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Mechanics - marble rolling in dish

  1. Jan 20, 2013 #1

    micromass

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    1. The problem statement, all variables and given/known data
    This is Problem 6.34 from Kleppner:
    A marble of radius b rolls back and forth in a shallow dish of radius R. Find the frequency of small oscillations. R>>b.

    Apparently, the answer is [itex]\omega=\sqrt{\frac{5g}{7R}}[/itex].


    2. Relevant equations



    3. The attempt at a solution
    Ok so I have my work so far shown in the uploaded picture. I chose my origin to be fixed to the center of the dish. You can see my expression for L and the problem is that term [itex]b^{2}\dot{\theta }[/itex]. If it were [itex](R - b)^{2}\dot{\varphi }[/itex] then I would have the answer but it isn't. I tried working with the geometry as much as possible e.g. considering areas of sectors swept out by the two different angles and seeing if they could be equal but I couldn't get anything useful. Any help would be appreciated.

    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-snc6/v/181191_324228377687326_892524912_n.jpg?oh=5761c9f6c173a7c7a5b8f99566ea0d3c&oe=50FF13AD&__gda__=1358899830_0b1d66b83972727376fa8fc0c558f7ce
     
  2. jcsd
  3. Jan 20, 2013 #2

    TSny

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    Hi, micromass.

    Don't forget the torque due to the friction force between the marble and the bowl.

    There is a kinematical constraint of rolling without slipping that will allow you to write ##\dot\theta## in terms of ##\dot\phi##.

    [Edit: Also, the algebra is less tedious if you use the theorem that the net torque about the center of mass of a system is equal to the rate of change of angular momentum about the center of mass even when the center of mass is accelerating.]
     
    Last edited: Jan 20, 2013
  4. Jan 20, 2013 #3

    ehild

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  5. Jan 20, 2013 #4

    micromass

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    Sorry ehild I didn't quite get to look at the thread because I got it (at least I think I got it) before I saw your response. Instead of taking the origin to be fixed to the center I took it to be instantaneously co - moving alongside the point of contact of the marble with the ground. This way I don't need to deal with friction. In this frame, [itex]L = R \times mv + I\omega = \frac{7}{5}mb^{2}\dot{\theta } [/itex]. We also have in this frame that [itex]\tau = R \times F = -mgbsin\varphi [/itex] using the same angles as before. Thus we have [itex]\frac{\mathrm{d} L}{\mathrm{d} t} = \frac{7}{5}mb^{2}\ddot{\theta } = -mgbsin\varphi [/itex] and using the fact that geometrically [itex]v = (R - b)\dot{\varphi }[/itex] along with the already used no slipping condition [itex]v = b\dot{\theta }[/itex] plus the fact that [itex]sin\varphi \approx \varphi [/itex] for small oscillations, [itex](R - b)\ddot{\varphi } = -\frac{5}{7}g\varphi [/itex]. [itex]R - b \approx R[/itex] therefore [itex]\ddot{\varphi } = \frac{-5g}{7R}\varphi [/itex] which gives the desired result. My only question is, if this is correct, why are we allowed to take a frame in which the origin is instantaneously co - moving with the point of contact without having to add any extra terms to the torque or angular momentum due to the acceleration of the center of mass itself etc.
     
  6. Jan 21, 2013 #5

    ehild

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    I do not quite understand your equations. You are right, you can write torque=angular acceleration times moment of inertia for the contact point, as fixed axis.
    I do not understand what you did further. What did you call v, and what frame of reference you used. If that, moving together with the contact point: it is not an inertial frame of reference. For me, it is hard to imagine what happens in such frame. As far as I can see, the CM of the marble is stationary, but gravity would turn round... I would not use such frame.
    If you use an instantaneous inertial frame with respect to the contact point, the speed of the CM is not equal to b(dθ/dt), linear speed of the perimeter of the marble. The rolling condition is R(dφ/dt)=b(dθ/dt) instead. And the angular velocity of the marble is d(θ-φ)/dt. Anyway, the same equation results you got at the end.

    ehild
     
    Last edited: Jan 21, 2013
  7. Jan 21, 2013 #6

    micromass

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    But if we don't take a frame co - moving with the contact point - that is an origin that is always with the contact point then we have to deal with friction.
     
