(solution) A 200 g, 24 cm diameter plastic disk is spun on an axle through

(solution) A 200 g, 24 cm diameter plastic disk is spun on an axle through

A 200 g, 24 cm diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 to 2000 rpm in 4. 9s? My solution: solve for moment of inertia: I = 1/2 x M x (R)^2 = 1/2 x (.2 kg) x ( .2 m)^2 = .004 kgxm^2 change rpm to rad/s: (1800 rev/min) x (1min/60 sec) x (2/rev) = 188.496 rad/s find : = / T = (188.496 rad/s / 4.00 s) = 47.124 rad/ s^2 find torque : = I = (47.124 rad/s^2) x (.004 kg x m^2) = .188496 N x m but the correct answer is .047 N x m can somone explain what i’m doing wrong?