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Rotationnal velocity and acceleration

  1. Nov 28, 2014 #1
    Hello,

    I would like to be sure about my rotationnal velocities w1(t) and w2(t):

    http://imagizer.imageshack.us/v2/800x600q90/912/c6X2du.png [Broken]


    The black arm is turning clockwise at w1. The blue disk is turning at w2 with w2 < w1. w1 and w2 are in labo frame reference. The disk is turning around itself counterclockwise at w2'=w2-w1 in the arm frame reference.

    I1: inertia of black arm+disk
    I2: inertia of the disk
    F: force from friction
    r: radius of the disk
    t: time.

    I guess no friction elsewhere than between disk and black arm. I guess that the force F from friction is constant even rotationnal velocities are changing, just for simplify calculations. Friction gives forces F1 and F2.

    For me, while w2<w1 and while the force F can be constant :

    w1(t)=w1-Frt/I1

    w2(t)=w2+Frt/I2-Frt/I1


    For you, is it ok ?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 28, 2014 #2

    Bystander

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    You are missing a radius in the problem statement, are you not?
     
  4. Nov 28, 2014 #3
    I defined the radius of the disk: 'r' and I take in account in w1(t) and w2(t) with Fr=F*r, I need another radius ?
     
  5. Nov 28, 2014 #4

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    An arm radius? You're trying to conserve angular momentum.
     
  6. Nov 28, 2014 #5
    I need it for the rotationnal velocities ? could you explain please ? There is friction, so energy of heating.
     
  7. Nov 28, 2014 #6
    What is this device?
     
  8. Nov 28, 2014 #7

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    You need it to calculate angular momentum, the quantity that must be conserved as the two angular velocities change.

    Edit: Hang on a minute, that's built into "I." Let me double-check myself --- I keep thinking you need it to get the angular accelerations.
     
    Last edited: Nov 28, 2014
  9. Nov 28, 2014 #8
    but there is energy of heating too, no ? angular momentum is conserved even there is energy from friction ?

    @zoki: just a disk with an arm, and friction between when w2 < w1.
     
  10. Nov 28, 2014 #9

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    The angular momentum is conserved. Total energy is conserved as that dissipated by friction plus the remaining rotational energy.

    ω1I1 + ω2I2 = constant;

    E = I1ω12 + I2ω22 - Frtω2.

    ... and, one radius is all you need. My apologies.



    You could call it a "tidal orrery" to demonstrate tidal friction --- doesn't change orbital radius of the moon, though.
     
  11. Nov 28, 2014 #10
    my expressions of w1(t) and w2(t) are correct, or not ? If I use your formula this would say that w2(t)=w2+Frt/I2, true ? w2(t) is increasing because it receives a clockwise torque from F2/F3, but in the same time w1(t) is decreasing of -Frt/I1, for me w2(t) depends of the change of w1(t) because the disk is on the arm, no ?

    Maybe I need to define moment of inertia:

    I1=Md²/3+md², for the arm+disk, 'd' is the length of arm, 'M' the mass of arm
    I2=mr²/2, for the disk, 'm' the mass of the disk
     
    Last edited: Nov 28, 2014
  12. Nov 28, 2014 #11

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    From conservation of angular momentum, ω1(t=0)I1 + ω2(t=0)I2 = constant = ω1(t)I1 + ω2(t)I2.

    Rearranging, ω1(t) = ω1(0) - (I2/I1)(ω2(t) - ω2(0)).

    The expression for conservation of energy is not strictly correct; the friction loss is a time dependent rate and must be integrated from t=0 to t of interest.
     
  13. Nov 28, 2014 #12
    I need to have 2 equations: w1(t) and w2(t), so w1(t)=w1-Frt/I1 is correct ? Your equation is for find w2(t) ?


    Edit: So like this: w2(t)=w2+Frt/I2
     
  14. Nov 28, 2014 #13

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    No, and no. That is ω1(t) in terms of ω1(0), ω2(t), ω2(0), I1, and I2.
     
  15. Nov 28, 2014 #14
    Ok, but I have only one equation, and I would like to find w1(t) and w2(t), how can I do ?
     
  16. Nov 28, 2014 #15

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    I've shown you a rearrangement for ω1(t); you can rearrange in similar fashion for ω2(t).
     
  17. Nov 28, 2014 #16
  18. Nov 28, 2014 #17

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    I1 and I2 are not necessarily equal from the information given.
     
  19. Nov 28, 2014 #18
    Sure, but why I can't write:

    w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1
    w2(t)=w2(0)-Torque/I2*t = w2(0)-Frt/I2

    Why the formula of angular acceleration is not correct in this case ?

    With:

    I1=Md²/3+md², for the arm+disk, 'd' is the length of arm, 'M' the mass of arm
    I2=mr²/2, for the disk, 'm' the mass of the disk, 'r' the radius of the disk
     
  20. Nov 28, 2014 #19

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    Nor, are the two torques equal.
     
  21. Nov 28, 2014 #20
    Sorry, it is:

    w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1
    w2(t)=w2(0)+Torque/I2*t = w2(0)+Frt/I2

    Why the formula of angular acceleration is not correct in this case ?
     
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