Question

A ball of mass \((\mathrm{m}) 0.5 \mathrm{~kg}\) is attached to the end of a string having length (L) \(0.5 \mathrm{~m}\). The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is \(324 \mathrm{~N}\). The maximum possible value of angular velocity of ball (in radian/s) is (A) 9 (B) 18 (C) 27 (D) 36

Short answer

The maximum possible value of angular velocity of the ball is 36 radian/s.

Step-by-step solution

Determine the formula for tension in the string

The tension in the string when the ball is rotated in a horizontal circle can be represented by the formula for centripetal force, which is the force needed to keep the object moving in a circular path. The formula for centripetal force (which is the same as the tension when there is no other vertical force) is given by: \(T = m \cdot r \cdot \omega^2\) where \(T\) is the tension, \(m\) is the mass of the ball, \(r\) is the radius of the circular path, and \(\omega\) is the angular velocity.

Substitute the given values into the centripetal force formula

Use the given maximum tension (T), mass of the ball (m), and length of the string (L, which is equal to the radius r) to find the angular velocity (\(\omega\)). The provided values are: \(T=324\,N\), \(m=0.5\,kg\), and \(r = L = 0.5\,m\). Therefore, the formula becomes \(324 = 0.5 \cdot 0.5 \cdot \omega^2\).

Solve for maximum angular velocity (\(\omega\))

To find \(\omega\), you rearrange the formula to solve for \(\omega\) and obtain \(\omega = \sqrt{\frac{T}{m \cdot r}}\). Substituting the known values gives \(\omega = \sqrt{\frac{324}{0.5 \cdot 0.5}}\).

Calculate the result

Carrying out the calculation yields, \(\omega = \sqrt{\frac{324}{0.25}} = \sqrt{1296} = 36\,\text{radian/s}\). The maximum possible angular velocity of the ball is 36 radian/s.

Key concepts

Centripetal Force

The concept of centripetal force is pivotal in understanding circular motion in physics. It is the force that acts on an object moving in a circular path and is directed towards the center of the circle. Imagine tying a ball to a string and whirling it around in a horizontal circle above your head. The inwards pull you feel on the string is a result of the centripetal force. Mathematically, it is expressed as
\[ F = m \cdot r \cdot \omega^2 \]
with \( F \) being the centripetal force, \( m \) the mass of the object, \( r \) the radius of the circle, and \( \omega \) the angular velocity. In our textbook problem, the centripetal force keeps the ball moving in its circular path and is provided solely by the tension in the string. This force is crucial as it determines the maximum angular velocity the object can attain without breaking the string, hence playing a fundamental role in problems related to circular motion.

Tension in Circular Motion

Tension in circular motion describes the force that is exerted by a string or rope when an object is moving along a circular path. In the context of our textbook example, as the ball rotates, the string undergoes tension due to the centripetal force required to maintain the circular motion. An important aspect of this tension is that it has an upper limit — the maximum tension the string can bear before breaking. The relationship of tension with angular velocity is given by the formula from our solution steps:
\[ T = m \cdot r \cdot \omega^2 \]
As the angular velocity increases, the tension increases exponentially, making it a limiting factor in how fast the object can rotate. Understanding the dynamics of tension is essential for real-life applications such as the design of bridges, the operation of amusement park rides, and for students preparing for competitive exams like the JEE Advanced Physics, where such concepts are frequently tested.

JEE Advanced Physics

JEE Advanced Physics is known for testing students on a deep conceptual level, often incorporating complex problems from various topics of physics, including mechanics, electromagnetism, and modern physics. Circular motion, and specifically concepts like angular velocity, centripetal force, and tension, form an integral part of the syllabus.
For students preparing for such competitive exams, our textbook problem is a perfect example of how to solve questions where maximum limits, like the maximum tension a string can bear in circular motion, come into play.
Aspiring candidates must be adept at quickly identifying the underlying physical principles and applying them to resolve intricate problems. They must also have a strong foundation in applying mathematical concepts to physics, which is evident in how we derived the maximum angular velocity of the ball using the given maximum tension. Concentrated practice in these areas can significantly enhance the problem-solving skills necessary for success in the JEE Advanced Physics examination.