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Question re. Conservation of angular momentum

  1. Aug 18, 2005 #1
    Um, I just saw this problem on another discussion forum and currently I'm stumped. So I decided to post this over here and see if anyone can help me with it:

    Two wheels with radii R1 and R2 (R1>R2) have rotational inertia I1 and I2, respectively. Initially the small wheel is at rest, while the big wheel has rotational velocity of W1 about its center. We gradually move the small wheel closer to the big one, until the wheels touch. Then friction between the wheels causes the small wheel to speed up, and the big wheel to slow down, until at last the two wheels spin with the same tangential velocity in opposite directions. What is this tangential velocity? (Hint: angular momentum is not conserved.)

    I don't understand why angular momentum would not be conserved, considering that the frictional forces are of the same magnitude and opposite direction. Where does the external torque come from? Could someone enlighten me on this?
     
  2. jcsd
  3. Aug 18, 2005 #2

    HallsofIvy

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    Are you sure the hint didn't say that the angular momentum of the large wheel is not conserved? Since this is a closed system the total angular momentum surely should be conserved.

    Let ω1' be the final angular speed of the large wheel, ω2' be the final angular speed of the small wheel. By conservation of angular momentum, I1ω1= I1ω1'+ I2ω2'.

    Since the tangential velocities of the two wheels are the same, R1ω1'= R2ω2'. Solving that for ω2', [itex]\omega_2'= \frac{R_1}{R_2}\omega_1'[/itex].

    The first equation then is: [itex]I_1 \omega_1= I_1\omega_1'+\frac{I_2R_1}{R_2}\omega_1'= \frac{I_1R_2+I_2R_1}{R_2}\omega_1'[/itex].
    The final angular velocity of the large wheel is
    [itex]\omega_1'= \frac{R_2I_1\omega_1}{I_2R_1+ I_1R_2}[/itex]
    The final angular velocity of the small wheel is
    [itex]\omega_2'= \frac{R_1}{R_2}\omega_1'= \frac{I_1R_1\omega_1}{I_2R_1+I_1R_2}[/itex]
    and the tangential velocity is
    [itex]R_1\omega_1'= R_2\omega_2'= \frac{I_1R_1R_2\omega_1}{I_1R_2+ I_2R_1}[/itex].
     
    Last edited: Aug 18, 2005
  4. Aug 18, 2005 #3

    mukundpa

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    First of all it should be clear that the two wheels are in the same plane and their axis of rotation (passes through their centers) are parallel. When the two wheels come closer, their rims are touching each other and the friction is acting against their relative motion on the two wheels in opposite direction. These forces (friction) will move the centers of the two wheels in opposite direction. To hold axes of the wheels at the same place we have to apply equal force at the center of each wheel parallel and opposite to the friction. These external forces creats an external torque on the system and hence angular momentum is not consurved.
     
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