  8. Jan 21, 2013 #7

    ehild

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    You can write the torque equation with respect to the point of contact instantaneously in rest: so no effect of friction as it acts at the axis.

    ehild
     
  9. Jan 21, 2013 #8

    micromass

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    But this origin has to be constantly moving with the contact point so why is this not an issue when calculating the torque?
     
  10. Jan 21, 2013 #9

    ehild

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    There are three different coordinate system: One fixed to the CM of the marble. The friction has to be included, and we need both equations, one for the motion of the CM and one for rotation. I understand that one most.

    The other is an inertial frame of reference at an instant, fixed to the instantaneous axis at the contact point. A bit confusing.

    The third is non-inertial frame of reference, moving together with the contact point, with all forces of inertia. That I can not handle.

    ehild
     
  11. Jan 21, 2013 #10

    micromass

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    So basically what I did to get the solution was use the second coordinate system?
     
  12. Jan 21, 2013 #11

    ehild

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    I think it was the third.
    What about writing up the Langrangian in a simple inertial frame of reference?

    ehild
     
  13. Jan 21, 2013 #12

    micromass

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    At a given instant if I fix an inertial frame to the point of contact, as you stated, then [itex]\frac{\mathrm{d} L}{\mathrm{d} t} = \frac{\mathrm{d} }{\mathrm{d} t}(R\times mv_{cm} + I_{cm}\omega )= \frac{\mathrm{d} }{\mathrm{d} t}(mb^{2}\dot{\theta } + \frac{2}{5}mb^{2}\dot{\theta }) = -mgbsin\varphi [/itex] is it not (I am using the same angles I defined in the picture I uploaded above)? This combined with the fact that the center of mass itself is traveling on a circle of radius R - b gives [itex]v_{cm} = (R - b)\dot{\varphi }[/itex] and the marble isn't slipping on the dish so [itex]v_{cm} = b\dot{\theta }[/itex] and that gives us what we want.
     
  14. Jan 21, 2013 #13

    ehild

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    The problem is that [itex]v_{cm} = b\dot{\theta }[/itex] is not true.

    ehild
     
  15. Jan 21, 2013 #14

    micromass

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    I'm not supposed to use lagrangians for this book. Why is that condition not true if I use a frame that is instantaneously at rest with respect to the contact point?
     
  16. Jan 21, 2013 #15

    ehild

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    The rolling condition [itex]v_{cm} = b\dot{\theta }[/itex] is valid if the marble rolls along a straight line. Not the case here. .

    ehild
     
  17. Jan 21, 2013 #16

    micromass

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    But we are saying at any given instant that the point of contact has to be at rest w.r.t the dish and the tangential velocity of the point of contact w.r.t the dish is [itex]v_{cm} - b\dot{\theta }[/itex] at the given instant is it not?
     
  18. Jan 21, 2013 #17

    ehild

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    Sorry I did not notice that theta was the angle with that the marble is rotated from the vertical. It is all right then.

    ehild
     
  19. Jan 21, 2013 #18

    micromass

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    Awesome!!! Thanks a lot ehild and TSny!! I'm glad there are people like you on PF :smile:
     
  20. Jan 21, 2013 #19

    ehild

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    I was confused as no slipping condition means that the length of the arc covered by the marble on the big circle is the same as the length of arc on its perimeter between two contact points. That is bβ, where β is the angle with respect to the line between the centres. bβ=Rφ. Meanwhile the marble rotates with θ=β-φ with respect to the vertical, so bθ=b(β-φ)=(R-b)φ.

    ehild
     
  21. Jan 21, 2013 #20

    micromass

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    Actually, what difference would it have made if we had taken theta with respect to the horizontal at every instant?
     
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