How to find Radial and Tangential velocity components from a position vector

Question

I am having trouble solving this problem: $\vec r = r\cos(\theta(t)) \hat i +r\cos(\theta(t)) \hat j$ and $\theta(t) = t^2$ (everything in MKS). I should find $\vec v_{\rm R}$ and $\vec v_{\rm T}$, the radial and tangential components of the velocity at $t=2$ seconds.

I didn’t find much information about the calculation of these velocity vectors on the internet and in the book (University Physics with Modern Physics), there’s information on how to calculate radial and tangential acceleration: $a_{\rm rad.} = v^2 / R = 4 \pi^2R/T^2$ and $a_{\rm tan.} = d|\vec v|/ dt$ (nonuniform circular motion).

I am thinking of getting the derivative of $\vec r$, but I am not sure.

Any help with the question or elaboration on the topic would be appreciated. :)

Answer 1

You've got the first step—namely to find the velocity $\ \vec{v}(t)\ $ by differentiating $\ \vec{r}\ $ with respect to $\ t\ $. By definition, its radial component $\ \vec{v}_R(t)\ $ is its projection onto the unit vector $\ \vec{u}_R(t)=$$\,\cos\theta(t)\hat{i}+$$\,\sin\theta(t)\hat{j}\ $ in the direction of $\ \vec{r}\ $: $$ \vec{v}_R(t)=\big(\vec{v}(t)\cdot \vec{u}_R(t)\big)\,\vec{u}_R(t)\ . $$ Likewise, its tangential component $\ \vec{v}_T(t)\ $ is its projection onto the unit vector $\ \vec{u}_T(t)=$$\,{-}\sin\theta(t)\hat{i}+$$\,\cos\theta(t)\hat{j}\ $ which is the anticlockwise rotation through the angle $\ \frac{\pi}{2}\ $ of the vector $\ \vec{u}_R(t)\ $: $$ \vec{v}_T(t)=\big(\vec{v}(t)\cdot \vec{u}_T(t)\big)\,\vec{u}_T(t)\ . $$

Answer 2

Complex numbers offer a compact way to represent circular motions in 2D, think of the real axis as $x$-coordinate and the imaginary axis as $y$-coordinate (base vectors $\hat{i}$ and $\hat{j}$ in your example).

Keeping Euler's formula, describing a unit circle in the complex plane, \begin{equation} e^{i\theta} = \cos\theta + i \sin\theta \end{equation} in mind, you obtain for the velocity of a position vector $\vec{r}=re^{i\theta}$ as in your example (just using exp instead of sin/cos) \begin{equation} \dot{\vec{r}} = \dot{r} e^{i\theta} + r\dot{\theta} i e^{i\theta}, \end{equation} where the dot denotes derivative with respect to time. The first term is the radial velocity and is zero $v_\text{R}=0$ for $r=\text{const.}$ (constant radius means circle), whereas the second term is the tangential velocity (note how multiplication with $i$ rotates $e^{i\theta}$ by $\frac{\pi}{2}$) and in your case \begin{equation} v_\text{T} = r\dot{\theta} = r 2t. \end{equation} The direction you get from $ie^{i\theta} = i\cos\theta - \sin\theta$.

Note that the angle $\theta$ has been given in units [s]^2, being strict, there should be a conversion factor to RAD (or time defined accordingly).