Question

A 250 g , 23-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 1900 rpm in 4.3 s ?

Answer 1

Torque is given by:

Torque = I*alpha

I = Moment of inertia of disk = m*r^2/2

m = mass of disk = 250 gm = 0.25 kg

r = radius of disk = 23 cm/2 = 11.5 cm = 0.115 m

So, I = 0.25*0.115^2/2 = 1.65*10^-3 kg-m^2

alpha = angular velocity

Using 1st kinematic equation

wf = wi + alpha*t

wi = 0 rad/sec

wf = 1900 rpm = (1900 rev/min)*(2*pi rad/1 rev)*(1 min/60 sec)

wf = 1900*2*pi/60 = 198.97 rad/sec

So,

alpha = (wf - wi)/t

alpha = (198.97 - 0)/4.3 = 46.3 rad/sec^2

So, Now torque will be

Torque = 1.65*10^-3*46.3

Torque = 0.076 N-m

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