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Given power, find speed

  1. Oct 11, 2006 #1
    The power output from a trolley motor depends upon velocity P(v) = av(b^2-v^2). The power is 0 for v^2>b and "a" and "b" are constants. Determine the speed of the maximum power output and the speed of the maximum force exerted by the motor.

    How do I begin this problem? I don't know how to start.
  2. jcsd
  3. Oct 11, 2006 #2
    In terms of the calculus, what do you know about power?
  4. Oct 11, 2006 #3
    It's the derivative of work with respect to time.
  5. Oct 11, 2006 #4
    Okay, good, but let's take it a step further. We know that [tex]P = \frac{dW}{dt} = \frac{\vec{F} \cdot \vec{ds}}{dt} = \vec{F} \cdot \frac{ \vec{ds}}{dt} = \vec{F} \cdot \vec{v} [/tex]. Does this give you a start or any ideas?
  6. Oct 11, 2006 #5


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    The first part is a calculus maximization problem (or maybe a graphing calculator problem??). You are given P as a function of v and asked to maximize with respect to v.

    I'm still thinking about the second part. The force is not constant here, so you cannot say P = Fv
    Last edited: Oct 11, 2006
  7. Oct 12, 2006 #6
    Oh, oops, I overlooked that, but you are right. The force is not constant and can't be taken out of the derivative. Hum, it would be easy to graph into mathematica or another graphical program to find the quantitative solution, but the analytical solution is tough. Yes, I will have to think about this some more too.
  8. Oct 13, 2006 #7


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    The more I think about it, the more I think you were right to begin with, at least for any physically reasonable force. The force related to an incremental displacement ds is not going to change while moving the distance ds. So it's not a product rule derivative when you write
    dW = Fds
    It's just dividing the incremental work by the incremental time

    dW/dt = Fds/dt
    P = Fv
    F = P/v = a(b^2-v^2) is maximum when v = 0

    That seems reasonable actually. I think electric motors typically have maximum torque at zero speed, so you would expect maximum force at zero speed.
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