Here, we solve the precession of the plane of oscillation of a pendulum. We do so under the following approximations:
(1) The precession rate of the plane of oscillation is slow compared with the oscillation of the pendulum.
(2) The pendulum oscillates with a small amplitude, such that its motion can be approximated as harmonic (a spring).
The derivation involves mapping the pendulum problem into the mass-on-spring problem in two dimensions, and then solving it in polar coordinates, to obtain the equation describing the
precession of the oscillation plane.
Other derivations of the precession rate of Foucault's Pendulum are abundant in the literature, these include for example: (1) A
geometric solution. This solution is not too rigorous but it certainly illuminates the physical origin of the precession. (2) Derivation of the precession equation using
spherical coordinates. This solution involves ugly mathematics arising from the time derivatives of the spherical coordinate system's unit vectors. (3) A
full solution in x-y coordinates, it is formally valid also for fast rotations, but requires the solution of two coupled second order differential equations. (Actually, it is not that valid in fast rotating systems because in systems where Coriolis is as important as the main dynamics of the pendulum, the centrifugal force will tend to be important as well, but in all the derivations mentioned here, the centrifugal force is assumed to be constant, and is therefore eliminated by redefining "g".)
Pendula as 2D springs systems
The equation of motion of a mathematical pendulum without Coriolis can be written as:
dLdt=mℓ2ϑ¨=Ngrav
where L is the angular momentum relative to the oscillation axis and Ngrav=−mgℓsinθ is the force moment. We can work here for the moment with scalars because without Coriolis, the motion of the pendulum is in a plane.
We now perform our first approximation. We assume small angle oscillations. That is, we assume that sinϑ≈ϑ. We find:
mℓϑ¨=−mgϑ.
Instead of ϑ, we can change our variable from ϑ to r=ℓϑ which is the distance from the axis (under the small angle limit. We find:
mr¨=−mgℓr.
In other words, the equations describing the pendulum are equivalent (at the small angle limit) to the motion of a spring, with a spring constant given by keff=mg/ℓ and indeed the oscillation frequency of the pendulum is ω=keff/m−−−−−√=g/l−−−√ as it should.
We are therefore allowed to describe the pendulum as a mass on spring. In general, the plane of the oscillation can have precession, we therefore write the motion of the pendulum as a 2D mass on spring problem:
mr¨=−keffr.
(Without any additional forces, the direction of r remains fixed, but with any non-radial force, such as of Coriolis, it will change).
Adding the Coriolis Effect
The Rotating coordinate system {x,y,z} is non-inertial since Earth is rotating. As a result, a Coriolis force is added when working in this frame of reference.
We now add the Coriolis force. Since the mass cannot exit the plane, the equation is valid on to the horizontal directions:
mr¨=−mgℓr−2mΩ×r˙.
For convenience, we will work with polar coordinates - r and θ. (Note that this angle θ is different from the oscillation angle ϑ). The vector rand its time derivatives in is polar coordinates, are given by:
r=rr^,r˙=drdtrr^+rdr^dt=r˙r^+rθ˙θ^
r¨=r¨r^+r˙dr^dt+r˙θ˙θ^+rθ¨θ^+rθ˙dθ^dt
=(r¨−rθ˙2)r^+(2r˙θ˙+rθ¨)θ^.
Next, we need to write the Earth's rotation vector Ω^ in the polar coordinates we use:
Ω^=Ωsinλz^+Ωcosλcosθr^−Ωcosλsinθθ^.
Note that the last term is with a minus sign, because Ω^ has a component opposite to the direction to which θ grows.
The pendulum and the equivalent spring system in polar coordinates. Once the problem becomes 2D, the oscillations are described by r and θ.
We are now in a position where we can take the θ^component of the equation of motion. It will gives us the dynamics of the oscillating plane, i.e., how its direction evolves with time. The component is:
mrθ¨+2r˙θ˙=−2m(Ω^×r˙)θ
where the last parenthesis implies that we take the θ^ component of the vector product. Note that {θ^,r^,z^} is a right handed system (as can be seen in the figure above). Therefore, the θ^ mcomponent of the product is:
(Ω^×r˙)θ=Ωrr˙z−Ωzr˙r=−Ωr˙sinλ,
after plugging in the expressions for r˙ and Ω^. The equation of motion we find is:
mrθ¨+2r˙θ˙=2mr˙Ωsinλ.
In general, this equation cannot be solved without knowing the time dependence of r. However, luckily for us, this equation can be simplified in the limit where the precession is much slower than the oscillations of the pendulum.
The Slow Precession Approximation
If the angular frequency of the pendulum's oscillations is ω then we typically have that r˙∼ωr. On the other hand, the evolution of the direction of oscillations is given by θ, we expect these variations to be of order a day, the time it takes Earth to complete a revolution, i.e., θ˙∼Ωθ and θ¨∼Ω2θ (note that at the end, we have to verify that our solution is consistent with this assumption!).
Since ω≫Ω then θ¨r is smaller than r˙θ˙ by a factor of ω/Ω (e.g., if the pendulum has a few sec period, and 2π/Ω is of order a day, we are talking about a very small correction, one of 1/10000 that we are neglecting.)
Under this limit, the rθ¨ term is neglected, and the equation we find is (after canceling out 2mr˙):
θ˙=Ωsinλ.
This solution satisfies that θ˙∼Ωθ, thus, our assumption is valid.
The time it takes the pendulum to rotate by π radians (i.e., 180°), after which the pendulum appears to have returned to its original state, is:
Tprec=πθ˙=2πΩ12sinλ=day2sinλ.
In Jerusalem, for example λ≈32∘, it is just under a day. In Paris, it will be shorter.