My question concerns these components of a question addressing the motion of a particle expressed in polar:

enter image description here

With that, here's the particular part of the question I'm trying to answer:

enter image description here

From my knowledge (which will surely have flaws in it due to the fact that I wrote this question), the angular momentum can be expressed using linear momentum from the following thought process:

$$\vec L = \vec r \times\vec p$$

$$\vec L = \vec r \times m \vec v$$

Which, I guess, answers the first bit?

Now, onto the next one.

I'm not sure how this can be true along with the vector expression simultaneously, but I know this is true as well:

$$L = rmv_{\theta}$$

And $$v_{\theta} = \frac{d\theta}{dt}$$

And recalling what $\theta$ is defined as in terms of time in the first part of the question -

$$\dot \theta = \beta t^2$$ $$r = \frac{\alpha}{t}$$

Thus, $$L = m\frac{\alpha}{t} \beta t^2$$

This is clearly not the result I'm trying to confirm. I think it has to about something with the $\hat z$ unit vector, which I'm not sure about what it is or what it's doing there, and that I'm using $v_{\theta}$, not $\vec v$, which is the vector sum $v=v_r + v_{\theta}$. Can someone explain to me what I ought to consider and do next here?

  • $\begingroup$ Try using $L=I\omega$ for angular momentum instead. ($I=mr^2$ for a point mass) As $L$ is a vector it requires a basis as well... $\endgroup$ Sep 27, 2017 at 20:24
  • $\begingroup$ Your formula for $v_{\theta}$ is incorrect. You should get $L=mr^2\dot\theta$. $\endgroup$ Sep 27, 2017 at 20:26

1 Answer 1


Consider the equations of motion in polar coordinates

$\vec{r} = r\hat{r}$

$\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$

where the unit vectors $\hat{r}$ and $\hat{\theta}$ are given by

$\hat{r} = \hat{i}\cos\theta + \hat{j}\sin\theta$

$\hat{\theta} = -\hat{i}\sin\theta + \hat{j}\cos\theta$

You were correct in noting that $L = \vec{r} \times m\vec{v}$ but the forms that you put $\vec{r}$ and $\vec{v}$ in were incorrect. Instead, we have

$L = \vec{r} \times m\vec{v}$

$ = r\hat{r} \times m(\dot{r}\hat{r} + r\dot{\theta}\hat{\theta})$

$ = r\hat{r} \times mr\dot{\theta}\hat{\theta}$

Working out the above cross product by substituting the definitions of $r$ and $\theta$ that were provided will give you your result. The unit vector $\hat{z}$ is simply the unit vector in the direction perpendicular to the plane of motion.

  • $\begingroup$ But isn't $\theta = -sin\theta \hat i + cos\theta \hat j$? $\endgroup$
    – sangstar
    Sep 27, 2017 at 20:35
  • $\begingroup$ Sorry that was a typo. Yes, $\hat{\theta} = -\hat{i}\sin{\theta} + \hat{j}\cos{\theta}$ $\endgroup$
    – J_Psi
    Sep 27, 2017 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.