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Physical pendulum oscillates with a period that can be described by the equation: $T=2\pi I^{\alpha}m^{\beta}g^{\gamma}d^{\delta}$, where the exponents are all non-zero. The units are: $T\,[s]$, $I\,\left[kg\cdot m^2\right]$, $m\,[kg]$, $g\,\left[\frac{m}{s^2}\right]$ and $d\,[m]$. To extract the unknown exponents I use dimensional analysis. Thus:

$$\left.\begin{matrix}T\\M\\L\end{matrix}\right.\left.\begin{matrix}:\\:\\:\end{matrix}\right.\left.\begin{matrix}1\\0\\0\end{matrix}\right.\left.\begin{matrix}=\\=\\=\end{matrix}\right.\left.\begin{matrix}-2\gamma\\\alpha+\beta\\2\alpha+\gamma+\delta\end{matrix}\right.$$

Obviously this system of equations is underdetermined, so the result is not unequivocal. Can anyone tell me what I am missing, if anything, because from physics I know the solution (and there is only ONE, namely $T=2\pi\sqrt{I/mgd}$) but I can't pin it down with maths. Or maybe the general form is ill posed?

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    $\begingroup$ You have four variables and three dimensions, your system is overdetermined and will yield a dimensionless factor as a result of the Buckingham pi theorem. $\endgroup$
    – Triatticus
    Commented Oct 15, 2023 at 3:58
  • $\begingroup$ This system is underdetermined (more unknowns than equations needed to find them). See wiki. $\endgroup$
    – Radek D
    Commented Oct 18, 2023 at 10:22
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    $\begingroup$ Yeah by the time I had realized it was too late to edit and the answer expanding on it had already gone up. $\endgroup$
    – Triatticus
    Commented Oct 18, 2023 at 15:22

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You can rewrite the equations to give $$\gamma = -\frac{1}{2}$$ $$\beta = - \alpha$$ $$\delta = \frac{1}{2} - 2\alpha$$ Substituting these into the equation, $$T = 2\pi \Big(\frac{I}{md^2}\Big)^{\alpha} \sqrt{\frac{d}{g}}$$ Now the problem is, $\frac{I}{md^2}$ is a dimentionless quantity, whose exponent can not be determined by dimension analysis. So this is how far you can get with it.

You could've reached the same conclusion quickly with Buckingham Pi theorem which states if you have N quantities with D inpendent dimensions, you'd have $N-D$ dimensionless quantities.

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    $\begingroup$ Thank you. I thought, however naively, that dimensional analysis is almighty tool for getting the right equation :D Anyways, good to know about Buckingham Pi theorem :) $\endgroup$
    – Radek D
    Commented Oct 18, 2023 at 8:15

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