Question

A smooth semicircular wire track of radius $R$ is fixed in a vertical plane as shown in the figure. One end of a massless spring of natural length $\frac{3}{4}R$ is attached to the lowest point $O$ of the wire track. A small ring of mass m, which can slide on the track, is attached to the other end of the spring. The spring makes an angle of ${60}^0$ with the vertical. The spring constant is $k=\frac{mg}{R}$. Consider the instant when the ring is making an angle of ${60}^0$ with the vertical. Find the tangential acceleration of the ring.

• A. $\frac{g5\sqrt{3}}{8}$
• B. $\frac{g3\sqrt{3}}{8}$
• C. $\frac{g5\sqrt{3}}{4}$
• D. $\frac{g3\sqrt{3}}{4}$

Answer 1

The correct option is A $\frac{g5\sqrt{3}}{8}$

Given:
Natural length of the spring = $\frac{3R}{4}$
Spring constant $k=\frac{mg}{R}$

First of all let us try to find the length by which spring is stretched or compressed at this moment (at ${60}^0$ with the vertical)
Length of spring = $R$ (from the figure)
$\therefore x=|R-\frac{3R}{4}|=\frac{R}{4}$

Let us now draw the free body diagram of the ring


Now resolve these forces in radial and tangential direction

The components of $Kx$ and $mg$ along radial and tangential directions will be as shown Net tangential force acting on the ring

= $kx\cos {30}^0+mg\cos {30}^0$

$⟹m{a}_t=\frac{mg}{R}\times \frac{R}{4}\times \frac{\sqrt{3}}{2}+mg\frac{\sqrt{3}}{2}$

$⟹m{a}_t=\frac{mg5\sqrt{3}}{8}$

$\therefore a_t=\frac{5\sqrt{3}}{8}g$