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UPSC 2017 Maths Optional Paper 1 Q7c — Step-by-Step Solution

17 marks · Section B

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

Question

A particle is free to move on a smooth vertical circular wire of radius $a$. At time $t=0$ it is projected along the circle from its lowest point $A$ with velocity just sufficient to carry it to the highest point $B$. Find the time $T$ at which the reaction between the particle and the wire is zero.

Technique

Bead “just reaches top” $\Rightarrow v_B=0\Rightarrow v_A^2=4ga$; radial Newton law gives $N$; quadrature $dt=\tfrac12\sqrt{a/g}\sec(\theta/2)d\theta$ gives $T$.

Solution

The particle is threaded on a wire (bead), so the wire can push either inward or outward; “just sufficient to reach $B$” therefore means the speed becomes zero at the top (not the string condition $v^2=ga$).

Step 1 — Projection speed

Energy from $A$ (lowest) to $B$ (highest, height $2a$), arriving with $v_B=0$:

Step 2 — Speed at angle $\theta$

Let $\theta$ be measured from the centre $O$, from the downward vertical $OA$. Height above $A$ is $a(1-\cos\theta)$, so by energy conservation

Step 3 — Reaction and where it vanishes

Resolve along the radius (centripetal, towards $O$). The component of gravity along the outward radius is $g\cos\theta$ (at $\theta$ from the bottom). Newton’s law towards the centre, with $N$ the reaction measured positive outward, gives

The reaction vanishes when

(This is above the horizontal level $\theta=\tfrac\pi2$, in the upper half — consistent with the bead pressing on opposite sides above and below.)

Step 4 — Time to reach $\cos\theta=-\tfrac23$

Since $v=a\dot\theta$ and $v^2=2ga(1+\cos\theta)=4ga\cos^2(\theta/2)$,

Integrate from $\theta=0$ to $\theta_1$ where $\cos\theta_1=-\tfrac23$:

Now $\cos\theta_1=-\tfrac23$ gives $\cos^2\tfrac{\theta_1}2=\tfrac{1+\cos\theta_1}{2}=\tfrac16$ and $\sin^2\tfrac{\theta_1}2=\tfrac{1-\cos\theta_1}2=\tfrac56$, so

Answer