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

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

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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 '17 at 20:24
  • $\begingroup$ Your formula for $v_{\theta}$ is incorrect. You should get $L=mr^2\dot\theta$. $\endgroup$ Sep 27 '17 at 20:26

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 '17 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 '17 at 20:37

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