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Help with cycloid

  1. Jul 3, 2006 #1
    I was working out some old putnam problems and one of them is as follows:

    "A circle radius r rolls around the inside of a circle radius 3r, so that a point on its circumference traces out a curvilinear triangle. Find the area inside this figure."

    Currently I'm just trying to find the equation of the path it traces. According to the following site:

    http://www.kalva.demon.co.uk/putnam/psoln/psol396.html [Broken]

    it is not hard to see that x/r = 3 - 2 cos θ - cos 2θ, y/r = 2 sin θ - sin 2θ. But I can't see it. Can anybody help explain why this is so?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 3, 2006 #2


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    Say the small circle starts at the top of the inside of the large one. Call the point on the small cirlce currently touching the big circle the marker. Now if the center of the small circle has moved through an angle t about the center of the big circle, then the small circle has rolled 3rt along the circumference of the big circle, which means that the marker, the center of the small circle, and the point where the big and small circle touch make an angle of 3t. I don't see immediately why the equations you've given for x/r and y/r are right, but given what I've said, you should be able to parametrise the path of the marker in terms of the angle t. It's then a matter of using first-year calculus to find the area of the "triangle" by integrating using polar co-ordinates.
  4. Jul 4, 2006 #3


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    In their equations, when theta = 0, (x, y) = (0, 0) which does not seem the simplest from one way of thinking about it. You can get equations similar to theirs if you consider the following facts, as the circle rolls from the rightmost position in the larger circle.
    1. The small circle rotates at a constant rate and its center moves at a constant rate along a circle C inside the larger circle. C has radius 3r - r = 2r.
    2. The small circle makes two revolutions per full roll around the large circle (this is slightly harder to see but you can check that when the point on the circumference goes from one contact point with the large circle to the next contact point, the small circle must make only 2/3 of a rotation. Consider the angle with the center of the small circle that the point must go through during the partial roll). Since the revolutions of the small circle are made in the opposite direction in which the small circle's center moves, they occur at a rate of -2 radians/sec if the small circle's center moves at 1 radians/sec.

    The equation for a point--in this case the small circle's center--revolving at rate of 1 radian/sec around a circle of radius 2r (namely C) is
    [tex](x, y) = (2r cos \theta, 2r sin \theta)[/tex]
    The equation for the circumference point on the small circle revolving at a rate of -2 radians/sec around a circle of radius 1r (namely the small circle) is
    [tex](x, y) = (r cos (2 \theta), - r sin (2 \theta))[/tex]
    Adding them together to get the resultant path,
    [tex](x, y) = (2r cos \theta + r cos(2 \theta), 2r sin \theta - r sin(2 \theta))[/tex]

    For some arbitrary reason, they flipped the figure over the y-axis relative to that equation and moved the center of the "triangle" to (3, 0), but these are just rigid transformations and do not affect the area.
    Last edited: Jul 4, 2006
  5. Jul 4, 2006 #4
    Thank you both. You've been very helpful.