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 lowest point
• B. inclined at an angle of $60^{\circ}$ from vertical
• C. the mass is at the highest point
• D. the wire is horizontal

Answer 1

The Correct Option is A

To solve this problem, let's analyze the forces acting on the mass when it is attached to a thin wire and whirled in a vertical circle.

1. The forces acting on the mass include gravitational force and tension in the wire.
2. When the mass is at the lowest point of the circle, both the gravitational force and the centripetal force (required to maintain circular motion) work in the same direction, thereby maximizing the tension in the wire.
3. The expression for the tension in the wire at any point in the vertical circle is given by: T = m \cdot g + \frac{m \cdot v^2}{r}, where
   • T is the tension in the wire,
   • m is the mass of the object,
   • g is the acceleration due to gravity,
   • v is the speed of the mass,
   • r is the radius of the circle.
4. At the lowest point, the speed v is typically at its maximum due to conservation of mechanical energy, resulting in a maximum centripetal force requirement.
5. Consequently, the tension is maximum at the lowest point, because both the gravitational component (m \cdot g) and the centripetal force component (\frac{m \cdot v^2}{r}) add up.

For the other options:

• When inclined at an angle of 60^{\circ} from vertical or when the wire is horizontal, the gravitational and centripetal force components don't add up to create maximal tension.
• At the highest point, the gravitational force actually subtracts from the centripetal force requirement, reducing overall tension.

Therefore, the point at which the wire is most likely to break is when the mass is at the lowest point in the vertical circle.

Correct Answer: The mass is at the lowest point.