Physics - W.E.P. Question with Solution | TestHub

$\text{In }\Delta \text{OCP},\text{ OC }=\text{ CP }=\text{ R}.$

$\therefore$        The triangle is isosceles

$\therefore$         $\angle COP=$  $\angle CPO=60°$  $\Rightarrow \angle OCP=60°$

$\therefore$         $\Delta OCP$ is an equilateral triangle

$\Rightarrow$        OP = R

$\therefore$      Extension of string = $R–\frac{\mathrm{}3R}{4}=\frac{R}{4}=x$

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

$=F\cos 30°+mg\cos 30°=\left[kx+mg\right]\cos 30°$

$\therefore F_t=\frac{\mathrm{}5mg}{8}\sqrt{3}$                               $\therefore$  Ft = mat

$\Rightarrow \frac{\mathrm{}5mg\sqrt{3}}{8}=m{a}_t$                              $\Rightarrow a_t=\frac{5\sqrt{3}}{8}g$

Also when the ring is just released

N + F sin30º = mg sin 30º

$=\left(mg–\frac{mg}{4}\right)\times \frac{1}{2}=\frac{\mathrm{}3mg}{8}$