← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q8c — Solution

20 marks · Section B

Question

A particle is projected inside a fixed smooth cylinder with circular cross-section in a vertical plane from the lowest point with initial horizontal velocity $u$. Show that for (i) ($u^2 \leq 2ag$): the particle oscillates about the mean position in the lower half, (ii) ($u^2 \geq 5ag$): the particle executes complete circular motion, and (iii) ($2ag < u^2 < 5ag$): the particle will leave the curve in a tangential direction, making an angle $\alpha$ with the horizontal such that $\cos\alpha = \dfrac{u^2 - 2ag}{3ag}$.

Technique

Use energy conservation (smooth surface) for the speed and the radial (centripetal) equation for the normal reaction $N$. The three cases follow from comparing the angle at which $v=0$ (turning) with the angle at which $N=0$ (leaving the track).

Solution

The cylinder is smooth, so the particle slides without friction on the inner surface; motion is in a vertical circle of radius $a$ (the circular cross-section), centre $O$. Let $\theta$ be the angular position of the particle measured from the lowest point. Height above the lowest point at angle $\theta$ is $a(1-\cos\theta)$.

Energy conservation (smooth surface), with speed $v$: \tfrac12 v^2 = \tfrac12 u^2 - g\,a(1-\cos\theta) \;\Longrightarrow\; v^2 = u^2 - 2ag(1-\cos\theta). \tag{1}

Radial equation. The inward normal reaction $N$ (the track can only push the particle toward the centre) and the radial component of gravity provide the centripetal force $\dfrac{mv^2}{a}$. The inward component of gravity is $-mg\cos\theta$, so N - mg\cos\theta = \frac{mv^2}{a} \;\Longrightarrow\; N = \frac{mv^2}{a} + mg\cos\theta. \tag{2}

Substituting (1) into (2): N = \frac{m}{a}\bigl[u^2 - 2ag(1-\cos\theta)\bigr] + mg\cos\theta = \frac{mu^2}{a} - 2mg + 3mg\cos\theta. \tag{3}

The two critical events are:

• Turning ($v=0$): from (1), $\cos\theta = 1 - \dfrac{u^2}{2ag}$.
• Leaving ($N=0$): from (3), $\cos\theta = \dfrac{2ag - u^2}{3ag} = -\dfrac{u^2 - 2ag}{3ag}$.

Case (i): $u^2 \le 2ag$ — oscillation in the lower half

$v=0$ occurs at $\cos\theta = 1 - \dfrac{u^2}{2ag} \ge 0$, i.e. $\theta \le 90^\circ$ (lower half). In the lower half ($\theta\le 90^\circ$, $\cos\theta\ge0$), the normal reaction (2) is $N=\frac{mv^2}{a}+mg\cos\theta\ge 0$ throughout, so the particle stays in contact and simply comes to rest, then swings back. It oscillates about the lowest point, confined to the lower half. (The boundary $u^2=2ag$ gives the turning point exactly at the horizontal level of the centre, $\theta=90^\circ$.)

Case (ii): $u^2 \ge 5ag$ — complete circular motion

For a complete circuit the particle must maintain contact even at the highest point $\theta=180^\circ$, where $N$ is least. From (3) at $\theta=180^\circ$ ($\cos\theta=-1$): $N_{\text{top}} = \frac{mu^2}{a} - 2mg - 3mg = \frac{mu^2}{a} - 5mg \ge 0 \;\Longleftrightarrow\; u^2 \ge 5ag.$ Then $N\ge0$ for all $\theta$ and $v^2>0$ everywhere (since $v^2_{\text{top}}=u^2-4ag\ge ag>0$), so the particle makes complete circular motion.

Case (iii): $2ag < u^2 < 5ag$ — leaves tangentially

Here the turning point would require $\cos\theta = 1-\dfrac{u^2}{2ag} < 0$ (upper half), but before reaching it the reaction vanishes. $N=0$ from (3) gives $\cos\theta = -\frac{u^2 - 2ag}{3ag},$ which lies in $(-1, 0)$ for $2ag<u^2<5ag$, i.e. an upper-half angle $90^\circ<\theta<180^\circ$. Since this happens while $v^2>0$, the particle leaves the track there, moving tangentially.

Leaving angle with the horizontal. At the departure point the velocity is tangent to the circle, i.e. perpendicular to the radius $OP$. The radius $OP$ makes angle $\theta$ with the downward vertical, so the acute angle $\alpha$ of the velocity with the horizontal satisfies $\alpha = 180^\circ - \theta$ (for $\theta$ in the upper half). Therefore $\cos\alpha = \cos(180^\circ - \theta) = -\cos\theta = \frac{u^2 - 2ag}{3ag}.$

(Indeed for $2ag<u^2<5ag$, $\dfrac{u^2-2ag}{3ag}\in(0,1)$, a valid cosine.)

Answer