UPSC 2017 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Work-energy theorem · Dynamics & Statics · asked 2× in 13 yrs · Read the full method →

Question

A fixed wire is in the shape of the cardioid $r=a(1+\cos\theta)$ , the initial line being the downward vertical. A small ring of mass $m$ can slide on the wire and is attached to the point $r=0$ of the cardioid by an elastic string of natural length $a$ and modulus of elasticity $4mg$ . The string is released from rest when the string is horizontal. Show by using the laws of conservation of energy that $a\dot\theta^2(1+\cos\theta)-g\cos\theta(1-\cos\theta)=0$ , $g$ being the acceleration due to gravity.

Technique

Energy conservation $T+V_g+V_{\text{el}}=\text{const}$ ; release condition fixes the constant at $\theta=\pi/2$ .

Solution

The ring sits at $\big(r,\theta\big)=\big(a(1+\cos\theta),\theta\big)$ , with $\theta$ measured from the initial line, which points vertically downward. The string runs straight from the pole $O$ to the ring, so its length equals $r$ .

Step 1 — Kinetic energy

With $r=a(1+\cos\theta)$ , $\dot r=-a\sin\theta\,\dot\theta$ . In plane polar coordinates the speed satisfies

v^2=\dot r^2+r^2\dot\theta^2=a^2\sin^2\theta\,\dot\theta^2+a^2(1+\cos\theta)^2\dot\theta^2.

Expand $(1+\cos\theta)^2=1+2\cos\theta+\cos^2\theta$ and add $\sin^2\theta$ :

v^2=a^2\dot\theta^2\big(\sin^2\theta+1+2\cos\theta+\cos^2\theta\big)=2a^2\dot\theta^2(1+\cos\theta).

T=\tfrac12 m v^2=m a^2\dot\theta^2(1+\cos\theta).

Step 2 — Gravitational potential energy

The initial line is the downward vertical, so the depth of the ring below $O$ (distance measured downward) is $r\cos\theta=a(1+\cos\theta)\cos\theta$ . Taking $O$ as datum (PE increases upward),

V_g=-mg\,(r\cos\theta)=-mg\,a(1+\cos\theta)\cos\theta.

Step 3 — Elastic potential energy

Natural length $a$ , modulus $\lambda=4mg$ . The stretched length is $r=a(1+\cos\theta)$ , so the extension is

e=r-a=a\cos\theta\qquad(0\le\theta\le\tfrac\pi2,\ \text{string taut}).

Elastic PE $=\dfrac{\lambda}{2a}e^2=\dfrac{4mg}{2a}\,(a\cos\theta)^2=2mg\,a\cos^2\theta.$

Step 4 — Apply conservation of energy

Released from rest ( $\dot\theta=0$ ) when the string is horizontal, i.e. $r$ horizontal $\Rightarrow\theta=\tfrac\pi2$ . There $\cos\theta=0$ , so $e=0$ , $V_g=0$ , $T=0$ ; total energy at release is $E_0=0$ . Conservation $T+V_g+V_{\text{el}}=E_0=0$ gives

m a^2\dot\theta^2(1+\cos\theta)-mg\,a(1+\cos\theta)\cos\theta+2mg\,a\cos^2\theta=0.

Divide by $am$ and collect the cosine terms:

a\dot\theta^2(1+\cos\theta)-g\cos\theta\big[(1+\cos\theta)-2\cos\theta\big]=0,

Answer

\boxed{\,a\dot\theta^2(1+\cos\theta)-g\cos\theta(1-\cos\theta)=0.\,}