User:Cleonis/Sandbox/Wheatstone-Foucault device

In the case of a polar Foucault pendulum, or a Foucault pendulum located somewhere on the equator, understanding the pattern of motion is straighforward. This article adresses the pattern of motion that is observed at all latitudes other than the poles and the equator.



A Foucault pendulum located at the latitude of Paris takes about 32 hours to complete a precession cycle. This means that after 24 hours, when the Earth has returned to its orientation of 24 hours before, the pendulum has precessed three quarters of a cycle. If it has started in north-south direction, it is swinging east-west when the Earth has returned to the orientation of 24 hours before. The direction of swing of the pendulum with respect to the inertial frame has changed, which implies that the pendulum bob must have been subject to a net torque. This article shows how this net torque arises.

The motion pattern of the Foucault pendulum involves two oscillations, at different scales of magnitude. First, there is the swing of the pendulum,which will be referred to as 'the vibration'. Ideally, the swing of the pendulum is as close to planar swing as possible. Secondly, there is the overall rotation of the Earth around its axis. Any rotation can be regarded as a linear combination of two harmonic oscillations. There is a coupling between the rotation and the vibration, which causes transfer of momentum from one direction of the plane of swing to the perpendicular direction. Due to the large difference in period of oscillation, the coupling is not strong. It appears to be negligable, but the coupling is significant nonetheless because its effect is cumulative.

See also the article rotational-vibrational coupling for a discussion of the most basic case of this kind of coupling in classical mechanics.

The Wheatstone-Foucault device


In 1851 a paper by Charles Wheatstone was read to the Royal Society, in which Wheatstone described the device that is shown in Image 1. A transcription of the paper by Wheatstone is available at wikisource: "Note relating to M. Foucault's new mechanical proof of the Rotation of the Earth"

The helical spring acts as a heavy string; when plucked, it vibrates. The circular platform can rotate around a vertical axis. This axis will be referred to as the central axis.



The Wheatstone pendulum


To simplify the analysis, and to underline the similarity with the Foucault pendulum, a slightly different version of the Wheatstone-Foucault device will be discussed. The device depicted in image 2 will be referred to as "the Wheatstone pendulum". The small sphere corresponds to the bob of the Foucault pendulum, the inside spring corresponds to the gravitational force that is being exerted on the bob, and the force that is exerted by the outside spring corresponds to the force that the pendulum wire exerts on the bob.



Physics principles
The assumptions for the purpose of simplifying the analysis of the Wheatstone pendulum
 * The Earth's gravity is not considered, as if the device is floating in an orbiting space station.
 * All of the mass of the bob is taken as concentrated in the midpoint of the bob
 * The springs are taken als massless.
 * The range of motion of the bob with respect to the rest state is taken to be within the limits of the Small angle approximation.



Images 4 and 5 show a detail of the Wheatstone pendulum, in different states of motion.

Image 4: The top side of the image shows the shape of the springs when the Wheatstone pendulum is not in motion. Both springs are equally stretched, and they are aligned. The bottom side of the image illustrates the outward displacement of the pendulum bob when the platform is rotating. The inner spring is more extended than in the rest state, the outside spring is less extended than in the rest state, and thus the springs provide the required centripetal force to make the pendulum bob move along a circular trajectory.

Image 5: The top side of the image shows the shape of the springs when the Wheatstone pendulum is vibrating, but not rotating around its central axis. Both springs are equally stretched, and at the equilibrium point they are aligned. The bottom side of the image shows that when the pendulum bob is vibrating while the platform is rotating the midpoint of the vibration is the outward displaced point of the pendulum bob.

In these images the amplitude of the vibration is slightly smaller than the outward displacement of the bob, whereas usually the amplitude of the vibration exceeds the magnitude of the outward displacement. This does not detract from the validity of this representation for the general case. One of the properties of the Foucault pendulum is that its precession does not depend on the amplitude of the swing; when the amplitude of a Foucault pendulum decays, the rate of precession remains the same. Therefore a representation with the outward displacement roughly equal to the amplitude of the swing is just as valid; the physics principles that are at play are the same in both cases.

Let the platform be rotating counterclockwise. To underline the comparison with the Foucault pendulum, let the bob's motion in a direction tangent to the rotation be called 'east-west motion'.



