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If we consider a cycloid made by a wheel. Then will the cycloid intersect the wheel when the wheel touches the topmost point of the cycloid? Thus will the radius of curvature be same to that of the wheel at that point?

Arnav Mahajan
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    The cycloid "intersects" the wheel everywhere. –  Jun 02 '20 at 08:48
  • @Yves Daoust But at the top of its cycle the circle does not cut the cycloid, but is tangent to it. While it is true that OP is perhaps using the incorrect terminology, it is a fair inference that this is what OP means. – Prime Mover Jun 02 '20 at 09:04
  • @PrimeMover But do you have any proof to say this? – Arnav Mahajan Jun 02 '20 at 11:12
  • a) When the circle is at the top, their radii of curvature are on the same straight line as the point of tangent, and that of the circle is smaller. Hence tangency. b) At any other point, the top of the circle is higher than the cycloid so they cut each other. – Prime Mover Jun 02 '20 at 20:02

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From the parametric equations

$$\begin{align}x&=R(\theta-\sin\theta),\\y&=R(1-\cos\theta)\end{align}$$

the radius of curvature is

$$\frac{R((1-\cos\theta)^2+\sin^2\theta)^{3/2}}{|(1-\cos\theta)\cos\theta-\sin^2\theta|}=2^{3/2}R\sqrt{1-\cos\theta}.$$