particle moving on a cardioid at constant frequency

This is another elementary example¹¹C.F. particle moving on the astroid at constant frequency about particle kinematics. In this case we will use polar coordinates. Let us consider the cardioid ²²the locus of the points of the plane described by a circle (or disc) boundary point which it is rolling over another one with the same radius $R$ .

r=4R\\cos^{2}\\frac{\\omega t}{2},

³³indeed the native polar equation of the cardioid is

r=2R(1+\\cos\\theta),\\quad\\theta=\\omega t.

with $R,\\omega>0$ given constants and $t\\in[0,\\infty)$ means time parameter. The position vector of a particle, respect to an orthonormal reference basis $\\{\\mathbf{\\hat{r}},\\mathbf{\\hat{\\theta}}\\},$ moving on the cardioid is

\\mathbf{r}=4R\\cos^{2}\\frac{\\omega t}{2}\\,\\mathbf{\\hat{r}},

and its velocity ⁴⁴in polar coordinates we have $\\mathbf{\\dot{r}}=\\dot{r}\\,\\mathbf{\\hat{r}}+r\\dot{\\theta}\\,\\mathbf{\\hat{\\theta}},$ because the base vectors $\\mathbf{\\hat{r}},\\mathbf{\\hat{\\theta}}$ are changing on direction and sense according the formulas $\\frac{d\\mathbf{\\hat{r}}}{d\\theta}=\\mathbf{\\hat{\\theta}},\\qquad\\frac{d\\mathbf{%\\hat{\\theta}}}{d\\theta}=-\\mathbf{\\hat{r}}.$ We are using the chain rule with $\\dot{\\theta}=\\omega.$ Overdot denotes time differentiation everywhere.

\\mathbf{v}=\\mathbf{\\dot{r}}=-4R\\omega\\sin\\frac{\\omega t}{2}\\cos\\frac{\\omega t}%{2}\\,\\mathbf{\\hat{r}}+4R\\omega\\cos^{2}\\frac{\\omega t}{2}\\,\\mathbf{\\hat{\\theta}}.

Therefore the speed is

v=4R\\omega\\cos\\frac{\\omega t}{2},

and the tangent vector

\\mathbf{T}=-\\sin\\frac{\\omega t}{2}\\,\\mathbf{\\hat{r}}+\\cos\\frac{\\omega t}{2}\\,%\\mathbf{\\hat{\\theta}}.

Next we use the formula

\\frac{v}{\\rho}:=\\big{\\|}\\mathbf{\\dot{T}}\\big{\\|}=\\bigg{\\|}-\\frac{\\omega}{2}%\\cos\\frac{\\omega t}{2}\\,\\mathbf{\\hat{r}}-\\sin\\frac{\\omega t}{2}\\,\\mathbf{\\dot{%\\hat{r}}}-\\frac{\\omega}{2}\\sin\\frac{\\omega t}{2}\\,\\mathbf{\\hat{\\theta}}+\\cos%\\frac{\\omega t}{2}\\,\\mathbf{\\dot{\\hat{\\theta}}}\\bigg{\\|},

and by using the time derivative of base vectors

\\frac{v}{\\rho}=\\bigg{\\|}-\\frac{3\\omega}{2}\\cos\\frac{\\omega t}{2}\\,\\mathbf{\\hat%{r}}-\\frac{3\\omega}{2}\\sin\\frac{\\omega t}{2}\\,\\mathbf{\\hat{\\theta}}\\bigg{\\|},

getting the equation

v=\\frac{3}{2}\\omega\\rho.