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Perpendicular force with a spiral

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  1. Jan 3, 2015 #1
    1. The problem statement, all variables and given/known data

    A red Archimedean spiral is fixed to the ground. An external motor turns a grey support clockwise at w, the support can only turn around itself. On the support there is one orange disk that doesn't turn around itself at start. A stem is on the support.

    That stem:

    - turns like the support clockwise at w
    - is attached on the spiral at one end
    - has no mass
    - is connected to the disk:
    ........... - if the spiral attrack the stem: like gears can be: no friction between the stem and the disk, no sliding
    ............- if the spiral block the stem, there is friction between the stem and the disk

    At start w=0.

    1) the spiral attack or block the stem ?
    2) drawn all forces
    3) find the work that the motor need to give
    4) find the work needed to move the stem around the spiral
    5) does the disk has a rotation around itself ?

    Datas:

    Inertia of the support: I1
    Inertia of the disk around main axis: I2
    Inertia of the disk around itself: I3

    There is no friction except between the stem and the disk

    http://imageshack.com/a/img910/1162/77armo.png [Broken]

    2. Relevant equations

    --

    3. The attempt at a solution

    1) The support turns clockwise so the disk turns counterclockise at -w in the support reference, I need to add 2piR/4 when the support has turned of 90°. I noted R the radius of the disk. I measured it with 3 positions:

    http://imageshack.com/a/img540/3440/9WNKka.png [Broken]

    I don't know how to calculate this, if you have a method because it's only a graphical solution. With the graphical method, the spiral block the stem.

    Edit:With a numerical method I found the the spiral attrack the stem. I found for my spiral, the distance pass from 11 to 13.194 and the tangent is at 1.3274, so the radius of the disk can be 1.3274, it become 1.3274*2*pi/4=2.0850 but the spiral move to 2.194, it confirms what I found with a graphic method. I take a=1.4 for the spiral. The parametric equations are : ##x=-1.4\theta cos\theta## and ##y=1.4\theta sin\theta##.

    2) The stem gives the forces F2 to the disk and receives F1. The spiral receives the force F4. The support receives the forces F3.

    http://imageshack.com/a/img673/3516/M03spN.png [Broken]

    the stem rotates the disk clockwise.

    3) the motor needs to give only the work for turn the support and the disk around the main axis, the work is :

    ##\frac{1}{2}I_1w^2+\frac{1}{2}I_2w^2##


    4) The force F4 is always perpendicular to the trajectory then the work needed is 0. The stem has no mass so its kinetic energy is always 0.

    5) Yes, I measured the distance for 3 positions and the stem rotates the disk around itself clockwise, and there is friction. In this case the support must receive a counterclockwise torque for keep constant the energy. So, I think I'm wrong in my forces or the spiral don't attrack or block the stem ?

    With a better image:

    http://imageshack.com/a/img673/7249/JcXIC5.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 3, 2015 #2

    haruspex

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    Are you not told the constant for the spiral in terms of R? Determining the relationship from measurements of the diagram seems unsatisfactory.
    Looks to me that the stem is always perpendicular to the spiral at point of contact. (In fact, you have assumed this in your force diagram.) That should give you a useful relationship.
    These questions are a bit confusing. If work is needed to move the stem then that's on-going, i.e. we're talking about power, or maybe work per radian; but unless the disk's rotation is accelerating the work to rotate the support and disk are only what was required to set it moving from rest. It's not clear what is being asked.
    Do you mean 'attract' the stem, i.e. pulls the stem out?
    If there's no friction then there is sliding. Which do you mean?

    In your attempt at solution diagram, the stem extension after a quarter turn is correct, but not the one after a half turn (2R).

    Your work equation is not right. Even if the disk were not rotating in itself, it would still be orbiting the centre of the support.
    Also, you've taken the angular velocity to be the same for support and disk. Have you shown they're the same?
     
  4. Jan 3, 2015 #3
    True, I don't understood that point, I corrected my last calculation, like I see the picture R is choose for have the stem perpendiculary to the spiral, always. So, the equation of my spiral is ##x=−1.4\theta cos\theta## and ##y=1.4\theta sin\theta##. I have some trouble for find a general relation between the spiral and R but for my example with a=1.4, I found the derivated of 1.4y' = ##1.4\theta sin\theta##, it's ##sin\theta + \theta cos\theta##, I'm looking for y'=0, I found ##\theta=2.0287 rad## and the radius is R=1.25. Each 90° the spiral is far away of ##1.4*5*2pi/4*sin(5*2pi/4)-1.4*6*2pi/4*sin(6*2pi/4)=11-13.194=-2.194##. With ##R*2pi/4 < 2.194##. Like the support turns the spiral is far away after 90°. But maybe it's not like that R can be find ?

    Yes, I need to find if the spiral push or pull the stem.

    imagine the link between the disk and the stem like 2 gears, no friction and no sliding. But it's only is the spiral pull the stem not push.

    Why ? I don't see where my drawing is false, I reported all distances I found.

    I think the question is the work when the angular velocity is at w.

    The disk can turn around itself but it turns with the support at w, the axis of rotation of the disk is on the support, so the disk is at w too. But I'm not sure it turns around itself. I drawn a grey part inside the orange disk for show its axis of rotation is on the support.
     
    Last edited: Jan 3, 2015
  5. Jan 3, 2015 #4

    haruspex

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    I've come to the conclusion that an Archimedean spiral cannot result in such constant perpendicularity.
    Let P = (r, θ) be a point on the spiral with r only a little greater than R. Let Q be the point where the stem touches the disk. So OQP is a right angled triangle, |OQ| = R, |OP| = r. The angle QOP is small. The tangent to the spiral at P is parallel to OQ, so almost parallel with OP. I.e. the curve is headed almost directly away from O, making ##\frac{dr}{d\theta}## very large. But in an Archimedean spiral it should be constant.
    So either it is not Archimedean or we must drop the perpendicularity assumption.
     
