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

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

Title	particle moving on a cardioid at constant frequency
Canonical name	ParticleMovingOnACardioidAtConstantFrequency
Date of creation	2013-03-22 17:14:20
Last modified on	2013-03-22 17:14:20
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	6
Author	perucho (2192)
Entry type	Topic
Classification	msc 70B05