Torque question

Question

A disk of mass 6 kg and radius 25 cm is rotating around horizontal axis that goes through its center with n=360 revolutions/min. When the disk was being stopped it stopped right after t=30 s. Calculate the stopping torque exerted on flywheel and number of revolutions N.

Answer 1

First find moment of inertia I of the disk of radius 0.25 m, I = 1/2 M R² = (0.5)(6)(0.25)² = 0.1875.

Convert revs/min into angular velocity in rad/sec, ω_o = (360 rev/min)(2π rad/rev)(1 min/60 s) = 12π = 37.70 rad/s.

Find angular acceleration α = Δω/Δt = (0 - 37.7)/30 = (-) 1.257 rad/s² (we can ignore negative sign hereafter).

The torque τ = Iα = (0.1875)(1.257) = 0.236 Nm.

To find N, get the angular displacement θ from rotational kinematics, θ = 1/2 (ω_o + ω_f) t = (0.5)(37.7 + 0)(30) = 565.5 radians. Then number of revolutions N = (θ rad)/(2π rad/rev) = 90.0 rev.