A smooth semicircular wire-track of radius R is fixed in a vertical plane. One end of a 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 of $60^\circ$ with the vertical. The spring constant $K=mg/R$. Consider the instant when the ring is released, and (i) draw the free body diagram of the ring, (ii) determine the tangential acceleration of the ring and the normal reaction.

Step 1: Analyze the geometry of the system.

Let $C$ be the center of the semicircular track and $O$ be the lowest point.

$$OC = CP = R$$ $$\angle COP = 60^\circ$$

Since $OC = CP$, the triangle $OCP$ is isosceles, which means the base angles are equal.

$$\angle OPC = \angle COP = 60^\circ$$

Thus, triangle $OCP$ is equilateral, and the length of the spring $OP$ is equal to the radius.

$$OP = R$$

Step 2: Calculate the spring force.

The extension $x$ of the spring is the difference between its stretched length and its natural length.

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

Using Hooke's Law with spring constant $K = \frac{mg}{R}$, the spring force $F_s$ is:

$$F_s = Kx = \left(\frac{mg}{R}\right)\left(\frac{R}{4}\right) = \frac{mg}{4}$$

This force is directed along the line $PO$ toward the lowest point $O$.

Step 3: Identify the forces for the Free Body Diagram (FBD).

At the instant of release, three forces act on the ring at point $P$.

• Weight $mg$ acting vertically downward.
• Spring force $F_s = \frac{mg}{4}$ acting along the chord $PO$ toward $O$.
• Normal reaction $N$ acting radially along the line $PC$ toward the center $C$.

Step 4: Calculate the tangential acceleration.

Resolve the forces along the tangent to the track at $P$ in the direction of $O$.

$$F_t = mg \cos 30^\circ + F_s \cos 30^\circ$$ $$F_t = mg \left(\frac{\sqrt{3}}{2}\right) + \frac{mg}{4} \left(\frac{\sqrt{3}}{2}\right)$$ $$F_t = \frac{5\sqrt{3}}{8}mg$$

Apply Newton's second law to find the tangential acceleration $a_t$.

$$a_t = \frac{F_t}{m} = \frac{5\sqrt{3}}{8}g$$

Step 5: Calculate the normal reaction.

At the instant of release, the velocity of the ring is $v = 0$, so the radial acceleration is zero.

$$a_r = \frac{v^2}{R} = 0$$

Set the sum of the radial forces toward the center $C$ to zero.

$$N + F_s \sin 30^\circ - mg \cos 60^\circ = 0$$ $$N = mg \cos 60^\circ - F_s \sin 30^\circ$$ $$N = mg \left(\frac{1}{2}\right) - \frac{mg}{4} \left(\frac{1}{2}\right)$$ $$N = \frac{3mg}{8}$$

Answer:

(i) Free Body Diagram forces:

• Weight $mg$ acting vertically downward
• Spring force $F_s = \dfrac{mg}{4}$ acting along the chord $PO$ toward $O$
• Normal reaction $N$ acting radially inward toward the center $C$

(ii) Acceleration and Normal force:

• Tangential acceleration: $a_t = \dfrac{5\sqrt{3}}{8}g$ (directed toward $O$)
• Normal reaction: $N = \dfrac{3mg}{8}$ (directed toward $C$)