Answer:
30.85 N-m
Explanation:
we can use the work energy theorem
that ΔK = 1/2Iω_f^2- 1/2Iω_i^2
= 1/2mR_eff^2×ω_f^2-0
= 0.5×192×0.62^2×2π×120/60
=0.463
b) If the torque is constant then so is the angular acceleration. Hence, ω_f= ω_i+αt
now α= 4π/30 = 0.4180
then we use the relationship τ= Iα = to calculate the torque
τ = mr^2 (α)
τ = 192×0.62^2×0.4180
= 30.85 N-m