The radius of a ring is 1 m and its mass is 10 kg. What torque will increase its angular velocity from 0 to 120 rotations/minute in 10 seconds ?π N-m2π N-m3π N-m4π N-m

To solve this problem, we need to follow these steps:

1. Convert the angular velocity from rotations per minute to radians per second. We know that 1 rotation = 2π radians and 1 minute = 60 seconds. So, 120 rotations/minute = 120 * 2π / 60 = 4π radians/second.

2. Calculate the moment of inertia (I) of the ring. The moment of inertia of a ring is given by the formula I = mr², where m is the mass and r is the radius. So, I = 10 kg * (1 m)² = 10 kg*m².

3. Calculate the final angular momentum (L) of the ring. The angular momentum is given by the formula L = Iω, where ω is the angular velocity. So, L = 10 kgm² * 4π rad/s = 40π kgm²/s.

4. Calculate the initial angular momentum of the ring. Since the ring starts from rest, its initial angular velocity is 0, so its initial angular momentum is also 0.

5. Calculate the change in angular momentum (ΔL). This is given by the formula ΔL = L_final - L_initial. So, ΔL = 40π kgm²/s - 0 = 40π kgm²/s.

6. Calculate the torque (τ). Torque is the change in angular momentum divided by the change in time. So, τ = ΔL / Δt = 40π kgm²/s / 10 s = 4π Nm.

So, the torque that will increase the ring's angular velocity from 0 to 120 rotations/minute in 10 seconds is 4π N*m.