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 1700 rpm in 4.8 s ?

Answer 1

first Convert all the values into SI units. The angular speed is:
rad/s = (rpm x 2pi) / (60 sec)
1700rpm = 1700rev/min(2πrad/1.00rev)(1.00min/60.0s)
= 178rad/s

The radius is:

r = 22.0cm/2
= 11.0 cm = 11/100 m = 0.11m

The mass is:

200g = 200g(1.00kg/1000g)
= 0.200kg

The angular acceleration of the disk is:

α = (ω - ω₀)/t
= (178rad/s - 0) / 4.8s
= 37.08rad/s²

Then the torque is:

torque = Iα

I, the moment of inertia of the disk is 0.5mr², so:

torque = 0.5mr²α
torque = 0.5(0.200kg)(0.11m)²(37.08rad/s²)
torque = 0.0.40788 N∙m