Problem 31 - Master Resource Book in JEE Main Physics

Question

A point \(P(R \sqrt{3}, 0,0)\) lies on the axis of a ring of mass \(M\) and radius \(R\). The ring is located in \(y-z\) plane with its centre at origin \(O .\) A small particle of mass \(m\) starts from \(P\) and reaches \(O\) under gravitational attraction only. Its speed at \(O\) will be (a) \(\sqrt{\frac{G M}{R}}\) (b) \(\sqrt{\frac{G m}{R}}\) (c) \(\sqrt{\frac{G M}{\sqrt{2} R}}\) (d) \(\sqrt{\frac{G m}{\sqrt{2} R}}\)

Step-by-Step Solution

Answer

The speed of the particle at the origin is \(\sqrt{\frac{G M}{R}}\), corresponding to option (a).

Step 1: Understand the Problem

We need to determine the speed of a small particle of mass \(m\) as it reaches the center \(O\) of a ring due to gravitational attraction. The point \(P\) is initially located at \((R \sqrt{3}, 0, 0)\), and the center of the ring is at the origin \((0, 0, 0)\) on the \(y-z\) plane.

Step 2: Gravitational Potential Energy at Point P

The gravitational potential energy \(U\) at a point due to a ring of mass \(M\) is given by \(U = - \frac{G M m}{\sqrt{R^2 + x^2}}\), where \(x\) is the distance from the center along the axis. Since the point \(P\) is at \((R \sqrt{3}, 0, 0)\), we have \(x = R \sqrt{3}\), leading to \(U = - \frac{G M m}{\sqrt{R^2 + (R \sqrt{3})^2}} = - \frac{G M m}{2R}\).

Step 3: Gravitational Potential Energy at Origin

At the center of the ring, the gravitational potential energy \(U_O\) is \(U_O = - \frac{G M m}{R}\) since the distance to the center in the plane is zero.

Step 4: Apply Conservation of Energy

The change in mechanical energy is zero, so the initial potential energy plus the initial kinetic energy equals the final potential energy plus the final kinetic energy. Initially, the particle is at rest, so the kinetic energy is zero, and the potential energy at \(P\) is \(- \frac{G M m}{2R}\). At \(O\), the kinetic energy is \(\frac{1}{2}m v^2\), and the potential energy is \(- \frac{G M m}{R}\).

Step 5: Solve for Final Speed

Applying the conservation of energy: \[- \frac{G M m}{2R} = \frac{1}{2}m v^2 - \frac{G M m}{R}\] Rearranging terms gives: \[\frac{1}{2} m v^2 = - \frac{G M m}{R} + \frac{G M m}{2R}\] \[\frac{1}{2} m v^2 = \frac{G M m}{2R}\] Thus, \[v^2 = \frac{G M}{R}\]The speed \(v\) is \(\sqrt{\frac{G M}{R}}\).

Step 6: Select the Correct Answer

From the options given, the expression for the speed \(\sqrt{\frac{G M}{R}}\) corresponds to option (a). Hence, the correct answer is (a).

Key Concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is a crucial concept in mechanics and physics. It represents the energy an object possesses due to its position in a gravitational field. For an object at a specific height above the ground or any reference point, the gravitational potential energy is the work done to bring it to that height against the force of gravity. In simpler terms, it's the energy stored due to an object's position relative to a gravitational source, like Earth or in our case, a ring.
Unlike kinetic energy, which involves motion, GPE is about position. In the context of the original exercise, a particle is influenced by the gravitational field of a ring. When at point \( P \), the particle has a specific gravitational potential energy computed using the formula \[ U = - \frac{G M m}{\sqrt{R^2 + x^2}} \] where \( G \) is the universal gravitational constant, \( M \) is the ring's mass, \( m \) is the particle's mass, and \( x \) is the distance from the center of the ring. The negative sign indicates that the gravitational force is attractive, pulling the particle towards the center.

• Gravitational potential energy is proportionate to the objects' masses and inversely proportional to the distance between them.
• A smaller distance means stronger gravitational attraction, hence more negative GPE.

Conservation of Energy

The principle of conservation of energy states that the total energy in an isolated system remains constant. This means energy can neither be created nor destroyed but can only change forms. In the realm of mechanics, this principle plays an essential role in understanding how energy transforms from potential to kinetic and vice versa.
Applying this to our exercise, the particle initially at rest at point \( P \) starts off with pure gravitational potential energy. In the absence of any external work, this energy will convert into kinetic energy as the particle moves towards the center \( O \) of the ring.
At the start, all energy is potential, described by \[ U = - \frac{G M m}{2R} \]
At point \( O \), some of this potential energy has converted into kinetic energy as the particle gains speed:

• Initial Energy: purely gravitational potential \( - \frac{G M m}{2R} \)
• Final Energy: a mix of kinetic \( \frac{1}{2}m v^2 \) and potential \( - \frac{G M m}{R} \)

Setting initial energy equals final energy allows us to solve for the particle's speed upon reaching the center. This transformation conserves the total mechanical energy, reaffirming the essential law of conservation.

Mechanics in Physics

Mechanics in physics focuses on motion and the forces that induce these changes. This fundamental branch is vital for understanding actions in the physical world, particularly when it involves interactions such as gravitational pull.
In the given exercise, mechanics principles help explain how a particle moves under the gravitational influence of a massive ring. The ring creates a gravitational field that affects the particle's motion, pulling it from rest to a state of motion.

• Newton's Law of Gravitation: This law shows how two masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
• Equations of Motion: These equations describe the state of the particle, its velocity, acceleration, and changing position over time under the influence of gravitational forces.

As the particle is drawn towards the ring's center, its potential energy decreases (becomes more negative) while its kinetic energy increases. The exercise beautifully illustrates how mechanical concepts such as energy transformation and gravitational forces come into play, enabling us to calculate motion parameters like velocity at the center \( O \). Mechanics in physics thus not only explains "how" but also helps us predict outcomes in gravitational systems.