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Problems with ideal fidget spinners

  1. May 27, 2017 #1

    arivero

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    Consider an ideal fidget spinner, composed of a central disk (a bearing, usually) plus some N disks, all of them of radius L distributed regularly with their centres along a circle of radius R. In the simplest case all of them have the same mass, M, and are able to turn freely without resistence. The central disk constitutes a pivot, attached to some fix equipment in a way parallell to gravity field, so the whole problem is one of rotations in a plane. Plus, the symmetry of the problem makes that total angular momentum is preserved.

    I think that some interesting problems could be proposed in this setup because it goes a little beyond the simplest planar rotation but, not having external force field, is is still simpler than other articulated problems as the double pendulum. So perhaps this thread could aggrupate them.

    A first question: percussion. If with start the spinner with a single impulse [itex]F \Delta t[/itex], applied say at a distance R+L of the rotation center, so that we transmit a total angular momentum [itex]F \Delta t (R+L)[/itex], ¿which will be the rotational speed of the spinner? ¿will it be the same if we block the disks so they are not able anymore to rotate freely?
     
    Last edited: May 27, 2017
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  3. May 27, 2017 #2

    arivero

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    For a second question, substitute the disks by pairs of masses, this is, dumb-bells of lenght 2L and mass M/2 in each weight, still cased into a sort of ball bearing casing, so that their centre is still constrained to be in the circle of radius R. Solve again the problem. Is it possible to transfer angular momentum from the orbiting system to each dumb-bell, altering thus the rotational speed?

    Now, for a more complicated thing, allow the dumb-bell to have assymetrical masses, say m1 and m2 with m1+m2=M. Will the rotational speed of the spinner, after the initial percussion, be constant in this case? If not, can you describe the variation?
     
  4. May 27, 2017 #3

    haruspex

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    The impulse is applied to the large disk, right?
    What will be the rotation rate of the smaller disks about their centres? If you are not sure, consider the impulse one of them gets, and what torsional impulse it exerts.
     
  5. May 27, 2017 #4

    arivero

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    I think that the in the simpler case yes, the impuls is on the casing (not needed to be a disk really, typically treefoiled in commertial spinners)

    But applying the initial impulse to the disks, or to the dum bell, makes also a nice variant. Interesting.
     
  6. May 27, 2017 #5

    arivero

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    To fix the notation, and provide some drawing, I think the the general case could be this one

    upload_2017-5-28_1-17-13.png

    In the simplifications one could consider uniform disks of mass M instead of the dumbars, and to start considering only one disk, so

    upload_2017-5-28_1-24-16.png
     
    Last edited: May 27, 2017
  7. May 27, 2017 #6

    haruspex

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    Applying the impulse to the common support (casing, large disk, whatever) is somewhat trivial, if youanswer my question in post #3. Applying it instead to one of the smaller disks/dumbbells is more interesting, but you need to specify the orientation and location of the impulse so that the torsional impulse around each axis can be determined.
     
  8. May 27, 2017 #7

    arivero

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    Yes I think that for the common support and with an uniform disk or a symmetric dumbar, the questions is the same that with a point mass M at distance R, is it? So if the impulse is I, the angular momentum would be (L+R) x I, and the angular speed [tex] (L+R) I \over M R^2[/tex]

    is it?

    for the less trivial case, we could consider the impulse at the same distance, R+L, and direction (tangential to the main casing) but applied on the internal disk. At this moment I am unsure of how the total angular momentum, say J, should divide between "orbital" and "spin" parts

    Could it easier, to study percussions, to consider first the "constrained dumbell" instead of the uniform disk?
     
  9. May 29, 2017 #8

    arivero

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    As for the equations of movement, not I look at them, it is more or less the same than the "horizontal" (g=0) double pendulum, with zero mass (and thus zero inertia) for the "shoulder" piece. So for instance from https://physics.csuchico.edu/ayars/427/handouts/AJP001038.pdf we can extract the study for the coordinate [itex]\phi_2=\theta_2-\theta_1[/itex], the rotation of the disk seen from the (also rotating) system of reference of the casing of the spinner.
    [tex]
    \dot \phi_2 = \pm \sqrt { 2 E (M R^2 + I_d + 2 M R l \cos \phi_2) - J^2 \over M I_d R^2 - M^2 R^2 l^2 \cos^2 \phi_2}
    [/tex]
    Where J is the total angular momentum of the system and [itex]l[/itex] is distance between the centre of rotation of the disk and its centre of mass, which is zero for a symmetrical disk or a symmetrical dumbell, so that in such case

    [tex]
    \dot \phi_2 = \pm \sqrt { 2 E (M R^2 + I_d) - J^2 \over M I_d R^2}
    [/tex]

    And particularly for the disk [itex]I_d=\frac 12 M L^2[/itex] so

    [tex]
    \dot \phi_2 = \pm \sqrt { 2 E M (R^2 + \frac 12 L^2) - J^2 \over \frac 12 M^2 L^2 R^2}
    [/tex]

    Just for check out, we can put [itex]J= M R^2 \dot \theta_1 + \frac 12 M L^2 \dot \theta_2[/itex], so

    [tex]
    \dot \theta_2 -\dot \theta_1= \pm \sqrt { 2 E/M (R^2 + \frac 12 L^2) - ( R^2 \dot \theta_1 + \frac 12 L^2 \dot \theta_2 )^2 \over \frac 12 L^2 R^2}
    [/tex]

    which fits with [itex]E=\frac 12 M R^2 \dot \theta_1^2 + \frac 14 M L^2 \dot \theta_2^2[/itex] as expected :-)

    By the way, a non null [itex]l[/itex] seems a lot funnier, as it will cause changes in the angular speed [itex]\dot \theta_1[/itex] From eqs (2) and (4) of the cited paper, we can see the extra corrections to energy and total angular momentum in such case.

    [tex]E_l = M R \ l \ \dot \theta_1 \dot \theta_2 \cos \phi_2[/tex]
    [tex]J_l= (\dot \theta_1 + \dot \theta_2) M R \ l \ \cos \phi_2[/tex]

    (btw, I wonder which is the advantage of using the relative angle [itex]\phi_2[/itex] instead of the absolute [itex]\theta_2[/itex] and so the difference [itex]\theta_2 - \theta_1[/itex])
     
    Last edited: May 29, 2017
  10. May 29, 2017 #9

    arivero

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    Fidgetter-wise, it could be thought that the player should want all its energy used to increase [itex]\dot \theta_1[/itex], so a greasy, well working bearing should be, for the same mass, more benefitial than a stuck one, where forcefully [itex]\dot \theta_2[/itex] is equal to [itex]-\dot \theta_1[/itex] and thus nonzero. But on the other hand, the player could want a higher momentum of inertia, and the stuck bearing, via https://en.wikipedia.org/wiki/Parallel_axis_theorem, grants it to be [itex]MR^2 + \frac 12 M L^2[/itex]
     
    Last edited: May 29, 2017
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