A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5 t²-8 t, where θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t=2 s ?

Solution

Physics sir · 4 steps

Power from a torque is just τω, so differentiate θ once for ω and twice for α.

1. Angular velocity. Differentiate once: ω = dθ/dt = 10t - 8. At t = 2 s, ω = 20 - 8 = 12 rad s⁻¹.
2. Angular acceleration. Differentiate again: α = dω/dt = 10 rad s⁻². It is constant, so the torque is constant too.
3. Torque on disc. For a disc about its central perpendicular axis, I = 1/2MR². So τ = Iα = 1/2MR² × 10 = 5MR².

   P = τω = 5MR² × 12 = 60MR²

4. So. The power delivered at t = 2 s is 60MR² watts, option (a).

The idea to keep. In rotation, power is P = τω, just like P = Fv in a straight line. Get ω and α by differentiating θ(t), and use the correct moment of inertia.

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