Question

A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

• A. the mass is at the highest point
• B. the wire is horizontal
• C. the mass is at the lowest point
• D. inclined at an angle of \(60\degree\) from vertical

Answer 1

The Correct Option is C

The problem involves analyzing when the tension in the wire is highest as a mass is whirled in a vertical circle. This is relevant in considering when the wire is most likely to break due to excessive tension.

1. In a vertical circular motion, the forces acting along the circular path include gravitational force and the tension in the wire. At any point, the centripetal force needed for circular motion is provided by the component of gravity and the tension.
2. At the highest point: The gravitational force acts downward, and the tension in the wire also acts towards the center. Hence, the total centripetal force at the top is the tension plus weight, T + mg = \frac{mv^2}{r}, where T is the tension, m is the mass, v is the speed, r is the radius, and g is the acceleration due to gravity.
3. When the wire is horizontal: The tension and the gravitational component perpendicular to the circle balance out to provide the centripetal force. Here, tension needs to counteract only the horizontal component as gravity acts vertically.
4. At the lowest point: The gravitational force acts against the tension, requiring the tension to be larger to provide the additional force needed for circular motion, T = \frac{mv^2}{r} + mg. Thus, the tension is maximal at the lowest point.
5. Inclined at an angle of 60\degree: The tension force must provide the necessary centripetal force, balancing part of the gravitational component and the necessary circular motion requirement.

Therefore, the wire is most likely to break at the lowest point due to the maximum tension required to sustain circular motion against the gravitational pull. This explains why the correct answer is: the mass is at the lowest point.