# Curved trajectory represented by Polar Cordinates

A particle is moving within a curved trajectory defined in the plane $(xoy)$, with the polar coordinates : $r(\theta) = r_0(1-cos\theta)$ with $\theta=\omega t$ with $r_0$ and $\omega$ positive constants.

First question is to give the positional vector $\vec{OM(t)}$ and velocity $\vec V(t)$ and acceleration $\vec a(t)$

I feel like what i did is wrong : what i did is i simply put $\omega t$ instead of $\theta$ in the equation $r(\theta)$ and we have $\vec {OM(t)} = r(t).\vec U_r$ ($\vec U_r$ unit vector).

and then i continue deriving to get $\vec V(t)$ and $\vec a(t)$., but is there a way to transform $r(\theta)$ to cartesian equations $x(t),y(t)$? i know that $x(t) = r(t).cos \theta$, but does it work in this case?

Second question is to give the algebraic value of $a_t$, now i worked with both unit vectors $\vec U_r,\vec U_\theta$, and i know that $a_t$ is the derivative of the magnitude of velocity, so all i have to do is derive $v(t) = \sqrt{V_r^2(t)+V_\theta^2(t)}$ and get it as a function of time? this looks wrong to me.

I'm having troubles with polar coordinates and i really hope to get some references to some good exercises/books to get my self used to working with them.

If the radius in polar coordinates is $r(\theta)$, then the position vector is $\mathbf{r}=(x,y)$, where \begin{eqnarray} x &=& r(\theta) \cos \theta \\ y &=& r(\theta) \sin \theta \end{eqnarray} I am assuming your problem is to consider $\theta=\omega t$ (if this is true, please edit your question to clarify). These sorts of calculations are always the most straightforward in Cartesian coordinates. Then we can take derivatives to yield

\begin{eqnarray} x &=& r_0 \left( 1-\cos\omega t \right) \cos\omega t \\ y &=& r_0 \left( 1-\cos\omega t \right) \sin\omega t \end{eqnarray} \begin{eqnarray} \dot{x} &=& r_0 \omega \left( \sin2\omega t -\sin\omega t \right) \\ \dot{y} &=& r_0 \omega \left( -\cos2\omega t +\cos\omega t \right) \end{eqnarray} \begin{eqnarray} \ddot{z} &=& r_0 \omega^2 \left(2\cos2\omega t-\cos\omega t \right) \\ \ddot{y} &=& r_0 \omega^2 \left(2\sin2\omega t -\sin\omega t \right) \end{eqnarray} Now, if you want to write these quantities as vectors, one can do so in terms of the Cartesian unit vectors $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$. For example, the velocity is $$\dot{\mathbf{r}} = r_0 \omega \left( \sin2\omega t -\sin\omega t \right)\hat{\mathbf{x}} + r_0 \omega \left( -\cos2\omega t +\cos\omega t \right)\hat{\mathbf{y}}$$

It is also possible to do the entire calculation with the polar unit vectors $\hat{\mathbf{r}}$ and $\hat{\mathbf{\theta}}$, where

\begin{eqnarray} \hat{\mathbf{r}} &=& \cos\theta \, \hat{\mathbf{x}} +\sin\theta \,\hat{\mathbf{y}} \\ \hat{\mathbf{\theta}} &=& \cos\theta \, \hat{\mathbf{y}} -\sin\theta \,\hat{\mathbf{x}} \end{eqnarray}

\begin{eqnarray} \hat{\mathbf{x}} &=& \cos\theta \, \hat{\mathbf{r}} -\sin\theta \,\hat{\mathbf{\theta}} \\ \hat{\mathbf{y}} &=& \sin\theta \, \hat{\mathbf{r}} +\cos\theta \,\hat{\mathbf{\theta}} \end{eqnarray}

For this problem, since $\theta = \omega t$, the derivatives of the polar unit vectors are simple:

\begin{eqnarray} \dot{\hat{\mathbf{r}}} &=& \omega \hat{\mathbf{\theta}} \\ \dot{\hat{\mathbf{\theta}}} &=& -\omega \hat{\mathbf{r}} \end{eqnarray}

And then, because of the simplicity of the derivatives of the unit vectors, we can show \begin{eqnarray} \mathbf{r}/r_0 &=& \left(1-\cos\omega t \right)\hat{\mathbf{r}} \\ \dot{\mathbf{r}}/(r_0\omega) &=& \sin\omega t \, \hat{\mathbf{r}}+(1-\cos\omega t ) \hat{\mathbf{\theta}} \\ \ddot{\mathbf{r}}/(r_0\omega^2) &=& (2\cos\omega t-1) \, \hat{\mathbf{r}}+2\sin\omega t \, \hat{\mathbf{\theta}} \end{eqnarray}