# On whether the motion of a Foucault pendulum bob is comparable to ballistics

• I
• Cleonis
In summary: The tangential velocity is perpendicular to the direction of the vehicle's velocity, so it does not contribute to the resultant circumnavigating velocity. As a result, the pendulum bob will swing forward and backward, but it will not experience any net movement in the circumnavigating direction.When the pendulum bob swings front-to-rear:Then the resultant circumnavigating velocity of the pendulum is the difference of two velocity vectors: 1) the velocity of the circumnavigating vehicle, and 2) the tangential velocity of the pendulum bob. The tangential velocity is again perpendicular to the direction of the vehicle's velocity, so it does not contribute
Cleonis
Gold Member
A recurring question is: while the motion of a polar Foucault pendulum is fairly straightforward, the case of a non-polar Foucault pendulum is quite difficult to visualize.

In 2020, on physics stackexchange someone submitted that question and I contributed an answer.

In a comment to another answer someone raised the question: "If I visualize the motion of the pendulum bob as that of a flying cannonball, does that work out for east-to-west motion?"

I replied to that comment, arguing that comparison with ballistic motion is wrong. (I will explain further down how to see that that is wrong.)

A year later someone replied to my comment, arguing I was in error. Some days ago I happened to notice that comment, and then back and forth comments started flying.

Now: stackexchange specifically states: comments are not for protracted discussion. I endorse that policy, so I proposed to migrate to physicsforums. My discussion partner agreed to that, so that is why I start this thread here.

I should mention: in my first comment on stackexchange I was pressed for space. (Stackexchance comments are 600 characters maximum)

With all of the above in place:
For a mathematically idealized Foucault pendulum setup: we have that that to a very good approximation the plane of swing of a Foucault pendulum is turning at an even pace. (Actually, Joe Wolfe points out there is a subtlety with the pace of the plane of swing turning, but it's not necessary to go into that level of detail.)

A Foucault pendulum bob must turn the same amount during east-to-west swing as during west-to-east swing, otherwise the pace of the plane of swing turning would be uneven. That would go against observation.

For a cannonball being fired east-to-west or west-to-east makes a difference.

For simplification: neglect the influence of air friction on the cannonball.
During ballistic trajectory the motion of a cannonball is effectively orbital motion. (It's just that that the cannonball does not have enough velocity to clear the Earth. Very quickly the point is reached where the "orbital" motion intersects with the Earth surface, and the cannonball impacts.)

First I will discuss the case of a celestial body with negligably slow rotation, then the case of a celestial body with significant rotation: the Earth.

(Instead of specifying zero rotation I specify a very slow rotation so that poles and latitude lines can be defined)

In order to avoid convoluted wording: all the description below is for a cannonball fired on the northern hemisphere. (For the southern hemisphere everything is mirrored.)

On the very-close-to-not-rotating-celestial body:
Let a cannonball be fired in west-to-east direction. The ballistic motion is orbital motion, and the ground track of that motion will coincide with a section of a great circle. The cannonball will proceed towards the equator.

Let a cannonball be fired in east-to-west direction. Again: the ground track will coincide with a great circle. The cannonball will proceed towards the Equator.

Now the Earth:
Let a cannonball be fired in west-to-east direction.
Once the cannonball is proceeding along its trajectory the only interaction the cannonball has with the Earth is the gravitational attraction to the Earth's center of mass: orbital motion. The impact of the cannonball will be south of the latitude line it was fired from. Since the Earth is rotating underneath that orbit: compared to the very-close-to-not-rotating-celestial body case the impact site will be shifted accordingly.

Let a cannonball be fired in east-to-west direction.
Orbital motion: the impact of the cannonball will be south of the latitude line it was fired from. There will be some shift of the impact site in accordance with the Earth rotation underneath the orbit, but there is no scenario such that the cannonball impacts inside of the latitude line it was fired from. The cannonball will still proceed to the Equator, and impact to the south of the latitude line it was fired from.Difference between cannonball motion and the motion of a Foucault pendulum bob

A Foucault pendulum bob is suspended, and the combination of suspension force and gravity provides the centripetal force that is required to make the pendulum bob circumnavigate the Earth's axis.

