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In polar co-ordinates we have
$\vec{r} = r\hat{r}$ and $\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$ and $\vec{a} = (\ddot{r}-r\dot{\theta}^2)\hat{r} + 2\dot{r}\dot{\theta}\hat{\theta}$
Now suppose we had $r(t) = be^{wt}$ and $\theta(t) = wt$. Then $\vec{v} = wbe^{wt}(\hat{r} + \hat{\theta})$ and $\vec{a} = w^2be^{wt}\hat{r} + 2w^2be^{wt}\hat{\theta}$
How could I show that the angle between $\vec{v}$ and $\vec{a}$ is constant?
One way would be to use the definition of the dot product for vectors
$$ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta, $$ hence $$ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}.$$
I've done the calculation quickly and you will definitely end up with a constant angle :)
Consider how you would compute the angle between two vectors in Cartesian coordinates, we would take $cos^{-1}$ of $\frac{\vec{v} \cdot \vec{a}}{|\vec{v}||\vec{a}|}$. Since $\hat{r}$ and $\hat{\theta}$ are mutually perpendicular, just like $\hat{i}$ and $\hat{j}$, we can think of the $\hat{r}$ and $\hat{\theta}$ components as a rotation of the $\hat{i}$ and $\hat{j}$ coordinates. As angles between vectors are preserved under rotation, this analysis should give the correct answer for the angle between the two vectors.
