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 = mg/R. The instant when the ring is released

• A. tangential acceleration of the ring is $\frac{5\sqrt{3}}{8}g$
• B. normal reaction of the ring is $\frac{3mg}{8}$
• C. tangential acceleration of the ring is $\frac{11\sqrt{3}}{7}g$
• D. normal reaction of the ring is $\frac{7\sqrt{11}}{8}mg$

Answer 1

Answer: (A), (B)

The correct options are
A tangential acceleration of the ring is $\frac{5\sqrt{3}}{8}g$
B normal reaction of the ring is $\frac{3mg}{8}$


i) CP = CO = Radius of circle = R

$\angle CPO=\angle POC={60}^{\circ }=\angle OCP$

OCP is an equilateral triangle.

$\therefore$ OP = R = Extended length of spring.

Natural length of spring $=\frac{3R}{4}$

$\therefore$ Extension $x=R-\frac{3R}{4}=\frac{R}{4}$

Let spring force = F

$F=Kx$, where $K=\left(\frac{mg}{R}\right)$ is given

or $F=\left(\frac{mg}{R}\right)\left(\frac{R}{4}\right)=\frac{mg}{4}$

The free body diagram of the ring is shown in the figure.



Here $F=Kx=\frac{mg}{4}$

N = Normal Reaction

(ii) Tangential acceleration $a_T$:

As soon as the ring is released it will move towards x-axis.

$\therefore F_x=Fsin{60}^{\circ }+mgsin{60}^{\circ }$

or $F_x=\left(\frac{mg}{4}\right)\left(\frac{\sqrt{3}}{2}\right)+mg\left(\frac{\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{8}mg$

$a_T=a_x$ or $a_T=\frac{F_x}{m}=\frac{5\sqrt{3}}{8}g$

$\therefore$ Tangential acceleration when the ring is released $=\frac{5\sqrt{3}}{8}g$

Normal reaction:

As soon as the ring is released, the net force along y-axis is zero.

$F_y=0$

$\therefore N+Fcos{60}^{\circ }=mgcos{60}^{\circ }$

or $N=mgcos{60}^{\circ }-Fcos{60}^{\circ }$

or $N=\frac{mg}{2}-\left(\frac{mg}{4}\right)\left(\frac{1}{2}\right)=\frac{3mg}{8}$