Question

A smooth semicircular wire track of radius $R$ is fixed in a vertical plane. One end of a massless spring of natural length $\frac{3R}{4}$ 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 $K=\frac{mg}{R}$ . Consider the instant when the ring is released. The normal reaction on the ring by the track is

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

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 $=R-\frac{3R}{4}=\frac{R}{4}=x$
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
$=Fcos30^{\circ }+mgcos30^{\circ }=\left[kx+mg\right]cos30^{\circ }$
$\therefore F_t=\frac{5mg}{8}\sqrt{3}$
$\therefore F_t=m{a}_t$
$\Rightarrow \frac{5mg\sqrt{3}}{8}=m{a}_t$
$\Rightarrow a_t=\frac{5\sqrt{3}}{8}g$
Also when the ring is just released
$N+Fsin30^{\circ }=mgsin30^{\circ }$
$=\left(mg-\frac{mg}{4}\right)\times \frac{1}{2}=\frac{3mg}{8}$