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Whenever it comes to radius of curvature of complex curves like cycloid, we all take the help of calculus. But I am still in high school and not that competent with calculus, so please do not answer using calculus.

Another interesting resource which I found was this: www.jstor.org/stable/2967957. But this makes use of geometry.

I found a problem while doing my physics(rotation) homework which asked to calculate radius of curvature of the path traced by a point on the circumference of a disc rotating without slipping on a surface. I think that this must be somehow related to the chapter.

So can someone tell how to do it using some rotation concepts?

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Hints


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A. Looking at Figure 01

  1. A cycloid is a curve traced by a point $\:\mathrm P\:$ on the rim of a circular wheel as the wheel rolls along a straight line without slipping.(Wikipedia)

  2. A brachistochrone curve$\rm 1$ or curve of fastest descent, is the one lying on plane between a point $\:\mathrm A\:$ and a lower point $\:\mathrm F$, where $\:\mathrm F\:$ is not directly below A, on which a bead $\:\mathrm P\:$ slides frictionlessly under the influence of a uniform gravitational $\:\mathbf{g}\:$ field in the shortest time.(Wikipedia)

To determine the radius of curvature $\:\rho\:$ of the cycloid using centripetal acceleration we note at first that for any curvilinear motion (not necessarily circular)

\begin{equation} \Vert\mathbf{a}_{\textrm{n}}\Vert=\dfrac{\Vert\boldsymbol{\upsilon}\Vert^{2}}{\rho}=\dfrac{\upsilon^{2}}{\rho} \tag{01} \end{equation} where $\:\mathbf{a}_{\textrm{n}}\:$ the centripetal acceleration, normal to the curve and $\:\boldsymbol{\upsilon}\:$ the velocity, tangent to the curve.


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B. Looking at Figure 02

  1. From the equilibrium of the bead on the direction normal to the cycloid you could determine the magnitude $\:\Vert\mathbf{a}_{\textrm{n}}\Vert\:$ of the centripetal acceleration as function of $\:\mathrm g, \theta$ \begin{equation} \Vert\mathbf{a}_{\textrm{n}}\Vert=\mathrm h(\mathrm g, \theta) \tag{02} \end{equation} In this Figure it seems that the straight line normal to the cycloid at point $\:\mathrm P\:$ passes through the point $\:\mathrm M\equiv$ contact point of the wheel with the straight line on which it is rolling. Why???

  2. From energy conservation of the bead between the starting point $\:\mathrm A\:$ and present point $\:\mathrm P\:$ you could determine the square of the speed $\:\Vert\boldsymbol{\upsilon}\Vert^{2}\:$ as function of $\:R, \mathrm g, \theta$

\begin{equation} \Vert\boldsymbol{\upsilon}\Vert^{2}=\mathrm f(R,\mathrm g,\theta) \tag{03} \end{equation}

Then from (01) you'll determine the radius of curvature $\:\rho\:$ as function of $\:R,\theta$ \begin{equation} \rho=\dfrac{\:\:\Vert\boldsymbol{\upsilon}\Vert^{2}}{\Vert\mathbf{a}_{\textrm{n}}\Vert}=\dfrac{\upsilon^{2}}{\Vert\mathbf{a}_{\textrm{n}}\Vert}=\rho(R,\theta) \tag{04} \end{equation}


After all these : what is the geometric locus of the centers of curvature ?

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The geometric locus of the centers of curvature $\:\mathrm P'\:$ of points of a cycloid $\:C\:$ is a cycloid $\:C'\:$ identical to $\:C\:$ but shifted by the vector $\:(-\pi R, -2R)\:$, see Figure 03.

The new cycloid $\:C'\:$ is the envelope of the family of normal to the cycloid $\:C\:$ straight lines, see Figure 04.

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$1\:\:$ For more details on the brachistochrone and tautochrone cycloids see my answer therein : What is the position as a function of time for a mass falling down a cycloid curve?

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