Question

A smooth semicircular wire track of radius $R$ is fixed in a vertical plane. One end of a massless spring of natural length $\frac{3 R}{4}$ 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 ring is held stationary at point $P$ such that the spring makes an angle of $60 ^\circ $ with the vertical. The spring constant $K=\frac{m g}{R}$ . Consider the instant when the ring is released. The normal reaction on the ring by the track is

• A. $\frac{3 mg}{8}$
• B. $mg$
• C. $\frac{m g}{4}$
• D. $\frac{3 mg}{4}$

Answer 1

Answer: (A)

ln $\Delta OCP,OC=CP=R.$
$\therefore $ The triangle is isosceles
$\therefore \angle COP=\angle CPO=60^\circ \Rightarrow \angle OCP=60^\circ $
$\therefore \Delta OCP$ is an equilateral triangle
$\Rightarrow OP=R$
$\therefore $ Extension of string $=R-\frac{3 R}{4}=\frac{R}{4}=x$
The forces acting are shown in the figure (i)
The free-body diagram of the ring is shown in fig (ii)
Force in the tangential direction
$=Fcos30^\circ +mgcos30^\circ =\left[k x + m g\right]cos30^\circ $
$\therefore F_{t}=\frac{5 mg}{8}\sqrt{3}$
$\therefore F_{t}=ma_{t}$
$\Rightarrow \frac{5 mg \sqrt{3}}{8}=ma_{t}$
$\Rightarrow a_{t}=\frac{5 \sqrt{3}}{8}g$
Also when the ring is just released
$N+Fsin30^\circ =mgsin30^\circ $
$=\left(m g - \frac{m g}{4}\right)\times \frac{1}{2}=\frac{3 mg}{8}$