Motion towards and away from the central axis
At every point in time, the bob has a particular angular momentum with respect to the central axis of rotation. When the inside spring pulls the pendulum bob closer to the central axis of rotation, the centripetal force is doing mechanical work. As a consequence of the centripetal force doing work, the angular velocity of the bob changes. (Compare a spinning ice-skater who pulls her arms closer to her body to make herself spin faster.) The amount of angular acceleration can be calculated with the help of the principle of conservation of angular momentum. When the pendulum bob is moving away from the central axis the centripetal force is doing negative work, and the bob's angular velocity decreases.



Motion tangent to the rotation


During a swing of the pendulum bob from west-to-east, the bob is circumnavigating the central axis faster than the equilibrium velocity, hence during a west-to-east swing the bob will swing wide. (Compare a car going round on a banked circuit, with the incline of the embankment designed for a particular speed. A car that goes faster than that will tend to climb up the embankment.) During a swing of the pendulum bob from east-to-west, the bob is circumnavigating the central axis slower than the equilibrium velocity, hence during an east-to-west swing the centripetal force will pull the bob closer to the central axis.



The dependency on the sine of the latitude


In the case of the Wheatstone pendulum, just as in the case of the Foucault pendulum, the amount of precession with respect to the rotating system depends on the angle between the central axis and the latitude of the pendulum.

Image 9 illustrates the physical reason for that dependency. The amount of work that the centripetal force does is proportional to the amount of motion towards or away from the central axis.

In the case of a pendulum located at 30 degrees latitude: when the swing from one extremal point to the opposite extremal point covers a distance of L, then the motion towards (or away from) the central axis is over a distance of 1/2 L. This determines to what extend the centripetal force will affect the plane of swing. The closer to the equator, the weaker the coupling between the rotation and the vibration.



Relations applying to motion with respect to the inertial frame
First, some relations will be derived that are designed to be a faithful representation ofthe physics taking place.

In the section 'The equation of motion for the Earthbound frame' it will be discussed how the equation of motion is set up for rotating coordinates. The advantage of using a rotating coordinate system is that in the equation of motion for rotating coordinates some of the terms drop away against each other, resulting in much simplified calculations.

Overview of symbols

Motion towards and away from the central axis
When there is motion under the influence of a central force we have that velocity in radial direction results in change of angular velocity.

One way of finding the relation between radial velocity and angular acceleration is as follows. Calculate how much force must be exerted in order to prevent angular acceleration when there is radial velocity. If that force is not provided, the amount of angular acceleration will be the same as that force requirement (but opposite in direction).

The moment of inertia: $$ I = m r^2 \,\!$$ The angular momentum: $$ L = I \omega = m r^2 \omega \,\!$$ Force in tangential direction $$F_t$$ torque: $$ \tau = F_t r \,\!$$

Let $$ v_r $$ be velocity in radial direction. There is the following relation between change in angular momentum and torque that is being exerted.


 * $$ F_t r = - \frac{dL}{dt} = \frac {d(m \omega r^2)}{dt} $$

When the change in angular velocity is small compared to the change in radial distance, then &omega; can be treated as a constant.


 * $$ \frac {d(m \omega r^2)}{dt} = 2m \omega r \frac{dr}{dt} $$

This gives the following relation:


 * $$ F_t r = - 2 m \omega v_r r \,\!$$

With a factor r dropping away on either side of the equation, this yields the following relation between radial velocity and required tangential force:


 * $$ F_t = - 2m \Omega v_r \,\!$$

If that required tangential force is provided, then angular velocity remains the same, even though there is change of radial distance. When that required tangential force is not provided, then the radial velocity results in acceleration in tangential direction, opposite to the direction of the required force.


 * $$ a_t = 2 \Omega v_r \,\!$$

Motion tangent to the rotation
When there is motion under the influence of a central force we have that velocity relative to the rotating system in tangential direction results in change of radial velocity.

When the bob is circumnavigating the central axis slower than the rotating system it will be subject to a surplus of centripetal force, and there will be acceleration in inward radial direction.

The magnitude of the radial acceleration can be calculated by subtraction: subtract the centripetal force that would be required for the slower tangentialv velocity from the actually exerted centripetal force. The following relation applies when the centripetal force is a harmonic force.


 * $$ a_r = \frac{v_{t,e}^2}{r} - \frac{v_t^2}{r} $$

Of course, the above relation also gives the outward radial acceleration when the bob is circumnavigating faster than the rotating system.

