2$. For $\theta>\pi/2$ (particle above pivot), gravity also points inward (downward, same direction as tension when string vertical-up), so $T+mg|\cos\theta|=mv^{2}/\ell$.

For the two cases asked:

Step 3 — Case (i): String horizontal, $\theta=\pi/2$

$1-\cos\theta=1$:

Gravity has no radial component at $\theta=\pi/2$:

In symbolic form: $T_1=mu^{2}/\ell-2mg=(2.5)(64)/0.9-2(2.5)(9.8)=177.78-49=128.78$ N ✓.

Step 4 — Case (ii): String vertically upward, $\theta=\pi$

$1-\cos\theta=2$:

(Sanity check: does the particle reach the top? Required $v_2^{2}\ge g\ell$, i.e. $28.72\ge 8.82$ ✓.)

At $\theta=\pi$, both tension (pulling down toward pivot) and gravity (also down) point inward toward pivot:

In symbolic form: $T_2=mu^{2}/\ell-5mg=177.78-122.5=55.28$ N ✓.

Summary

Answer