# How to determine radius of curvature of cycloid using centripetal acceleration?

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?

Hints

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)

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

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 $\:\mathrm g, \theta$ $$\Vert\mathbf{a}_{\textrm{n}}\Vert=\mathrm h(\mathrm g, \theta) \tag{02}$$ 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 $\:R, \mathrm g, \theta$

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

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

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

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.

$1\:\:$ 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?