Question

A smooth semicircular wire track of radius $R$ is fixed in a vertical plane. One end of the massless spring of natural length $3R/4$ 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 ${60}^{\circ }$ with the vertical. Spring constant $K=mg/R$. The spring force is :

• A. $\frac{mg}{3}$
• B. $mg$
• C. $\frac{mg}{2}$
• D. $\frac{mg}{4}$

Answer 1

The correct option is D $\frac{mg}{4}$

Here, $CP=CO=R$ and $\angle CPO=\angle POC$
Thus , $\Delta CPO$ is an equilateral triangle. So, $OP=R$
Extension of the spring , $x=OP-$neutral length of spring$=R-\frac{3R}{4}=\frac{R}{4}$
Therefore spring force $=kx=\left(mg/R\right)\times \left(R/4\right)=\frac{mg}{4}$