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.

Here, $OP = r, PM = r\cos\theta = a(1+\cos\theta)\cos\theta$

KE = $\frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) \\ \frac{1}{2}m(a^2\sin^2\theta + a^2(1+\cos\theta)^2)\dot{\theta}^2 \\ ma^2(1+\cos\theta)\dot{\theta}^2$

PE due to gravity $= -mg (PM) = -mga(1+\cos\theta)\cos\theta$

PE due to extension in string $= \frac{1}{2} \frac{4mg}{l} (r-a)^2 \\ = \frac{1}{2} \frac{4mg}{a} (a\cos\theta)^2 \\ = 2mga \cos^2\theta$

Thus, by conservation of energy,

KE + PE due to gravity + PE due to extension $= 0$

$\implies ma^2(1+\cos\theta)\dot{\theta}^2 - mga(1+\cos\theta)\cos\theta + 2mga \cos^2\theta = 0$

$\implies a(1+\cos\theta)\dot{\theta}^2 - g(\cos\theta + \cos^2\theta - 2\cos^2\theta) = 0$

$\implies a(1+\cos\theta)\dot{\theta}^2 - g\cos\theta(1-\cos\theta) = 0$

Hence, proved.

answered Dec 23, 2017 by (1,920 points)