Let me take the case of the actual Foucault pendulum in the Pantheon.

Foucault describes that on rare occasions there was time for long uninterrupted runs. Over the course of such a long run the amplitude of the swing would decay to a mere 10 centimeters, but Foucault reports that the plane of swing was still turning at the same rate.

The Foucault pendulum in the Pantheon has a 67 meter cable. At the latitude of Paris the distance to the Earth's axis is such that the required centripetal force to sustain circumnavigation of the Earth's axis corresponds to a weight on a 67 meter cable to be displaced about 10 centimeter. The angle is 0.1 of a degree

That is a very small angle, but it is still the same phenomenon as a pendulum suspended in a vehicle. If that vehicle is slowly circumnavigating a roundabout then the suspended pendulum bob is shifted outward slightly. The angle of outward shift is the angle such that the required centripetal force is provided.

Now to the crucial difference:
The gravitational force acting on the cannonball in flight is directed towards the Earth center of mass. The centripetal force acting on the pendulum bob is directed perpendicular to the Earth's axis. Those are distinct directions.

Back to the example of a moving vehicle, with a pendulum suspended from in it. We go to a banked circuit, and we have the vehicle circumnavigating at precisely the velocity where the slope of the bank make gravity provide the required centripetal force.

With the vehicle circumnavigating at precisely that velocity: a pendulum that is stationary with respect to the vehicle will hang perpendicular to the vehicle.

Now let the pendulum be swinging.
We examine the case of swing tangent to the instantaneous direction of velocity. Then the swing (relative to the vehicle) is from rear-to-front and from front-to-rear.

When the pendulum bob swings rear-to-front:
Then the resultant circumnavigating velocity of the pendulum is the sum of two velocity vectors: 1) the velocity of the circumnavigating vehicle, and 2) the velocity of the bob relative to the vehicle. The pendulum bob will swing wide.

When the pendulum bob swings front-to-rear:
Then the resultant circumnavigating velocity is the difference of two velocity vectors. Now the pendulum bob is circumnavigating slower than the vehicle is circumnavigating. But the pendulum bob is still subject to the centripetal force that sustains the vehicle circumnavigation. So now the pendulum bob is subject to a surplus of centripetal force, and that surplus pulls the pendulum bob to the inside of the circumnavigating motion.

Incidentally: some years ago I created a Java simulation for that case, the title of that simulation is 'circumnavigating pendulum'. (So that is for the case of true planar motion.) The simulation is available on my website. The simulation shows side-by-side the motion relative to the inertial coordinate system, and a co-rotating point of view. (I haven't checked whether those simulations still run on recent versions of the JRE. I will have to port the simulations I created to a more accessible platform.)

The considerations from the planar case (circumnavigating vehicle with a pendulum in it) carry over to the terrestrial Foucault pendulum. The added factor, of course, is the angle between the Earth surface and the Earth's rotation axis.

Last edited:
Cleonis said:
The centripetal force acting on the pendulum bob is directed perpendicular to the Earth's axis.
I don't see why that is. Does the string tension not contribute to the centripetal force of the bob? Can you draw a diagram?

Also, I find the title of this thread vague. When you say "comparable", what specific aspect of each object's motion are you comparing?

Last edited:
russ_watters
Cleonis said:
Let a cannonball be fired in east-to-west direction.
Orbital motion: the impact of the cannonball will be south of the latitude line it was fired from.
You appear to be assuming the battle is held in the Northern Hemisphere.

kuruman said:
Does the string tension not contribute to the centripetal force of the bob? Can you draw a diagram?

The above diagram is for the case of a plumb line. The arrows in the diagram represent the forces acting on the plumb.

To bring out the angle: the oblateness of the Earth is much exaggerated.

Blue arrow: gravity
Red arrow: tension of the plumb line
Green arrow: the resultant of red and green, resolved in the direction parallel to the local Earth surface.

So indeed it is the resultant of gravitational force and wire tension that provides the required centripetal force.

The plumb bob is co-moving with the Earth, circumnavigating the Earth's axis. The required centripetal force lies in the plane of circumnavigation. In other words: the actual required centripetal force is in the direction perpendicular to the Earth's axis.

