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Weird Circuit with Wheel

  1. May 6, 2008 #1
    FOR PICTURE of CIRCUIT: community.webshots.com/photo/2479536900103460081IWuFAY
    A wheel consiting of a large number of thin conducting spokes is free to rotate about its axle. The brush makes contact one spoke at a time. There is a uniform B field pointing into the page. The wheel is initially at rest.
    R = radius of wheel
    J = moment of inertia of wheel
    L = inductance of circuit
    V = voltage

    Find an expression relating ω(t) with I(t) by calculating the torque on the wheel.

    I tried using τ = dL/dt and L=Jω to get that τ = J*dω/dt. For the other half of the expression I use τ = r x F. I am unclear how to do that part. Any help would be awesome, thanks!
  2. jcsd
  3. May 6, 2008 #2
    bump b/c im still confused
    also, picture is at community.webshots.com/photo/2479536900103460081IWuFAY
  4. May 6, 2008 #3
    You know there's a current going from the center of the wheel down to the brush. I think the inductor is just there to hint at the fact that the current remains constant (dI/dt for a conductor is always finite). Then you can use the force law:

    [tex]\vec{F}=I\vec{L} \times \vec{B}[/tex]

    In this case, L would be the radius of the spinning thing and points down. Then the torque is:

    [tex]\vec{\tau} = \int_0^L{d \vec{r} \times \vec{F}}[/tex]

    F is constant here, so this pretty easy. (NOTE: I don't know why, but I keep looking at the above equation, and it seems wrong, but I'm pretty sure it's right. Anyway, if something doesn't work out, maybe I just lied to you above)

    That's probably enough to get you started.

    NOTE: these are reducible to scalar equations (e.g. F=ILB).
  5. May 6, 2008 #4
    This might help you but I am also little confused about that:

    Divide the thing into two parts (wheel and wire) - two wings of the wheel close the wire connection (so they are just switches that are turned on and off after every few seconds)

    And as misho said because V = L.di/dt so the current doesn't make sudden jumps - stay constant

    Problem that I am facing:
    one force arrow goes up (for the wing that is connected to the top part of the circuit), and one force arrow points left (for the left side wire)
    so, torque = 0?
  6. May 7, 2008 #5
    Ok, when I use the above equations I get a constant torque, but the torque is not constant along the spoke (I got τ = I*B*R^2). Or is that not constant b/c I(t) changes w/ time?
  7. May 7, 2008 #6
    Well, the current shouldn't change with time. I'm pretty sure about that.
  8. May 7, 2008 #7
    I think you're confused just because the diagram isn't very good. I think that there is a wire that goes to the middle of the spinner and then the current goes down to the brush (please tell me if I'm wrong about that).
  9. May 7, 2008 #8
    You are right - the current travels from the wire through the spoke, and to the brush. Sorry for the crappy diagram.

    EDIT: also, i think the current does change with time b/c on a latter part of this question it asks for me to solve for I(t).
  10. May 7, 2008 #9


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    Yes, V = L di/dt
    V is constant, as is L.
    But i is not constant, according to the equation.
  11. May 7, 2008 #10
    because789, you wouldn't happen to be taking a take-home test for a certain Physics 217 course at Cornell, would you? Keep in mind that your email address is linked to your forum account.
  12. May 7, 2008 #11
    V = L di/dt <- the V here is the voltage across the inductor, which could be zero. Having said that, dI/dt is not zero, I was wrong about that.

    I decided to look at this in terms of conservation of power:

    Source power: [tex]P_s (t) = VI(t)[/tex]

    Inductor power: [tex]P_L (t) = V_L I_L = L I(t) \frac{dI}{dt}[/tex]

    Wheel power: [tex]P_w = \tau (t) \omega (t)[/tex]

    At some point, it hit me that if we have a force acting on an object the way it does on a spoke of the wheel here, we would take it to act at the center of mass, so:

    [tex]\tau = \frac{1}{2}RF[/tex]

    And since the force is magnetic:

    [tex]F = RBI(t)[/tex]

    So: [tex]\tau = \frac{1}{2}R^2 B I(t)[/tex]

    Conservation of power says:

    [tex]VI(t) = L I(t) \frac{dI}{dt} + \frac{1}{2}R^2 B I(t) \times \omega (t) [/tex]
    [tex]V = L \frac{dI}{dt} + \frac{1}{2}R^2 B \omega (t) [/tex]

    Hopefully, I didn't screw anything up this time.
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