  6. Jan 4, 2015 #5
    I guessed it was a Archimedean spiral, the exercice don't speak about the Archimedean but the image shows the perpendicularity in 2 positions, so it must be perpendicular always. I think it's possible to have a shape always perpendicular to the stem, in this case is it a Theodorus of Cyrene's spiral ?

    http://en.wikipedia.org/wiki/Spiral_of_Theodorus

    If yes, the distance is ##d=\frac{\sqrt3}{cos(\frac{\pi}{2}-acos(\frac{1}{\sqrt(2)})-acos(\frac{\sqrt(2)}{\sqrt(3)})}+R=1.75+R## far away for an angle of 90° for the spiral. Like the disk turns counterclockwise the stem moves of 1.57R, the distance is 1.75+R far away, so the stem can push or pull the spiral, it depends of the diameter of R ?
     
    Last edited: Jan 4, 2015
  7. Jan 4, 2015 #6

    haruspex

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    no. Try http://en.m.wikipedia.org/wiki/Involute
     
  8. Jan 4, 2015 #7
    Thanks ! I calculated some examples with the disk in the center of the support, and the distance is always 1.57R for 90°, like the disks turns counterclockwise, the stem moves of pi/2*R=1.57R, so the stem moves like the spiral, the spiral don't push nor pull the stem ? It is the same result if the disk is not in the center ?
     
    Last edited: Jan 4, 2015
  9. Jan 4, 2015 #8

    haruspex

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    No, I think it will be different for the off-centre disk.
    But instead of thinking of it as a straight straw, now, I think it will help to view it as a string wound onto a disk of radius R at the centre of the support.
    Consider three points on the string at some instant: A where the string touches the off-centre disk, B where it forms a tangent to the central disk and C, another 90 degrees around the central disk. AB is the distance between the centres of the disks.
    When the support turns 90 degrees, unwinding the string, C is now the tangent point. Where have A and B moved to?
     
  10. Jan 4, 2015 #9
    I try to do what you asked, I'm not sure to understand all your message, I drawn an image:

    http://imageshack.com/a/img661/4720/81Snwo.png [Broken]

    I noted A', B', C', points after unwinding the string. It's correct ? But I don't understand why the distance is not the same.
     
    Last edited by a moderator: May 7, 2017
  11. Jan 4, 2015 #10

    haruspex

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    No, that's not what I tried to describe.
    In keeping with the involute model, I've put another disk radius R at the centre of the support, but it doesn't rotate. We can think of the straw as a string unwinding from this disk as the support rotates. To avoid confusion, I'll call it the drum and reserve 'disk' for the disk in the statement of the problem.
    A, B and C are points on the string and move with the string (to A', B', C' say). A is initially touching the disk, B touches the drum at the point where the string is currently unwinding from the drum, and C is a quarter turn further around the drum. So the line AB is initially tangential to both disk and drum.
    After the support has turned 90 degrees, A', B' and C' will be in a straight line. The string now touches the disk at D. What is the distance B'D?
     
    Last edited by a moderator: May 7, 2017
  12. Jan 5, 2015 #11
    Sorry if I don't understand. Like that ?

    http://imageshack.com/a/img913/8661/42rQHN.png [Broken]

    B'D = AB -1.57 R ? It's the additionnal "length" (the spiral is far away) of the spiral after 90° ?
     
    Last edited by a moderator: May 7, 2017
  13. Jan 5, 2015 #12

    haruspex

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    Yes. Can you figure out from that how the disk rotates?
     
    Last edited by a moderator: May 7, 2017
  14. Jan 5, 2015 #13
    The disk rotates counterclockwise at 1.57R each quarter, like I drawn so the distance travelled is greater than the spiral ? The spiral push the stem ?

    Now, it depends of the distance AB ?

    If I take AB=1.57R, the spiral don't increase its "length" ?
     
    Last edited: Jan 5, 2015
  15. Jan 5, 2015 #14

    haruspex

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    Relative to the support or about its own centre?
    I don't understand what you mean. The distance travelled by what, and what do you mean by distance 'travelled' by the spiral?
    I don't think so.

    Sorry, i just realised I asked the wrong question at the end of post #10. I meant to ask for the distance A'D. Is this consistent with the disk not having changed its orientation? Imagine the disk staying still and the stem rolling around it from the first position to the second position your post #11 diagram. Does A still land at A'?
     
  16. Jan 5, 2015 #15
    Relative to the support.

    For resume, I would like to know if the spiral push or pull the stem. I need to know the distance D1 to reach the spiral, because the spiral increase more and more and I need to know what the distance D2 the stem move because the stem is on the support and the stem is push by the disk (and turn with the support turn in the same time).

    I'm ok with that.

    I understood ! yes, A'D is not 1.57R it's lower. A'D = R ? But A'D is D1 or D2 I defined just above ? A'D is used for build the spiral ?
     
    Last edited: Jan 5, 2015
  17. Jan 5, 2015 #16

    haruspex

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    Right. The support has rotated clockwise a quarter turn, but the disk has rotated anticlockwise a quarter turn relative to the support. So...?
    AB=A'B'=CD=distance between centre of disk and centre of support. BC (measured around the arc) = B'C = pi R/2. You can deduce A'D.
     
  18. Jan 5, 2015 #17
    it don't turn in the lab frame reference

    A'D = 1.57R, but what is that distance ?
     
  19. Jan 5, 2015 #18

    haruspex

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    To be accurate, it's pi R/2. This just confirms that the stem rolls around the disk without turning it.
     
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