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The slowing down of a Farady disc

  1. Mar 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A homopolar generator consists of a metal disc of radius ##a## and a central axle which has radius ##a/4##. The disc has resistivity ##\rho## and thickness ##t##. It is rotated in a uniform magnetic field ##B## about an axis through the centre, which is parallel to ##B## and perpendicular to the plane containing the disc, at an angular frequency ##\omega##. Thin ring brushes make good electrical contact with the disc near the axle and near the outer rim of the disc as shown.

    (a) Calculate the resistance of the disc ##R_D## measured between the brushes.

    (b) Show that the potential difference between the brushes is ##(15/32)ωBa2## .

    (c) A load resistance ##R_L## is connected across the generator and the drive is removed. Show that, in the absence of mechanical friction, the time ##\tau## taken for the disc to slow down to half its initial angular speed is
    $$\tau =( 32/15)^2 \frac{m(R_L + R_D) ln(2)}{ 2a^2B^2} $$.
    2. Relevant equations

    Inertia of a disc:
    $$I=1/2 ma^2$$
    Kinetic energy of rotating disc:
    $$E=1/2 I\dot{\theta}^2$$
    Electrical Power Dissipated:
    $$ P=\frac{dE}{dt}=\frac{v^2}{R_l+R_D}=(\frac{15Ba^2}{32})^2\frac{\dot{\theta}^2}{R_l+R_D}$$

    3. The attempt at a solution
    I have done part (a) and (b) - its only part C I need:

    Energy at full speed:
    $$E_1=\frac{ma^2\omega^2}{4}$$
    Energy at half speed:
    $$E_2=\frac{ma^2\omega^2}{16}$$
    Energy lost:
    $$E_2-E_1=\frac{3}{16}ma^2\omega^2=\int^\tau_0pdt=(\frac{15Ba^2}{32})^2\int^\tau_0\frac{\dot{\theta}^2}{R_L+R_D}dt$$
    Basically, how do I solve that integral? I know the start and end values of ##\dot{\theta}##/

    thank you
     
  2. jcsd
  3. Mar 26, 2017 #2

    TSny

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    To do the final integral, you would need to know how ##\dot{\theta}## depends on time.

    You have an expression for the rate at which electrical energy is produced. Can you come up with an expression for the rate at which mechanical energy is lost? How are these two rates related?
     
  4. Mar 26, 2017 #3
    Mechanical energy is lost at the same rate that energy is dissipated through the resistances.

    I could go into forces and go right back to first principles but I feel that would be far too long
     
  5. Mar 26, 2017 #4

    TSny

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    You don't need to go into forces. Just work with the electrical power and mechanical power.
     
  6. Mar 26, 2017 #5
    I thought about that but I couldnt see how to do it
     
  7. Mar 26, 2017 #6

    TSny

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    You already have an expression for the kinetic energy of the disk. Can you use that to obtain a general expression for the rate of change of the kinetic energy?
     
  8. Mar 26, 2017 #7
    Yes, I think :)
    $$\frac{dE}{dt}=\frac{15Ba^2}{32}^2\frac{\dot{\theta}^2}{R_l+R_d}=\frac{15Ba^2}{32}^2\frac{4E}{ma^2(R_l+R_d)}$$

    Which is seperable.


    Edit - I have worked through it all and it works, thank you
     
  9. Mar 26, 2017 #8

    TSny

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    OK. Good work.
     
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