Question

A 200 g , 22-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 1500 rpm in 4.7 s ?

Answer 1

Mass of the disk = M = 200 g = 0.2 kg

Diameter of the disk = D = 22 cm = 0.22 m

Radius of the disk = R = D/2 = 0.11 m

Moment of inertia of the disk = I

I = MR²/2

I = (0.2)(0.11)²/2

I = 1.21 x 10^-3 kg.m²

Initial angular speed of the disk = ₁ = 0 rpm = 0 rad/s

Final angular speed of the disk = ₂ = 1500 rpm = 2(1500)/60 rad/s = 157.08 rad/s

Time period = T = 4.7 sec

Angular acceleration of the disk =

₂ = ₁ + T

157.08 = 0 + (4.7)

= 33.42 rad/s²

Torque supplied by the motor = T

I = T

(1.21x10^-3)(33.42) = T

T = 4.044 x 10^-2 Nm

Torque supplied by the motor = 4.044 x 10^-2 Nm