Exploring the Relationship Between Rotational Rates and Area in Calculus

  • Thread starter Duhoc
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In summary: If you want the area under the curve, you need to integrate the curve from the starting position to the ending position. The result is a function of the radii and the rates of rotation. The specific relationship depends on the specific values of R1, R2, and \omega.
  • #1
Duhoc
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Let's say that someone is drawing a circle with a compass. As that circle is being drawn a second compass is attached to the first such that the needle leg is attached to the pencil of the first. Only instead of a needle it is a small wheel. As the first compass inscribes its circle the second compass is tracing out the circumference of the first and drawing a circle at twice the rate of rotation of the first. An analogy would be someone at the edge of a merry-go-round that is rotating above a piece of paper. The rider endeavors to trace out a circle on the paper as the merry-go-round rotates at twice the rate of rotation of the merry-go-round. Is there an exponential relation between the relative rates of rotation and the area incribed under the curve created by the second compass? Additionally, if a third compass rode on the path inscribed by the second attempting to sketch a circle at some ratio of the rate of rotation of the first and second could that area be expressed? (Please private message as well as post.)

Thank you,
Duhoc
 
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  • #2
Doesn't look all that hard. If you take (0,0) as the center of the circle described by the compasses and [itex]\omega[/itex] as the "rate of rotation", assuming radius R1, then the circle itself is described by the equations [itex](R_1cos(\omega t), R_1 sin(\omega t))[/itex]. A second circle with center at that point, radius R2 and "rate of rotation" [itex]2\omega[/itex] is given, relative to that point, by [itex](R_2cos(2\omega t), R_2 sin(2\omega t))[/itex].

That point, relative to the original coordinate system is just the sum:
[itex](R_1 cos(\omega t)+ R_2cos(2\omega t), R_1 sin(\omega t)+ R_2 sin(2\omega t))[/itex]
 
  • #3
Are you sure?

Remember that the second circle is not a circle. It is rotating around the circumference of circle 1 as it attempts to inscribe a circle forming a curve. I am seeking an expression for the area under that curve relative to the rotation rates of the circle and the would-be circle. And I would like to know, is that area in some way related to the relative rates of rotation of the two compasses. Again, to clarify, compass two has a wheel that is riding along the track of the circumference of circle one as it is trying to inscribe circle 2 at twice the rate of rotation of circle one. This is exactly analagous to an elbow joint rotating in relation to a shoulder joint.
 
  • #4
Yes, I understand all that. The equations I gave are correct.
 

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