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Rotating rubber wheels made to contact. Find final angular veloctiy

  1. Jul 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid rubber wheel of radius [itex]R[/itex] and mass [itex]M[/itex] rotates with angular veloctiy [itex]\omega_0[/itex] about a frictionless pivot. A second rubber wheel of radius [itex]r[/itex] and mass [itex]m[/itex], also mounted on a frictionless pivot, is brought into contact with it. What is the final angular velocity of the first wheel?


    2. Relevant equations

    Conservation of angular momentum?

    3. The attempt at a solution



    I'm not entirely sure what this question is asking. I think it means that the wheels are held together and you must find the final angular velocity of the first one. If that is the case then you would use the relations [itex]R\omega = r\omega'[/itex] and [itex] I\omega_0 = I\omega + I'\omega' [/itex], noting that the initially stationary wheel would have negative angular velocity. Is this the way to do it?

    2rfpxs7.jpg
     
    Last edited by a moderator: Jul 18, 2014
  2. jcsd
  3. Jul 18, 2014 #2
    Where did you read about this conservation law ?

    No . This is not the way .

    Initially only wheel 1 is rotating about its center . Then a second stationary wheel is mounted on top of it . Due to friction between the surfaces ,the angular velocity of first wheel reduces whereas that of second one increases .

    Is there any external torque acting on the system of two wheels about the central pivot ?
     
  4. Jul 18, 2014 #3

    Sorry I meant momentum.

    I will post an image of the problem but there is no external torque that I can see.
     
  5. Jul 18, 2014 #4
    Okay ,this problem is different from what I was perceiving before you attached the picture.

    Here is how you would solve this problem.

    When the two wheels are brought in contact then friction acts at the point where they touch each other .It is equal in magnitude but opposite in direction on the two wheels . The friction on wheel 1 provides an angular impulse to the wheel, such that its angular momentum changes . Similarly angular impulse due to friction on wheel 2 causes the wheel's angular momentum to change .The friction acts till there is slipping between the wheels .

    So you need to write two separate equations for the two wheels and an additional equation pertaining to the no slipping condition.

    These three equations will give you the answer .But please be careful with the signs .
     
    Last edited: Jul 18, 2014
  6. Jul 18, 2014 #5

    ehild

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    You are on the right track. The wheels interact and change the angular velocity of the other till they slip on each other. When they move with the same velocity at the contact, the angular velocities do not change any more. Be careful with the signs: the velocity at a point is ##\vec v = \vec ω \times \vec r ##

    ehild
     
  7. Jul 18, 2014 #6
    The first equation I gave was the condition for no slipping. The second was conservation of angular momentum as there as I can't see an external torque on the two wheels.
     
  8. Jul 18, 2014 #7

    ehild

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    It is better to solve the equations of motion for the wheels. There is kinetic friction between the wheels and that decelerates the rotation of the first wheel and accelerates the rotation of the second one. Assume that friction is constant in time. How would the angular velocities change with time?

    ehild
     
    Last edited: Jul 18, 2014
  9. Jul 18, 2014 #8
    Right.

    You have conserved angular momentum about which point ? Just think about this .
     
  10. Jul 18, 2014 #9
    Ok continuing on from what I first thought, that would be the magnitudes of the angular velocities multiplied by the radii are equal, i.e. [itex]R\omega = r\omega'[/itex]. The angular velocities have opposite. Using conservation of angular momentum [itex] \frac12 MR^2\omega_0 = \frac12 MR^2\omega - \frac12 mr^2\omega' [/itex] and subsituting [itex] \dfrac{R\omega}{r} = \omega'[/itex] gives you the final angular velocity of the first wheel.

    This, however, gives a angular velocity greater than the initial one which doesn't make sense considering that the torque was slowing the wheel down. I am missing something here with the signs.
     
  11. Jul 18, 2014 #10

    ehild

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    I would not say that the angular momentum is conserved. It is not a simple rotation about a fixed axis, but there are two fixed axes. One exerts torque with respect to the other. Try to solve the problem the other way.


    ehild
     
  12. Jul 18, 2014 #11
    Ok so letting the friction force be [itex]f[/itex] and using the relation [itex]\tau = I\alpha[/itex], we see the equation of angular speed for the first wheel is [itex] \omega_0 -\dfrac{2f}{MR}t[/itex] and the second is [itex] \dfrac{2f}{mr}t [/itex]. With the other relation we have [itex] R \left(\omega_0 -\dfrac{2f}{MR}t\right) = r \left(\dfrac{2f}{mr}t\right).[/itex] Solving for [itex]2ft[/itex] we find [itex]\omega = \dfrac{M\omega_0}{M+m} [/itex]
     
  13. Jul 18, 2014 #12

    ehild

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    It looks good.

    ehild
     
  14. Jul 18, 2014 #13
    Thank you.
     
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