The above relation can also be expressed as an equation that shows how the radial acceleration depends on tangential velocity relative to the rotating system. vt = vt,e + vt,r



\begin{align} a_r & = \frac{v_{t,e}^2}{r} - \frac{(v_{t,e} + v_{t,r})^2}{r} \\ & = \frac{v_{t,e}^2 - v_{t,e}^2 - 2v_{t,e}v_{t,r} - v_{t,r}^2}{r} \\ & = \frac{- 2v_{t,e}v_{t,r} - v_{t,r}^2}{r} \\ \end{align} $$

At all distances to the central axis, vt,e/r=&Omega;, so that can be substituted. When the velocity relative to the rotating system is small the above expression simplifies to:


 * $$ a_r = - 2 \Omega v_{t,r} \,\!$$

Overview
In the above discussions, the following two relations have been derived in a rough manner



\begin{cases} a_t = 2 \Omega v_r \\ a_r = -2 \Omega v_{t,r} \end{cases} $$

This can be reformulated in terms of position derivatives in a rotating coordinate system with x-direction corresponding to the tangential direction, and the y-direction corresponding to radial direction.



\begin{cases} \dfrac{d^2x}{dt^2} = 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - 2 \Omega \dfrac{dx}{dt} \end{cases} $$

In the next section it will be shown that the equation of motion for the Earthbound frame confirms and generalizes the above relations.

The equation of motion for the Earthbound frame
The centrifugal term and the coriolis term are computational tools that enable the equation of motion to be formulated for motion with respect to a rotating coordinate system. The coriolis term is facultative; it is only necessary when the object that is being described has a velocity with respect to the rotating system. The centrifugal term is always present, for it is proportional to the distance to the center of rotation.

In the absence of any force the following set of equations give the equation of motion for motion with respect to a rotating coordinate system. In the form presented here, the x and y coordinate are position with respect to the central axis; the central axis is the zero point of the coordinate system. &Omega; is the angular velocity of the rotating system.



\begin{cases} \dfrac{d^2x}{dt^2} = \Omega^2 x + 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = \Omega^2 y - 2 \Omega \dfrac{dx}{dt} \end{cases} $$

For a derivation of the centrifugal term and the coriolis term, see the fictitious force article.

The two forces in the equation of motion
In the case of a Foucault setup, such as the Foucault pendulum or the Wheatstone pendulum, two forces must be included in the equation of motion: the centripetal force that sustains the rotation, and the restoring force that sustains the vibration. Of course, both the centripetal force and the restoring force are exerted by the springs as a single action. Separating this single action into two components (centripetal force and restoring force) facilitates the calculation.

The centripetal force of the rotation, for a coordinate system with the zero point at the central axis, is given by:



\begin{cases} F_{c,x} = - m \Omega^2 x \\ F_{c,y} = - m \Omega^2 y \end{cases} $$

The restoring force acts towards the equilibrium point of the vibration; the point where the bob is when the pendulum hangs down like a plumb line. The coordinate system can be chosen in such a way that central axis and the equilibrium point are both on the y-axis. Let the y-coordinate of the equilibrium point be called y e



\begin{cases} F_{r,x} = - m \omega^2 x \\ F_{r,y} = - m \omega^2 (y - y_e) \end{cases} $$

The full equation of motion of the pendulum bob contains four terms: the centrifugal term, the coriolis term, an expression for the centripetal force, and an expression for the restoring force. The vector sum of those terms gives the acceleration of the bob with respect to the rotating system.



\begin{cases} \dfrac{d^2x}{dt^2} = \Omega^2 x - \Omega^2 x - \omega^2 x & \!\!\! + \, 2 \Omega \dfrac{dy}{dt}\\ \dfrac{d^2y}{dt^2} = \Omega^2 y - \Omega^2 y - \omega^2 (y - y_e) & \!\!\! - \, 2 \Omega \dfrac{dx}{dt} \end{cases} $$

It is immediately clear why it is so convenient to use a co-rotating coordinate system (Earthbound system) to perform the calculations: the expression for the centripetal force and the centrifugal term drop away against each other.

