Problem 20 - Master Resource Book in JEE Main Physics

Question

An object is tied to a string and rotated in a vertical circle of radius \(r .\) Constant speed is maintained along the trajectory. If \(T_{\max } / T_{\min }=2\), then \(v^{2} / r g\) is (a) 1 (b) 2 (c) 3 (d) 4

Step-by-Step Solution

Answer

(c) 3

Step 1: Understanding the Problem

We need to find the ratio \( \frac{v^2}{rg} \) where \( v \) is the speed of the object, \( r \) is the radius of the circle, and \( g \) is the acceleration due to gravity. We have \( \frac{T_{\max}}{T_{\min}} = 2 \), where \( T_{\max} \) and \( T_{\min} \) are the maximum and minimum tension in the string, typically occurring at the bottom and top of the circle respectively.

Step 2: Identify Forces at the Top and Bottom

At the top of the circle, the tension \( T_{\text{top}} \) and gravitational force act along the same direction towards the center, so the net force equation is:\[ T_{\text{top}} + mg = \frac{mv^2}{r} \]At the bottom of the circle, the tension \( T_{\text{bottom}} \) acts towards the center, while gravity acts in opposite, so:\[ T_{\text{bottom}} - mg = \frac{mv^2}{r} \]

Step 3: Relate Maximum and Minimum Tension

From the problem statement, we have \( \frac{T_{\text{bottom}}}{T_{\text{top}}} = 2 \). This relationship will help us find the expression for the speed squared, \( v^2 \), in terms of \( rg \).

Step 4: Express Tensions in Terms of Speed and Weight

Rewrite the tension equations:At the top: \[ T_{\text{top}} = \frac{mv^2}{r} - mg \]At the bottom:\[ T_{\text{bottom}} = \frac{mv^2}{r} + mg \]Now substitute expressions into \( \frac{T_{\text{bottom}}}{T_{\text{top}}} = 2 \) :\[ \frac{\frac{mv^2}{r} + mg}{\frac{mv^2}{r} - mg} = 2 \]

Step 5: Solve the Equation

Simplify and solve the equation:Cross-multiply to eliminate the fraction:\[ mv^2 + mgr = 2(mv^2 - mgr) \]Expand and simplify:\[ mv^2 + mgr = 2mv^2 - 2mgr \]Combine like terms:\[ mv^2 - 2mv^2 = -2mgr - mgr \]\[ -mv^2 = -3mgr \]Divide by \(-m\):\[ v^2 = 3rg \]

Step 6: Find the Required Ratio

We are asked to find \( \frac{v^2}{rg} \). Given \( v^2 = 3rg \), we find:\[ \frac{v^2}{rg} = \frac{3rg}{rg} = 3 \]

Key Concepts

Tension in a String

Understanding tension in a string during vertical circular motion is crucial. Tension is the force exerted by the string on the object to keep it moving in a circle. As the object moves along the vertical circular path, the tension varies at different points.
In the case of an object moving vertically in a circular path, the tension in the string depends on two main factors:

• The speed of the object
• The gravitational force acting on the object

At the top of the circle, the tension is at its minimum because both the tension and gravity act towards the center. Conversely, at the bottom, the tension is maximum since gravity acts against the direction of the tension. This variation in tension is what balances the forces to maintain the object's circular motion.

Circular Motion Dynamics

Circular motion dynamics explores the forces that keep an object moving in a circular path. In vertical circular motion, an object tied to a string and moving in a circle experiences a dynamic interplay of forces that must be understood.The key points of consideration in circular motion include:

• Centripetal Force: The force required to keep the object moving in a circle.
• Net Force at Each Point: How the sum of gravitational force and tension equate to provide the necessary centripetal force.

The centripetal force is expressed as:\[ F_c = \frac{mv^2}{r} \]where:

• \( m \) is the mass of the object,
• \( v \) is its speed,
• \( r \) is the radius of the circle.

This equation tells us the amount of force needed to change the object's direction continuously, thus maintaining the circular motion.

Gravitational Force

Gravitational force is a fundamental concept influencing the motion of the object in vertical circular motion. It acts downwards, contributing to the dynamics of tension within the string.Gravitational force is expressed as:\[ F_g = mg \]where:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity, roughly equal to \( 9.8 \text{ m/s}^2 \) on Earth.

In a vertical circle, gravitational force affects the object differently at various points:

• At the top, it adds to the tension, making the net force greater for maintaining circular motion.
• At the bottom, it reduces the effective tension force since it acts in the opposite direction of the tension.

Understanding this force helps explain why the tension changes and how it balances with the gravitational pull to keep the object in motion.