Question

A 240 g , 25-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 1800 rpm in 4.8 s ?

Answer 1

The moment of inertia of the disc about the center is
I = m r² / 2
Where m is the mass and r is the radius of the disc, m = 240 g = 0.24 kg, r = 0.25 m / 2 = 0.125 m
I = 0.24 kg x (0.125)² / 2
I =  0.001875 kg m²
Next we need to find the angular acceleration.
The initial angular velocity
_i = 0 rpm = 0 rad/s
The final angular velocity
_f = 1800 rpm = 1800 x 2 / 60 rad/s
_f = 188.4 rad/s
The angular acceleration of the disc
= _f - _i / t
Where t is the time taken, t = 4.8 s
= 188.4 rad/s - 0 rad/s / 4.8 s
= 39.25 rad/s²
The relation between the torque and the angular acceleration,
= I
=  0.001875 kg m² x 39.25 rad/s²
=  0.0736 N m