The fact that the centrifugal term and the expression for the centripetal force drop away against each other is not a coincidence, it is a property of the physical system that is considered here. At every distance to the center of rotation the springs provide the amount of centripetal force that is required to sustain rotation with angular velocity Ω. In effect the system is self-adjusting. For example, when the angular velocity of the system increases, the springs deform a little more until the point is reached where the springs provide the required amount of centripetal force.

Letting the centrifugal term and the expression for the centripetal force drop away against each other:



\begin{cases} \dfrac{d^2x}{dt^2} = - \omega^2 x & \!\!\! + \, 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - \omega^2 (y - y_e) & \!\!\! - \, 2 \Omega \dfrac{dx}{dt} \end{cases} $$

This can be simplified further by shifting the zero point of the coordinate system. The coriolis term only contains velocity with respect to the rotating system, so it is independent of where the zero point of the coordinate system is positioned. In the following equations x and y are not the distance to the central axis but the distance to the center point of the vibration.



\begin{cases} \dfrac{d^2x}{dt^2} = - \omega^2 x + 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - \omega^2 y - 2 \Omega \dfrac{dx}{dt} \end{cases} $$



The above equations apply for only a single case, the case when the springs are vertical, which corresponds to the case of a polar Foucault pendulum.

For the general case, what needs to be included is the aspect that is illustrated in image 9. The amount of work that the centripetal force is doing is a function of latitude. The closer to the equator, the smaller the velocity component perpendicular to the central axis, the less work is done. Thus the expression for the deflection must be multiplied with the sine of the latitude.



\begin{cases} \dfrac{d^2x}{dt^2} = - \omega^2 x + 2 \Omega \dfrac{dy}{dt} sin(\phi) \\ \dfrac{d^2y}{dt^2} = - \omega^2 y - 2 \Omega \dfrac{dx}{dt} sin(\phi) \end{cases} $$



The image on the right shows a solution to the above equation of motion with a ratio of &omega; to &Omega; of 11 to 1 (Usually the ratio of &omega; to &Omega; is in the order of thousands to one). The image depicts the case of releasing the bob in such a way that on release it has no velocity with respect to the rotating system.

Solving the equation of motion
Switching to complex coordinates $$z=x+iy$$ the equations read $$ \dfrac{d^2z}{dt^2} + 2i\Omega \dfrac{dz}{dt} \sin(\phi)+\omega^2 z=0. $$

To first order in $$\Omega/\omega$$ this equation has the solution

$$ z=e^{-i\Omega \sin(\phi)t}\left(c_1 e^{i\omega t}+c_2 e^{-i\omega t}\right). $$

If we measure time in days, then $$\Omega=2\pi$$ and we see that the pendulum rotates by an angle of $$-2\pi\,\sin(\phi)$$ during one day.

Overview: invariance under transformation
It is of crucial importance to distinguish between the physical effects that are at play, and the calculational devices that are used to do computations. The Coriolis term in the equation of motion is a computational device, and in itself it has no physics content. The physics content is in the action of the centripetal force: doing work when the bob moves closer to the central axis of rotation, and doing negative work when the bob moves away from the central axis of rotation.

For motion with respect to the inertial frame, we have the following relation between radial velocity and angular acceleration:


 * $$ a_t = 2 \Omega v_r \,\!$$

We have that velocity in radial direction is the same in the inertial system and in the rotating system. That is, radial velocity is invariant under coordinate transformation to the rotating coordinate system. (If the origin of the rotating coordinate system is placed on the central axis.) The same goes for angular acceleration: angular acceleration is invariant under coordinate transformation to the rotating coordinate system (if the rotating coordinate system has a constant angular velocity).

Correspondingly: the relation at=2&Omega;vr is invariant under coordinate transformation to a rotating coordinate system, and so is the relation ar=2&Omega;vt,r.

It is tempting to think that letting the centrifugal term and the expression for the centripetal force drop away against each other means that the centripetal force is not involved in the physics taking place. In actual fact, letting the centrifugal term and the expression for the centripetal force drop away against each other means that the remaining expression in the equation of motion becomes the relation that is invariant under transformation.



\begin{cases} \dfrac{d^2x}{dt^2} = 2 \Omega \dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2} = - 2 \Omega \dfrac{dx}{dt} \end{cases} $$

This relation can be referred to as the generalized Coriolis effect. The generalized Coriolis effect applies when the amount of centripetal force that is exerted is always the amount that is required for co-rotating motion.