For practicality the usual approach is to obtain the component of force parallel to the local surface. That is because if the plumb line starts to swing the motion is parallel to the local surface. That is why in the diagram the green arrow gives the force component in the direction parallel to the local surface.

Of course: to account for the required centripetal force completely one must also obtain a value for the force component perpendicular to the local surface. Providing that component comes at the expense of gravitational acceleration.

Example of required centripetal acceleration coming at the expense of gravitational acceleration:
For an object to co-rotate with the Earth along the Equator a centripetal acceleration of 0.0339 m/s^2 is required. At the Equator the measured gravitational acceleration is 9.7805 m/s^2 Hence we can infer that at the Equator the amount of gravitational acceleration due to the Earth's gravitational mass must be 9.8144 m/s^2
(9.7805 + 0.0339 = 9.814)

The actual oblateness of the Earth:
For an object located at 45 degrees latitude: to provide the required centripetal force the corresponding angle is about 0.1 degree
(More precisely: 0.1 degree away from being 180 degrees opposite in direction.)

kuruman said:
When you say "comparable", what specific aspect of each object's motion are you comparing?

Well, I have argued that the two are not comparable.

Motion of the bob of a Foucault pendulum (during swing from one cusp to the next cusp) is not comparable to ballistic motion.

I suppose that people who think they are in some sense comparable are thinking in terms of the ground track of the respective motions. The motion of a fired cannonbal has a particular ground track. You take the plane of the motion of the cannonball, and you intersect that plane with the Earth's surface; that is the ground track of the cannonball.

Baluncore said:
You appear to be assuming the battle is held in the Northern Hemisphere.
I will edit the original post to make that explicit.

Cleonis said:
Well, I have argued that the two are not comparable.
You did not answer my question. To decide whether two entities are comparable or not, you put them side by side and you examine some common aspect of them. For example, when you compare apples and oranges and the aspect is how they come about, you decide that they are comparable because they both grow on trees. If the aspect is how you eat them, you decide that they are not comparable because you can eat an apple without peeling it whereas the orange will not taste good if you don't peel it. So apples and oranges are not comparable as far as how you eat them is concerned.

My question to you is what aspect(s) of the motion of the bob and the cannonball did you put side by side and decided that they are not comparable?

Cleonis said:
The above diagram is for the case of a plumb line. The arrows in the diagram represent the forces acting on the plumb.
Which arrow represents the centripetal force that you say is perpendicular to the Earth's axis? It would help if you put some identifiers on your drawing like the angular velocity vector of the Earth.

Also, why do you think it is necessary to invoke the oblateness of the Earth? Would the plumb bob at, say, a ##45^{\text{o}}## latitude point straight at the center of a perfectly spherical rotating Earth?

Baluncore said:
You appear to be assuming the battle is held in the Northern Hemisphere.
Unlike the Battle of the Falkland Islands.

You can also say a pint is smaller than a quart. I just comparfed them - even though they are not equal.

Despite all the words, I have no idea what you are asking, if you are asking anything. I don't know what you are trying to teach us - if you are trying to teach us anything.

kuruman said:
Also, why do you think it is necessary to invoke the oblateness of the Earth? Would the plumb bob at, say, a ##45^{\text{o}}## latitude point straight at the center of a perfectly spherical rotating Earth?

In the hypothetical case of a rotating celestial body, with perfectly spherical shape: A plumb line would not point straight to the geometric center of that celestial body, due to the fact that the plumb line is co-rotating with the celestial body.

There is a connection with the oblateness of the Earth, it goes as follows: at every latitude a plumb line hangs perpendicular to the local Earth surface. This is because the spinning Earth as a whole is very close to a state of hydrostatic equilibrium.

Over geologic time scale the Earth is ductile enough such that as a whole the Earth is in hydrostatic equilibrium.

• Introductory Physics Homework Help
Replies
9
Views
766
• DIY Projects
Replies
53
Views
7K
• Mechanics
Replies
10
Views
2K
• Mechanics
Replies
5
Views
2K
• Mechanics
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• New Member Introductions
Replies
1
Views
110
• Classical Physics
Replies
86
Views
4K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Mechanics
Replies
10
Views
4K