The angular momentum of a flywheel having a rotational inertia of 0.150 kg·m2 about its axis decreases from 3.00 to 1.400 kg·m2/s in 0.80 s.(a) What is the average torque acting on the flywheel about its central axis during this period? N·m(b) Assuming a uniform angular acceleration, through what angle will the flywheel have turned?

(a) The average torque can be calculated using the formula for angular acceleration (α), which is the change in angular momentum (ΔL) divided by the change in time (Δt), and then using the formula for torque (τ), which is the product of rotational inertia (I) and angular acceleration (α).

First, calculate the change in angular momentum:

ΔL = L_final - L_initial = 1.400 kg·m²/s - 3.00 kg·m²/s = -1.60 kg·m²/s

Then, calculate the angular acceleration:

α = ΔL / Δt = -1.60 kg·m²/s / 0.80 s = -2.00 rad/s²

Finally, calculate the average torque:

τ = I * α = 0.150 kg·m² * -2.00 rad/s² = -0.30 N·m

(b) The angle turned by the flywheel can be calculated using the formula for angular displacement (Δθ), which is the average of the initial and final angular velocities (ω_initial and ω_final) multiplied by the change in time (Δt).

First, calculate the initial and final angular velocities:

ω_initial = L_initial / I = 3.00 kg·m²/s / 0.150 kg·m² = 20 rad/s ω_final = L_final / I = 1.400 kg·m²/s / 0.150 kg·m² = 9.33 rad/s

Then, calculate the average angular velocity:

ω_avg = (ω_initial + ω_final) / 2 = (20 rad/s + 9.33 rad/s) / 2 = 14.67 rad/s

Finally, calculate the angular displacement:

Δθ = ω_avg * Δt = 14.67 rad/s * 0.80 s = 11.73 rad