Rotating Shapes

  • #1

Main Question or Discussion Point

As a means of curiosity, I am working on a problem involving the rotation of various shapes. The thing that I am interested in is the cross-sectional length of the shapes as they cross the x axis. So imagine that you have a circle with a pivot point at its bottom. You then rotate it about the origin. I'm interested in the length of the x axis that is enclosed in the circle at a time t. This turns out to be easy as it can be expressed in polar coordinates as r=2(radius)sin(ωt).

That's not too bad, but what if the circle is rotating about a point that is h units above the origin? Clearly where h≥d, it never touches, but what if h≤d? How can we think about this? Further, what if we want to consider different shapes?
 

Answers and Replies

  • #2
Simon Bridge
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It is the same problem as rotating a line and finding where it intersects the circle.
 
  • #3
It is the same problem as rotating a line and finding where it intersects the circle.
Close. It is the same as rotating a line and finding how the normal of that line intersects the circle. I worked on this quite a bit today and I figured out the formula for a circle. See, I started by finding an equation for the chord length with respect to the circle's height (that is the length of the line that is normal to the chord and extending to the bound of the circle). Then I found an equation relating ωt, the fixed height above the origin and this varying height. Putting all of that together, I found the equation for the length to be 2[itex]\sqrt{r2-(rsin(ωt)-h)2}[/itex]

If h is set to 0, then the equation becomes 2rcos(ωt), which is what I expect. If h is greater than or equal to 2r, then the equation becomes undefined, which is what should be expected. I am pretty happy with this result and will now have to work with some different shapes.
 
  • #4
Simon Bridge
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Isn't the normal to a line just another line?
 

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