Question

A mass 'M' is attached at the end of a string whose length is 'L' and whirled at a constant speed in vertical circle. Generally, the tension in the string is minimum when the mass is

• A. at the top of the circle
• B. half way down from the top
• C. at the bottom of the circle
• D. quarter way down from the top

Hint:

In vertical circular motion, tension varies due to gravity. At the top, gravity aids the centripetal force, reducing the required tension from the string. At the bottom, gravity opposes the centripetal force, increasing the tension. Therefore, tension is minimum at the top and maximum at the bottom.

Answer 1

The Correct Option is A

Step 1: Analyze the forces acting on the mass at different points in the vertical circle.
Let \( T \) be the tension in the string and \( mg \) be the weight of the mass.
When a mass is whirled in a vertical circle, the forces acting on it are the tension in the string (towards the center of the circle) and gravity (vertically downwards). The net force towards the center provides the necessary centripetal force, \( \frac{Mv^2}{L} \).
Step 2: Consider the forces at the top of the circle.
At the top of the circle, both the tension \( T_{top} \) and the weight \( mg \) act downwards, in the same direction as the centripetal force. So, the equation of motion is: \[ T_{top} + mg = \frac{Mv^2}{L} \] \[ T_{top} = \frac{Mv^2}{L} - mg \] In this position, the tension is reduced by the full weight of the mass. This is where the tension is minimum, and if the speed is too low, the string might go slack ($T_{top}=0$).
Step 3: Consider the forces at the bottom of the circle.
At the bottom of the circle, the tension \( T_{bottom} \) acts upwards (towards the center), while the weight \( mg \) acts downwards (away from the center).
So, the equation of motion is: \[ T_{bottom} - mg = \frac{Mv^2}{L} \] \[ T_{bottom} = \frac{Mv^2}{L} + mg \] In this position, the tension is increased by the full weight of the mass. This is where the tension is maximum.
Step 4: Compare tension at different points.
Comparing the expressions for tension:
\( T_{top} = \frac{Mv^2}{L} - mg \)
\( T_{bottom} = \frac{Mv^2}{L} + mg \)
Clearly, \( T_{top} \) is less than \( T_{bottom} \). For points "halfway down from the top" or "quarter way down from the top", the component of gravity acting radially would be less than \( mg \) and would be perpendicular to gravity. The general equation for tension at an angle \( \theta \) from the top (measured clockwise) is:
\( T + mg \cos\theta = \frac{Mv^2}{L} \)
So, \( T = \frac{Mv^2}{L} - mg \cos\theta \).
At the top, \( \theta = 0^\circ \), \( \cos 0^\circ = 1 \), so \( T = \frac{Mv^2}{L} - mg \).
At the bottom, \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \), so \( T = \frac{Mv^2}{L} + mg \).
At horizontal positions (halfway down from the top and halfway up from the bottom), \( \theta = 90^\circ \) or \( \theta = 270^\circ \), \( \cos 90^\circ = 0 \), so \( T = \frac{Mv^2}{L} \).
The term \( -mg \cos\theta \) is minimized (most negative) when \( \cos\theta \) is maximized (positive), which occurs at \( \theta = 0^\circ \) (top). Thus, the tension is minimum at the top of the circle. The final answer is $\boxed{\text{1}}$.