Problem 14 - Master Resource Book in JEE Main Physics

Question

Three particles each of mass \(m\) rotate in a circle of radius \(r\) with uniform angular speed \(\omega\) under their mutual gravitational attraction. If at any instant the points are on the vertex of an equilateral of side \(L\), then angular velocity \(\omega\) is (a) \(\sqrt{\frac{2 G m}{L^{3}}}\) (b) \(\sqrt{\frac{3 G m}{L^{3}}}\) (c) \(\sqrt{\frac{5 G m}{L^{3}}}\) (d) \(\sqrt{\frac{G m}{L^{3}}}\)

Step-by-Step Solution

Answer

Angular velocity \( \omega \) is \( \sqrt{\frac{G m}{L^3}} \), option (d).

Step 1: Identify Forces Acting on Each Particle

Each particle is subjected to gravitational forces due to the other two particles. Since the particles are at the vertices of an equilateral triangle, we need to consider gravitational force exerted by one point-mass on another.

Step 2: Calculate Gravitational Force Between Two Particles

The gravitational force between two particles of mass \(m\) at a distance \(L\) is given by Newton's gravitational law: \( F = \frac{G m^2}{L^2} \), where \(G\) is the gravitational constant. This force acts along the side of the equilateral triangle.

Step 3: Resultant Force Towards Center

For each particle, the resultant force due to the other two particles is directed towards the center of the circle. Since the angle between two gravitational forces is 60°, use the law of cosines for vectors to find the resultant force magnitude as \( F_{ ext{res}} = 2 F imes \cosrac{60°}{2} = F \).

Step 4: Centripetal Force Requirement

The net gravitational force provides the centripetal force necessary for circular motion. Set the centripetal force required: \( F_c = m r \omega^2 \). The side of the triangle \( L \) satisfies that \( L = 2 imes r \). Include this relationship in the equation.

Step 5: Express Angular Velocity in Terms of Known Variables

Substitute \( F_{ ext{res}} = F_c \): \( \frac{G m^2}{r^2} = m r \omega^2 \). Simplify to express \( \omega \): \( \omega = \sqrt{\frac{G m}{L^3}} \).

Step 6: Determine Correct Option

The derived expression for \( \omega \) matches option (d): \( \sqrt{\frac{G m}{L^3}} \). Therefore, the angular velocity \( \omega \) is given by option (d).

Key Concepts

Gravitational Force

Gravitational force is the attractive force that acts between any two masses. It's a fundamental force described by Newton's law of gravitation. According to this law, the gravitational force \( F \) between two masses is calculated using the formula: \[ F = \frac{G m_1 m_2}{r^2} \] where

• \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
• \( m_1 \) and \( m_2 \) are the masses of the two objects
• \( r \) is the distance between the centers of the two masses

When the particles are on the vertices of an equilateral triangle, each particle feels the gravitational pull due to the other two particles. It's important to note that even though gravitational force is relatively weak on small scales, it is essential for maintaining the equilibrium and motion within large systems, like planets and astronomical bodies.

Circular Motion

Circular motion occurs when an object moves along a circular path. In this exercise, each particle follows circular motion due to both gravitational forces and the resultant force acting towards the center of the circle formed. Characteristics of circular motion include:

• Uniform speed, where the magnitude of the velocity is constant but its direction changes.
• Radius \( r \), the distance from the center of the circle to the particle.
• Angular velocity \( \omega \), the rate of change of the angular position with respect to time.

The concept of circular motion is pivotal in identifying how forces are balanced, especially when considering how gravitational influences govern the path each particle takes while orbiting along a circle. The stability of such motion highlights the equilibrium achieved by the forces involved.

Centripetal Force

Centripetal force is the inward force required to keep an object moving in a circular path. This force is necessary to constantly change the direction of the object's velocity, ensuring it remains on a curved trajectory rather than moving off tangent.In our scenario, the centripetal force \( F_c \) required for circular motion is \[ F_c = m r \omega^2 \] where

• \( m \) is the mass of the particle
• \( r \) is the radius of the circular path
• \( \omega \) is the angular velocity

The gravitational force between the particles acts as the centripetal force needed to maintain circular motion. By setting the gravitational force equal to the expression for centripetal force, we can derive expressions such as the angular velocity \( \omega \) in terms of known properties of the system. This shows how closely these types of forces interrelate and rely on each other to define the motion involved.

Equilateral Triangle Geometry

An equilateral triangle is a triangle where all three sides are of equal length, and all internal angles are 60 degrees. This symmetry is key to simplifying the force analysis in our exercise.When particles are located at the vertices of an equilateral triangle, the geometry allows for specific vector relationships and simplifications:

• Each gravitational force between particles can be easily calculated due to equal side lengths.
• The geometric symmetry helps in determining the resultant forces, using the law of cosines for vectors since each angle is 60 degrees, simplifying to use relationships like \( \cos(60^\circ) = \frac{1}{2} \).
• The distance from the center to any vertex (which is part of determining the circle's radius) can be found using geometric insights.

This geometric setup not only makes the mathematics more manageable but also provides insights into how symmetry and geometric properties can be leveraged in physical systems to predict behaviors and outcomes, such as angular velocity.