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Homework Help: Help in Newton's Laws of Rotation in Electric Motor and Electric Motor Dynamics

  1. Feb 23, 2010 #1
    Here is a copy of my notes on the motor. Basically it's just some of Newton's Law of Rotation combined with Kirchoff's Laws. Then we put the model into state space form.

    Could someone just please explain to me the meaning of the equations? It's been a while since I've had physics. Thanks alot for any help!

    I would just love to be able to derive these equations instead of just copying them down, but they're not explained in my book.
    Last edited: Feb 24, 2010
  2. jcsd
  3. Feb 24, 2010 #2
    Sorry there must have been a problem with the image linking. I am putting it in an attachment instead.

    A quick explanation of the motor and load models would be helpful.

    In particular, I'm having problems understanding the derivations of the differential equations of the model. I'm an electrical engineer in school and have had little exposure to mechanical modeling. A thorough explanation of the differential equations would be appreciated. I understand how to put the equations in state-space form.

    Thanks to all for looking.

    Attached Files:

    Last edited: Feb 24, 2010
  4. Feb 24, 2010 #3


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    The equation [tex]T = J \stackrel{..}{\theta}[/tex] is basically just the equivalent of "F=ma" for rotational systems. Mass is replaced with "moment of inertia" (J) and acceleration is replaced by angular acceleration ([tex]\stackrel{..}{\theta}[/tex]).

    The torque available for angular acceleration is equal to the electrical torque [itex]k_m \omega[/itex] minus the mechanical loss term [itex]B \omega[/itex]. That's pretty much all there is to the mechanical part of the model.
  5. Feb 24, 2010 #4


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    Re the electrical part of the system. Notice in the DC motor model there is a back EMF that is proportional to angular speed. This is actually where all the energy conversion takes place. The electrical power being absorbed by this back EMF is actually the raw mechanical output power. So just put together P=VI and T=P/w and you'll get the expression, [itex]T = k_m I[/itex], for the torque produced by the motor.
  6. Feb 24, 2010 #5
    Thank you very much. That actually made it easy! It's just been so long since I've seen rotational motion that I forgot where the basis for the equations came from. But now I see [tex]T = J \stackrel{..}{\theta}[/tex] should be the starting point for rotational systems...High school physics, I know.

    Using your equations, I put P=VI and P=Tw. Setting them equal yields VI=Tw meaning T=(VI)/w. Does that mean that Km=V/w? If so, what's the meaning of this?

  7. Feb 24, 2010 #6
    I did a quick search of DC motor constant and found that the units can be V/rad/sec, so yes Km=V/w. This is for Advanced Control Systems by the way. I really like control systems but my mechanical modeling ability must be greatly improved.
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