A smooth semicircular wire track of radius R is fixed in a vertical plane (figure). One end of a mas

Q: Wait, why is the triangle CPO equilateral?

Since $CO$ and $CP$ are both radii of the semicircular track, they are equal in length. This makes $\Delta CPO$ an isosceles triangle. We are given that the angle between the spring ($OP$) and the vertical ($CO$) is $60^\circ$. In an isosceles triangle, if one of the base angles is $60^\circ$, the other must also be $60^\circ$, leaving $180^\circ - 120^\circ = 60^\circ$ for the third angle. Thus, all angles are $60^\circ$, making it equilateral.

Q: Why is the centripetal acceleration zero?

Centripetal acceleration is given by the formula $a_c = \frac{v^2}{R}$. The problem asks for the forces and acceleration *just after* the ring is released. At this exact instant, the ring hasn't started moving yet, so its velocity $v = 0$. Consequently, $a_c = 0$.

Q: How do we know the normal reaction acts inwards?

We initially assume a direction for the normal reaction. If we assume it acts inwards (towards the center $C$), our force balance equation is $N + F_s \cos 60^\circ = mg \cos 60^\circ$. Solving this gives $N = \frac{3mg}{8}$. Since the result is positive, our initial assumption was correct. If it had been negative, it would mean the track is pushing the ring outwards.

Q: Why do we resolve forces along the tangent and normal instead of horizontal and vertical?

The ring is constrained to move along the circular track. Its acceleration will always be purely tangential (since $v=0$ initially). By resolving forces along the tangent and normal, we directly align our equations with the ring's possible motion, making the math much simpler.

Q: Why does the spring force have a component along the tangent?

The spring pulls the ring towards the lowest point $O$. The line $PO$ is not purely radial; it cuts across the circle. Because it's at an angle to the normal (radius $CP$), it has a component pulling the ring inwards (normal) and a component pulling it down the track (tangential).

Question

A smooth semicircular wire track of radius $R$ is fixed in a vertical plane (figure). 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 $60^\circ$ with the vertical. The spring constant $k = \frac{mg}{R}$. Consider the instant when the ring is making an angle $60^\circ$ with the vertical.

The spring is released, (a) Draw the free body diagram of the ring. (b) Determine the tangential acceleration of the ring and the normal reaction. (1996)

Sigma Percentile · Accepted Answer

### The Geometry of Forces: Unraveling the Semicircular Track and Spring Problem

Imagine you are standing on a massive semicircular track, holding a stretched spring attached to a ring. The moment you let go, a beautiful interplay of geometry, Hooke's Law, and Newton's Laws of Motion dictates exactly how that ring will move. I know this geometry looks terrifying at first glance, but let's take a breath and break it down step by step.

### Decoding the Hidden Geometry

Before we even think about forces, we must understand the spatial reality of our setup. We have a semicircular track with its center at $C$ and its lowest point at $O$. The ring is located at point $P$, and the spring connects $O$ to $P$.

We are given a crucial piece of information: the spring makes an angle of $60^\circ$ with the vertical. Since the vertical line passes through both $C$ and $O$, this means the angle $\angle COP = 60^\circ$.

Now, look at the triangle formed by $C$, $P$, and $O$. The sides $CO$ and $CP$ are both radii of the semicircular track, meaning $CO = CP = R$. This makes $\Delta CPO$ an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides must also be equal. Since one of the base angles is $60^\circ$, the other must be $60^\circ$ as well. This leaves exactly $60^\circ$ for the third angle.

Suddenly, the geometry simplifies beautifully: $\Delta CPO$ is an **equilateral triangle**! This means the length of the spring, $OP$, is exactly equal to the radius $R$.

### The Spring's Awakening

Now that we know the current length of the spring ($OP = R$), we can determine how much it has been stretched. The problem states the natural length of the spring is $l_0 = \frac{3R}{4}$.

The extension in the spring is simply the difference between its current length and its natural length:

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

With the extension known, we can unleash Hooke's Law. The spring constant is given as $k = \frac{mg}{R}$. The magnitude of the spring force $F_s$ pulling the ring towards $O$ is:

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

### The Art of Resolving Forces

To predict the ring's motion, we must resolve all acting forces. But standard horizontal and vertical axes will only make our lives miserable here. Because the ring is constrained to move along a circular track, we must use a local coordinate system: one axis **tangent** to the track, and one **normal** (radial) to it.

Let's start with gravity. Gravity acts straight down with a force of $mg$. The angle between the vertical and the normal line $CP$ is $\angle OCP = 60^\circ$. Therefore, gravity has a component pushing the ring *outward* against the track ($mg \cos 60^\circ$) and a component pulling the ring *down* the track ($mg \sin 60^\circ$).

Next, we resolve the spring force. The spring pulls the ring along the line $PO$. Because $\Delta CPO$ is equilateral, the angle between $PO$ and the inward normal $PC$ is $\angle CPO = 60^\circ$. Thus, the spring force has a component pulling the ring *inward* ($F_s \cos 60^\circ$) and a component pulling it *down* the track ($F_s \sin 60^\circ$).

### The Master Equations

Now we assemble our equations of motion. Let's look at the tangential direction first. Both gravity and the spring are working together to accelerate the ring down the track. The net tangential force $F_t$ is:

$$F_t = mg \sin 60^\circ + F_s \sin 60^\circ$$

Substituting our value for $F_s$:

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

Using Newton's Second Law ($F = ma$), the tangential acceleration $a_t$ is simply $\frac{5\sqrt{3}}{8}g$.

Finally, we determine the normal reaction $N$. The problem asks for the forces *just after* release. At this exact instant, the ring hasn't started moving, so its velocity $v = 0$. This means the centripetal acceleration ($a_c = \frac{v^2}{R}$) is zero, and the net force along the normal direction must perfectly balance out.

Assuming the track pushes the ring inward with a normal reaction $N$, we balance the inward and outward forces:

$$N + F_s \cos 60^\circ = mg \cos 60^\circ$$

Rearranging to solve for $N$:

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

Because our result for $N$ is positive, our assumption was correct: the track indeed pushes the ring radially inward. The harmony of geometry and physics in this problem is truly spectacular!

Visualized Solution

Visualizing the Setup

Geometry of the Triangle

The Equilateral Triangle

Calculating the Spring Extension

Magnitude of Spring Force

Setting up the Axes

Resolving Gravity

Resolving the Spring Force

Tangential Acceleration

Normal Reaction Equation

Calculating Normal Reaction

The Sigma Insight: Dynamics of Circular Motion

The Geometry of Forces

Decoding the Hidden Geometry

The Spring's Awakening

The Art of Resolving Forces

The Master Equations

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Comprehension Passage

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