Problem 40 - Master Resource Book in JEE Main Physics

Question

A stone of mass \(2 \mathrm{~kg}\) is tied to a string of length \(0.5 \mathrm{~cm}\). If the breaking tension of the string is \(900 \mathrm{~N}\), then the maximum angular velocity, that stone can have in uniform circular motion is [Kerala CET 2010] (a) \(30 \mathrm{rad} / \mathrm{s}\) (b) \(20 \mathrm{rad} / \mathrm{s}\) (c) \(10 \mathrm{rad} / \mathrm{s}\) (d) \(25 \mathrm{rad} / \mathrm{s}\)

Step-by-Step Solution

Answer

The maximum angular velocity is \(30\mathrm{~rad/s}\).

Step 1: Understanding the Problem Setup

We have a stone of mass \(m = 2\mathrm{~kg}\) tied to a string of length \(r = 0.5\mathrm{~m}\) (note the conversion from cm to m). The string has a breaking tension of \(T = 900\mathrm{~N}\). The problem asks us to find the maximum angular velocity, \(\omega_{\text{max}}\), that the stone can have while moving in a circle without breaking the string.

Step 2: Relating Tension and Centripetal Force

In circular motion, the tension in the string provides the centripetal force necessary to keep the stone moving in a circle. The formula for centripetal force is given by:\[ F_{c} = \frac{m v^2}{r} \]Since the tension is the maximum force, set \(F_{c} = T\):\[ T = \frac{m v^2}{r} \]

Step 3: Expressing Velocity in Terms of Angular Velocity

The linear velocity \(v\) can be expressed in terms of angular velocity \(\omega\) as:\[ v = r\omega \]Substituting this into the centripetal force equation gives:\[ T = \frac{m (r \omega)^2}{r} \]

Step 4: Simplifying the Equation

Simplify the equation by canceling out \(r\):\[ T = m r \omega^2 \]

Step 5: Solving for Maximum Angular Velocity

Rearrange the equation to solve for \(\omega\):\[ \omega = \sqrt{\frac{T}{mr}} \]Substitute the given values:\( T = 900\mathrm{~N}, \ m = 2\mathrm{~kg}, \ r = 0.5\mathrm{~m} \):\[ \omega = \sqrt{\frac{900}{2 \times 0.5}} \]

Step 6: Calculating the Angular Velocity

Evaluate the expression:\[ \omega = \sqrt{\frac{900}{1}} = \sqrt{900} = 30\mathrm{~rad/s} \]

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates or revolves relative to a point or axis. It's denoted by the symbol \( \omega \) and is expressed in radians per second (rad/s). When dealing with circular motion, angular velocity provides crucial information about the rotational speed of the object.
In the exercise, the maximum angular velocity of a stone tied to a string is calculated to determine the highest speed at which the stone can travel without breaking the string.
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by the formula:

• \( v = r \omega \)

where \( r \) is the radius of the circular path. This formula shows that angular velocity is proportional to linear velocity, meaning as one increases, so does the other. Angular velocity helps us understand how quickly the stone can complete a circle.

Centripetal Force

Centripetal force is the force required to keep an object moving in a circular path and is directed towards the center of the circle. This force is crucial in understanding circular motion because, without it, an object would not follow a circular path but instead move off in a straight line due to inertia.
In our stone-and-string scenario, the tension in the string acts as the centripetal force. The formula for centripetal force \( F_{c} \) is:

• \( F_{c} = \frac{m v^2}{r} \)

Here, \( m \) is the mass of the object, \( v \) is its linear velocity, and \( r \) is the radius of the circular path. The exercise illustrates how the tension, which is the maximum force the string can withstand, determines the maximum centripetal force available, and consequently limits the stone's speed to ensure it stays in circular motion.

Tension in String

The tension in a string is the pulling force exerted by the string when it is subjected to forces that may cause it to break. In a circular motion scenario, like the stone tied to the string, tension is crucial because it provides the necessary centripetal force to keep the object moving in a circle.
The breaking tension of the string is the maximum force it can handle before snapping. For this exercise, the breaking tension is given as \( T = 900 \text{ N} \). This means that before the string breaks, the force exerted by the swinging stone cannot exceed 900 N.
When you consider the relationship:

• \( T = m r \omega^2 \)

the tension is directly related to the mass, radius, and angular velocity of the stone. Calculating with this formula helps determine the maximum angular velocity at which the stone can move without breaking the string, ensuring the tension does not exceed the breaking point.