Question

A small ball of mass m is suspended from the ceiling of a floor by a string of length L. The ball moves along a horizontal circle with constant angular velocity ω, as shown in the figure. The torque about the center (O) of the horizontal circle is:

• A. mgL sin θ
• B. mgL cos θ
• C. 0
• D. mgl cos θ

Hint:

The torque in a rotational system can be calculated by multiplying the force by the perpendicular distance from the axis of rotation.

Answer 1

The Correct Option is C

1. The ball's horizontal circular motion implies balanced vertical forces: gravity (\(mg\)) and the string's vertical tension component. The ball experiences two forces:
Gravity: \(mg\) and String Tension: \(T\)
2. The horizontal tension component (\(T \sin \theta\)) provides the centripetal force for circular motion. The vertical component (\(T \cos \theta\)) counteracts the ball's weight.
3. Torque about the circle's center (O) is the force's cross product with the radius vector. Torque \(\tau\) is calculated as:
\(\tau = r \times F\)
where \(r\) is the radius (length \(L\)) and \(F\) is the force (tension). However, the tension acts along the string, generating no torque about the horizontal circle's center.
4. The total torque at the center is zero; the tension, causing circular motion, doesn't rotate the ball around the center.