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hacivat
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- TL;DR Summary
- An alternative method to solve Foucault's pendulum. But how can I justify this approach?
In many standard texts the behavior of Foucault’s pendulum is solved by adding the Coriolis force term to equation of motion and deriving two coupled differential equations. Here’s an alternative approach:
4 assumptions are made:
$$ \theta=\theta_0\sin{\omega t} $$
where ##\theta## is the pendulum’s angle with the vertical, ##\theta_0## is the amplitude of the oscillation and
$$ \omega=\sqrt{\frac{g}{d}} $$
##g## being the gravitational acceleration and ##d## being the length of the rod.
This is all standard stuff.
Now, after this point I write the angular momentum with respect to the hanging point and focus on the behavior of the angular momentum.
$$ L=mvd=md^2\dot{\theta}=md^2\omega\theta_0\cos{\omega t} $$
During oscillations angular momentum vector oscillates back and forth in a periodic manner. The vector changes its magnitude and sign as seen from the above equation but stays on a line which is perpendicular to the oscillation plane.
However, in each oscillation this line will be slightly changing its direction because of the torque created by the Coriolis force. Here I take the time average of the magnitude of the angular momentum vector during one oscillation which can be written as
$$ \langle L \rangle = md^2\omega \theta_0\frac{2}{\pi}$$
the last term coming from the time average of the absolute value of the trigonometric function over a period. I did this because I want to “create” a vector representing the plane of oscillation which I can act the slight change caused by the Coriolis torque and observe its rotation.
It is fairly easy to calculate the magnitude of ##\Delta L## by integrating the Coriolis torque over one period. And since we made the assumption 4, the circumference of the circle where the tip of the average angular momentum vector is moving could be approximated by the sum of these small ##\Delta L##’s.
To calculate ##\Delta L## we need to write the torque caused by the Coriolis term and integrate it over one oscillation period.
$$ \tau=2\Omega vmd =\frac{dL}{dt}$$
## \Omega ## being the rotation rate of the coordinate frame. If we substitute ## v ## and integrate over one period we get the following expression for ##\Delta L##:
$$ \Delta L=2\Omega md^2\theta_0\omega \int_{\frac{T}{2}}^{-\frac{T}{2}} \cos{\omega t} \, dt = 4 \Omega md^2\theta_0 $$
Now if we divide the circumference of the above circle by ## \Delta L ## we can get the number of oscillations to complete one rotation of the plane which is:
$$ \frac{md^2\omega\theta_0\frac{2}{\pi}2\pi}{4\Omega m d^2 \theta_0}=\frac{\omega}{\Omega} $$
This expression could also be written in terms of periods and is basically the same result from what we get from the orthodox solutions of the problem.
So my question: is this method justifiable in terms of mathematical rigor since use of time “average” angular momentum vector was purely coming from my physical insight?
4 assumptions are made:
- Mathematical pendulum (point mass attached to massless rigid rod)
- No dissipative terms
- Small angle oscillations
- Rotation period of the oscillation plane is very large with respect to the period of the oscillations.
$$ \theta=\theta_0\sin{\omega t} $$
where ##\theta## is the pendulum’s angle with the vertical, ##\theta_0## is the amplitude of the oscillation and
$$ \omega=\sqrt{\frac{g}{d}} $$
##g## being the gravitational acceleration and ##d## being the length of the rod.
This is all standard stuff.
Now, after this point I write the angular momentum with respect to the hanging point and focus on the behavior of the angular momentum.
$$ L=mvd=md^2\dot{\theta}=md^2\omega\theta_0\cos{\omega t} $$
During oscillations angular momentum vector oscillates back and forth in a periodic manner. The vector changes its magnitude and sign as seen from the above equation but stays on a line which is perpendicular to the oscillation plane.
However, in each oscillation this line will be slightly changing its direction because of the torque created by the Coriolis force. Here I take the time average of the magnitude of the angular momentum vector during one oscillation which can be written as
$$ \langle L \rangle = md^2\omega \theta_0\frac{2}{\pi}$$
the last term coming from the time average of the absolute value of the trigonometric function over a period. I did this because I want to “create” a vector representing the plane of oscillation which I can act the slight change caused by the Coriolis torque and observe its rotation.
It is fairly easy to calculate the magnitude of ##\Delta L## by integrating the Coriolis torque over one period. And since we made the assumption 4, the circumference of the circle where the tip of the average angular momentum vector is moving could be approximated by the sum of these small ##\Delta L##’s.
To calculate ##\Delta L## we need to write the torque caused by the Coriolis term and integrate it over one oscillation period.
$$ \tau=2\Omega vmd =\frac{dL}{dt}$$
## \Omega ## being the rotation rate of the coordinate frame. If we substitute ## v ## and integrate over one period we get the following expression for ##\Delta L##:
$$ \Delta L=2\Omega md^2\theta_0\omega \int_{\frac{T}{2}}^{-\frac{T}{2}} \cos{\omega t} \, dt = 4 \Omega md^2\theta_0 $$
Now if we divide the circumference of the above circle by ## \Delta L ## we can get the number of oscillations to complete one rotation of the plane which is:
$$ \frac{md^2\omega\theta_0\frac{2}{\pi}2\pi}{4\Omega m d^2 \theta_0}=\frac{\omega}{\Omega} $$
This expression could also be written in terms of periods and is basically the same result from what we get from the orthodox solutions of the problem.
So my question: is this method justifiable in terms of mathematical rigor since use of time “average” angular momentum vector was purely coming from my physical insight?