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If a point-mass body is moving in a plane $z=0$, its angular momentum can be taken to be a scalar, and from the vector product formulas (assuming $m=1$ as I'm only interested in geometry), its value is $\dot{x}{y}-x\dot{y}$.

Can this formula be intuitively explained or showcased on its own? Is there a simple argument that does not go through $\vec{r}\times\vec{p}$ 's expansion, but after which one goes "oh, of course the angular momentum must be $\dot{x}{y}-x\dot{y}$"?

For example, it occurred to me that this equals $y^2(\frac{x}{y})'$, and that $\frac{x}{y}$ can be seen as the slope of the line connecting the body to the origin, so it sort of makes sense that the angular momentum is the derivative of the slope, measuring the speed with which the body turns around the origin, although what is the $y^2$ factor doing there then... is that a helpful direction at all?

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Here's one:

Consider an object revolving about the $z$ axis with a constant angular velocity $\omega=\dot{\theta}$. We know that $\theta=\tan^{-1}(y/x)$, so:

\begin{align} \omega=\dot{\theta}&=\frac{d}{dt}\left[\tan^{-1}\left(\frac{y}{x}\right)\right]\\ &=\frac{1}{1+\left(\frac{y}{x}\right)^2}\frac{d}{dt}\left(\frac{y}{x}\right)\\ &=\frac{x^2}{x^2+y^2}\left(\frac{\dot{y}}{x}-\frac{y}{x^2}\dot{x}\right)\\ &=\frac{1}{x^2+y^2}(x\dot{y}-y\dot{x}) \end{align}

The moment of inertia will eliminate the factor of $\frac{1}{x^2+y^2}$ here. So you see that the $x\dot{y}-y\dot{x}$ factor comes specifically from the fact that you're taking a time derivative of an arctangent of a ratio.

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  • $\begingroup$ Exactly what I lacked, thank you! $\endgroup$ – AnatolyVorobey Sep 30 '19 at 15:40

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