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Find radial imprint on rotating disk

  1. Aug 7, 2014 #1
    This problem came up after drawing a line on the spinning rotor of a food processor. I was idly musing about relativity (parallel motion and perpendicular motion). Maybe some ancient mathematician found the solution while working clay on their potting wheel! Here it is:

    A flat disk rotates about (0, 0) on the (x, y) coordinate system. A point on the edge of the disk has angular velocity ω. During this rotation, a straight pencil line is drawn from the centre (0,0) along the y-axis toward the edge at linear velocity, v

    Let the radius=1,
    ω=1 rad^-2
    v =1^-1

    [Essentially, you draw the straight pencil line at the same velocity that each point on the disk's edge rotates.]

    The disk stops rotating when the pencil line reaches the disk’s edge. Interestingly, the pencil line is revealed as curled, not straight.

    Obviously when the line is uncurled it is longer than 1. Then what is the method to find out by how much?
  2. jcsd
  3. Aug 7, 2014 #2

    Simon Bridge

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    You do a line integral using the parameterization of the resulting spiral.

    It helps to use polar coordinates:
    If you moved the pencil at a constant speed in the y direction, then ##r=vt## and ##\theta = \omega t## and the line is the trajectory ##\vec r(t) = (r(t),\theta(t))##.

    If the disk has radius R, then ##v=R\omega##, and ##0\leq t \leq R/v##

    The distance along the line between t and t+dt is dL ... which you can work out by pythagoras.
    The total distance L is found by ading up a;; the dL's - i.e. integrating from t=0 to t=R/v.

    This is a good application BTW.
    Someone mass-producing cakes may want to put a spiral pattern in icing on each one and they would want to know how much icing to buy. Therefore they need to know how long the spiral will be - better, they would want to know how to rig the machines so the icing use is most economic. i.e. how fast to move the "pencil" to get a saleable spiral without it costing too much ;)
  4. Aug 7, 2014 #3
    Thanks Simon

    I did end up drawing spirals. And even better, when the pencil mark is repeatedly drawn out to the edge and back to the centre you get all kinds of wonderful ornate patterns. I just need to figure out how to program my robot to do it...:)
  5. Aug 7, 2014 #4


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    That's the whole idea of the "spirograph (TM)", isn't it?
  6. Aug 7, 2014 #5

    Simon Bridge

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    I had one of those sets. I don't think spirograph draws spirals - lots of loopy patterns though.
    It seems to me that a robot can draw a spiral on a rotating surface pretty much as described - you can do all kinds of fun stuff with record turntables. Failing that it is a matter of mapping the parametric equation for the spiral trajectory into commands the robot understands.
  7. Aug 16, 2014 #6
    Just talking to myself here; when the line is uncurled it is 0.147794 times longer than 1.
  8. Aug 16, 2014 #7

    Simon Bridge

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    ... do you mean that the length of the line is 0.147794units or that the length is 1.147794units or something else?

    It is important to include your working. Using the statement:
    ... for a disk radius ##R##, turning at speed ##\omega##, the pencil speed is ##v=R\omega## ...
    $$r(t)=v t = R\omega t,\; \theta(t)=\omega t\\
    \implies r(\theta) = R\theta:0<\theta<1\\
    L=\int_0^1 \sqrt{r+\frac{dr}{d\theta}}\;d\theta = R\int_0^{1} \sqrt{\theta^2 + 1}\;d\theta = (1.14779)R$$ ... in your example, R=1.

  9. Aug 17, 2014 #8
    The workings are below the dotted line. But more importantly, why the heck am I doing this? It relates to an obsession with spinning disks. Just in case you're interested.
    Using proportions of one, my goal is to find several spiral lengths corresponding to their corresponding pencil radial velocities when they are [itex]≤ [/itex] angular velocity (which is always one and equals one radian). Next I divide each spiral length by the length of the spiral when the radial velocity equals one, which is my base value (and I think it's 1.1477936575). When I plotted various values in a graph, I found the proportions obeyed a hyperbolic form. This is also the case when you observe [itex]y=1/(1-x^2)^{1/2}[/itex], which is a hyperbolic relation (thanks for your comment in another thread). It is also the the equation for velocity and time dilation. But mathematically, it comes from the equation of a unit circle [itex]y=(1-x^2)^{1/2}[/itex]. So I want to discover if the spiral proportions bear any resemblance to something like worldlines. I feel some function will relate the two ideas: a rotating disk with a radial imprint of a pencil and the rotating disk of a relativity model. It could be just a crazy idea. So the analogue I'm visualizing is that a rotating disk always rotates at angular speed of one ( like constant speed of light). And the pencil's radial motion can only approach zero speed, but never reach it, because the spiral would need to be infinitely long at zero speed which I declare is impossible in this model. However it can reach one, (and 1.1477936575/1.1477936575 =1)which is the analogue of zero relative speed in relativity, if that makes sense.

    I "phoned a friend", but the jury is out.

    Radial velocity of pencil line drawing is constant and equals pen’s speed [itex]V[/itex]

    [itex]Vr = V[/itex]

    If disk rotates at angular speed [itex]ω[/itex] and pen is at distance [itex]r[/itex] from center, tangential velocity equals

    [itex]Vt = ωr[/itex]

    So, the net speed of pencil line drawing is

    [itex]V_{net} = (V^{2} + ω^2r^2)^{1/2}[/itex]

    Then, the length of the pencil line drawn during a small time [itex]dt[/itex] is

    [itex]dl = V_{net}dt = dt(V^{2} + ω^2r^2)^{1/2}[/itex]

    Now, since the pencil moves at constant radial speed [itex]V[/itex], displacement of pencil line from center after time [itex]t[/itex] is

    [itex]r = Vt[/itex]

    Substituting value of [itex]r[/itex],

    [itex]dl = dt(V^{2} + ω^2V^2t^2)^{1/2}[/itex]

    Integrating as time reaches from [itex]0[/itex] to [itex]t[/itex], and length reaches from [itex]0[/itex] to [itex]L[/itex], we get the following result

    [itex]L = (V/2ω)∗ln(ωt+(1+ω^{2}t^{2})^{1/2})+(Vt/2)(1+ω^{2}t^{2})^{1/2}[/itex]

    If we draw pencil line at same speed as disk’s edge speed, [itex]V= ωR[/itex]

    The time taken to rotate one radian [itex]t = 1/ω[/itex]

    Replacing these in above equation, [itex]L = (R/2)ln(1+2^{1/2})+(R/2)(2^{1/2})[/itex]

    As [itex]R=1[/itex]

    [itex]L = (1/2)*ln(1+2^{1/2})+(1/2)(2^{1/2})=1.1477936575[/itex]

    ........FOR OTHER PENCIL SPEEDS.........

    If [itex]k[/itex] is the ratio of speeds, then [itex]V=kωR[/itex]

    Time taken for pencil to reach edge

    [itex]t = R/V = 1/(kω)[/itex]

    Replacing in equation

    [itex]L = (kR/2)*ln(1/k+(1+1/k^2)+(R/2)(1+1/k^2)^{1/2}[/itex]
  10. Aug 17, 2014 #9

    Simon Bridge

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    The line integral was faster and involved less numerology.
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