Calculating acceleration time of a motor driven disk.

The angular version of Newton’s law of motion, F=ma, is:


 
 

a is the angular acceleration, and so is the rate of change of angular velocity:


 


hence


Permanent magnet motor

The relationship between the torque and speed of the motor is shown in the graph:

This graph is of the equation:

Therefore, rearrangement gives:

Integrating both sides:

Performing integration of RHS:

Performing integration of LHS using the standard integral

Gives:

We want to know the time taken for w to reach 63% of w1. The 63% value is actually (1 – 1/e) where ‘e’ is the base of the natural logarithm, 2.7182818…..

Rearranging:

set w = (1 – 1/e) w1      (63% of w1):


Series wound motor

The relationship between the torque and speed of the motor is shown in the graph:

This graph is of the equation:

Therefore, rearrangement gives:

dw / dt = T1(1 - w2/w12) / I

\   1/(1 - w2/w12) dw = T1 / I dt

Integrating both sides:

1/(1- w2/w12) dw = T1 / I dt

Performing integration of RHS:

1/(1- w2/w12) dw =T1/I t

Performing integration of LHS using the standard integral

1/(a2 - x2) dx = 1/(2a) ln [(a + x) / (a –x)]

Gives:

1/2 w1 ln[(w1 + w) / (w1 - w)] = T1/I t

We want to know the time taken for w to reach 63% of w1. The 63% value is actually (1 – 1/e) where ‘e’ is the base of the natural logarithm, 2.7182818…..

Rearranging:

t = 1/2 w1 ln[(w1 + w) / (w1 - w)] I/T1

set w = (1 – 1/e) w1   (63% of w1), then after some cancelling out:

t = 1/2 w1 ln(2e + 1) I/T1

t = 0.931 w1 I/T1