I have been calculating the power required to accelerate a small solid rotor up to 2000 RPM using two different but equally valid methods of calculation, and yet they give different results by a factor of 2. Where am I going wrong?

I will lay out both methods:

The first is calculating the kinetic energy of the rotor at 2000 RPM and dividing that by the time to reach those revs, that being 22 seconds.

Method 1: P = KE/t = ยฝ I ๐Ž2/t

Rotor mass = 0.685 kg, radius: 0.055 m so I = 0.001 kgm2

๐Ž = 2000/60 x 2๐žน = 209.5 rad/s,

โˆด ๐Ž2 = 43,920

P = 0.5 x 0.001 x 43,920 / 22 = 0.995 W

Method 2 is calculating the Torque and multiplying that by the angular velocity, where the Torque is derived from the inertia times the angular acceleration.

i.e. P = T ๐Ž and T = I๐œถ therefore P = I๐œถ ๐Ž = I๐Ž2/t

I = 0.001 kgm2 (from above)

๐Ž = 2000/60 x 2๐žน = 209.4 rad/s

๐œถ = ๐Ž/t = (2000/60 x 2๐žน)/22 = 9.52 rad/s2

so T = 0.001 x 9.52 = 0.0095 Nm

and P = 0.0095 x 209.4 = 1.99 W

A factor of two different!

The only explanation I can come up with is that method 2 is a 'dynamic' calculation using angular acceleration and method 1 is not, being instead of sort of 'averaging' calculation; and so I should use the average time of (22-0)/2 = 11sec. So in effect, the KE of the rotor at 2000 RPM is derived from the area under the line as per the graphic and so it's half the area.

Energy as integral of power graph

That would bring the two answers in line with each other but I'm having difficulty justifying using the average time.

Can anyone help?


2 Answers 2


I've just edited your question an your equations are inconsistent.

In Method 1, you state the equation is P = ยฝIฯ‰2/t

In Method 2, you state, P = Tฯ‰ and that T = Iฮฑ therefore P = Iฮฑฯ‰

you also state ฮฑ = ฯ‰/t

Substituting this back into the final equation for Method 2, P = Iฯ‰2/t, which produces twice the value of the equation you used for Method 1 - it doesn't have the ยฝ that is in the equation for Method 1.

One of your equations is incorrect.

  • $\begingroup$ So Power is not 0.5 I ๐Ž^2 but instead I ๐Ž^2 ? If so then I will have to tell my source of their mistake :) $\endgroup$ Commented Jun 20, 2023 at 14:17
  • $\begingroup$ In Method 1, P is derived from KE/t and where the KE is 1/2 I ๐Ž^2 so P must equal 1/2 I ๐Ž^2/t. That being so then there must be something wrong with the equation in Method 2? Perhaps P = 1/2 T๐Ž? $\endgroup$ Commented Jun 20, 2023 at 14:21
  • $\begingroup$ I have checked that P = T๐Ž is correct so you can see my dilemma :| $\endgroup$ Commented Jun 20, 2023 at 14:28

In your second equation, you are assuming that $\omega$ is constant even though it is not. Both equations are assuming that power is constant, even though it is not.

Assuming constant torque for the interval $0 < t < t_{end}$:

$P(t) = T \omega(t)$

$\omega(t) = \alpha t$

The energy put into the flywheel in the interval is then

$$W = \int_0^{t_{end}} T\omega(t) dt = \int_0^{t_{end}} T\alpha t\ dt = \left . \frac 1 2 T\alpha t^2 \right |_0^{t_{end}} = \frac 1 2 T\alpha t_{end}^2$$

This is where your first equation comes from. But it is not the power, which is always changing.

If you wanted to calculate the average power of the event, then you would calculate $W / t_{end}$. That would be valid for, say, a plant that's always accelerating such wheels, and you need to know how much your energy expenditure will be. It would be totally invalid, however, for sizing drive electronics for a motor to accelerate the wheel, which would need to be sized for the maximum power -- which happens right as $t \to t_{end}$.

  • $\begingroup$ The second half of the above is my response. Thereโ€™s not enough space in this box. $\endgroup$ Commented Jun 20, 2023 at 20:32
  • $\begingroup$ I think you misunderstand how Stackexchange works. It is not a discussion site; discussion in comments is only encouraged in furtherance of clarifying questions or answers, and discussion in answers isn't sensible, because then they wouldn't be answers. Questions should be self-contained, and answers should respond to the question. To the degree that your additional information in your proposed edits were clarifications to your question, then you should edit that question with those clarifications. Then I may edit my answer to match your newly-improved question. $\endgroup$
    – TimWescott
    Commented Jun 20, 2023 at 22:13

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