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Faraday's Dynamo & kinetic energy

  1. May 2, 2014 #1
    1. The problem statement, all variables and given/known data
    (The picture is not from my textbook)

    A Faraday Dynamo is used as a brake. The brushes are connected with an external resistance [itex]R_\Omega[/itex]. The circular disk has mass m, and angular velocity [itex]\omega_0[/itex] when t = 0.

    a) How long does it take until the kinetic energy of the disk is halved?

    2. Relevant equations

    [itex]emf =\frac{B R^2 \omega}{2}[/itex]
    [itex]dW=P dt[/itex]

    3. The attempt at a solution
    I have no idea how to figure this problem out.
    I've tried integrating dW from [itex]\frac{1}{2}m \omega^2_0 R^2[/itex] to [itex]\frac{1}{4}m \omega^2_0 R^2[/itex] setting that equal to the integral from t=0 to t of P dt (where I substitute P for the emf)

    Looking at the answer, [itex]t=\frac{m R_\Omega ln(2)}{B^2 R^2}[/itex], I figured I need to integrate over [itex]\frac{1}{x}[/itex], in order to get the ln(2), but I've got no clue how to find that x. Any help much appriciated.
  2. jcsd
  3. May 2, 2014 #2


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    Using energy ideas, how is the rate of decrease of the KE of the disk related to the electrical power produced?
  4. May 3, 2014 #3
    Sorry, I don't quite know what you mean. The KE and the electrical power produced is linked via the angular velocity with the formula: [itex]W=\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}t=\Delta KE[/itex]
  5. May 3, 2014 #4


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    Note that ##\omega## decreases with time. So, the electrical power is not going to be constant. That means the link between electrical power P and KE must be written in terms of an instantaneous rate of change of KE (with care given to signs).
  6. May 3, 2014 #5
    [itex]\frac{dKE}{dt}=-\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}[/itex]
    Last edited: May 3, 2014
  7. May 3, 2014 #6


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    Good. What if you express ##\omega^2## on the right side in terms of KE and moment of inertia ##\small I##? If you let K stand for KE, can you express the equation as follows? $$\frac{dK}{dt} = -b K$$
    Here ##b## is a collection of constants.
  8. May 5, 2014 #7
    Thank you so much! Here's what I got:

    [itex]\frac{dK}{dt}=-\frac{B^2 R^4 \omega^2}{4 R_\Omega}=-\frac{B^2 R^2}{2 m R_\Omega}K[/itex], used that [itex]K=\frac{1}{2}m \omega^2 R^2[/itex]
    Solution for diff. equation: [itex]K=K_0 e^{-\frac{B^2 R^2}{2 m R_\Omega}t}[/itex]
    [itex]\frac{1}{2}=\frac{K}{K_0}=e^{-\frac{B^2 R^2}{2 m R_\Omega}t}[/itex]
    [itex]ln(\frac{1}{2})=-ln(2)=-\frac{B^2 R^2}{2 m R_\Omega}t[/itex]
    [itex]t=\frac{2 m R_\Omega ln(2)}{B^2 R^2}[/itex]

    But it's not quite right, there's a 2 here
  9. May 5, 2014 #8


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    Check this. ##K = \frac{1}{2}I\omega^2##. What is ##I## for a solid disk?
  10. May 5, 2014 #9
    [itex]I=\frac{1}{2}m R^2[/itex], thank you so much!
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