# 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 ParticleMovingOnACardioidAtConstantFrequency 2013-03-22 17:14:20 2013-03-22 17:14:20 perucho (2192) perucho (2192) 6 perucho (2192) Topic msc 70B05