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 w_{1}_{.}
The 63% value is actually (1 – ^{1}/_{e}) where ‘*e*’ is
the base of the natural logarithm, 2.7182818…..

_{Rearranging:}

**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 =
T_{1}(1 - w^{2}/w_{1}^{2}) / I

\ 1/(1 -
w^{2}/w_{1}^{2}) dw =
T_{1 }/ I dt

Integrating both sides:

ò 1/(1- w^{2}/w_{1}^{2}^{)
}dw = ò T_{1}
/ I dt

Performing integration of RHS:

ò 1/(1- w^{2}/w_{1}^{2}^{)
}dw =T_{1}/I ´ t

Performing integration of LHS using the standard integral

ò 1/(a^{2 }^{-
}x^{2}) dx = 1/(2a) ln [(a + x)
/ (a –x)]

Gives:

1/2 w_{1} ln[(w_{1} + w) / (w_{1} - w)] = T_{1}/I
´ t

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

_{Rearranging:}

t = 1/2 w_{1} ln[(w_{1} + w) / (w_{1} - w)] ´ I/T_{1}

set w = (1 – ^{1}/_{e})
w_{1 }(63% of w_{1}), then after some cancelling out:

**t = _{ }0.931 w_{1 }´
I/T_{1}**