I've been given an object with $r(t)=|v|t\hat{x}+b\hat{y}$ and been asked to find the specific angular momentum and see if angular momentum is conserved. I found $L=r\times p=-m|v|b\hat{z}$. My inclination is to say this is not conserved, both because $|v|$ is increasing with time (so $L$ is not constant). However, this would mean torque is zero, which seems to be wrong. I think it is because $p=m\dot{r}=m|v|\dot{|v|}\hat{x}$ and I forgot the chain rule? Also, it seems weird to ask about the specific angular momentum when I have only one object. I suppose if $L=m|v|b\hat{z}$, then specific angular momentum would just be $h=|v|b\hat{z}$.

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    $\begingroup$ If L is not constant it means there is a nonzero torque $\endgroup$ – Triatticus Sep 19 '18 at 17:35
  • $\begingroup$ Right I misspoke. I meant nonzero. My thought is that if I forgot the chain rule I'd have $a=\dot{v}=\frac{d}{dt}\dot{r}=|v|||\dot{v}|$ which is nonzero. If that's the case, I calculated $v$ wrong when I was evaluating $L$. Is this the case? $\endgroup$ – Radagast Sep 19 '18 at 18:26

If we suppose $v$ not to be constant, saying it is some real and differentiable function, then the momentum p is equal to

p $= m\frac{dr(t)}{dt} = m(|\dot v(t)|t+|v(t)|)i$

where $i$ is the x-axis unitary norm vector. Let's now compute angular momentum L $ = $ r $\times$ p, expressing it as the determinant of the following matrix

$\begin{vmatrix}i & j & z\\|v|t & b & 0 \\ m(|v|+|\dot v|t) & 0&0\end{vmatrix} = -m(|v|+|\dot v|t)bz$

As you can see, L contains a non conserved quantity, $|v(t)|$, thus the angular momentum is not conserved. This means that the torque $M$ is non zero, taking the derivative of L we get

$\frac{d}{dt} $ L= $ -2m|\dot v|+|\ddot v|tbz$

Then, multiplying by the moment of inertia $I$ we find the expression for the torque.

EDIT: The derivative of $r(t)$ is $(|\dot v(t)|t+|v(t)|)i$, thus the torque is $ -2m|\dot v|+|\ddot v|tbz$

  • $\begingroup$ Why exactly do you have a sum in the expression for the momentum instead of a product? It seems you take the derivative of vt, getting v, then you chain rule and multiply it by v dot, getting v vdot t (times m). $\endgroup$ – Radagast Sep 19 '18 at 20:05
  • $\begingroup$ I assumed $r(t) = |v(t)|ti+bj$, differentiating it twice gives you dL/dt $\endgroup$ – Matteo Campagnoli Sep 19 '18 at 21:15

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