A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes (3/π) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is __________ ML.

Solution

Physics sir · 3 steps

In a conical pendulum the angle cancels: T sinθ gives Mω²L sinθ, so T is just Mω²L.

1. Horizontal component. The mass moves in a horizontal circle of radius R = L sin θ. Only the horizontal part of the tension, T sin θ, points towards the centre, so it supplies the centripetal force.

   T sin θ = Mω² R = Mω² L sin θ ⇒ T = Mω² L

2. Angular speed. The frequency is f = 3/π rev s⁻¹, so ω = 2π f = 2π × 3/π = 6 rad s⁻¹ and ω² = 36 s⁻².
3. So. T = 36 ML, and the blank is 36.

The idea to keep. For a conical pendulum, the tension is always Mω² L, whatever the angle. Convert rotations per second to ω with 2π f before you square it.

Answer from the NTA final answer key. Solution by PrepWiser.