In my animation
https://www.geogebra.org/m/uucnkfhy
there is the chord OD which exerts a blue centripetal force on the
rotating body A and there is an opposite red centrifugal force which is
present only in the rotating reference (and not in the inertial one).

Having specified this, is it correct to say that the point of
application of the two opposing forces is point D?

[[Mod. note -- No, not really.

The *centrifugal* "force" acts on A's entire mass, so if we're going to
model it as a force acting at a single point, it (the centrifugal "force")
must have A's center of mass as its point of application.

Where the *centripetal* force is applied depends on how A is teathered to
the central point O. E.g. if A is attached to O by a string which connects
to A at point D (the innermost point of A), then that point (D) is where
the centripetal force would be applied. But if (e.g.) the string goes
into a hole drilled into the body A and the string is actually only attached
to A at the center of the body A, then the centripetal force would be applied to A there (the center of the body A). And if the centripetal force is supplied by the Newtonian gravity field of a spherical mass located at O,
then the point of application is approximately the center of A.
-- jt]]

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On Monday, 4 July 2022 at 08:05:58 UTC+2, Luigi Fortunati wrote:

In my animation
https://www.geogebra.org/m/uucnkfhy
there is the chord OD which exerts a blue centripetal force on the
rotating body A and there is an opposite red centrifugal force which is present only in the rotating reference (and not in the inertial one).

Having specified this, is it correct to say that the point of
application of the two opposing forces is point D?

[[Mod. note -- No, not really.

The *centrifugal* "force" acts on A's entire mass, so if we're going to
model it as a force acting at a single point, it (the centrifugal "force") must have A's center of mass as its point of application.

Where the *centripetal* force is applied depends on how A is teathered to
the central point O. E.g. if A is attached to O by a string which connects
to A at point D (the innermost point of A), then that point (D) is where
the centripetal force would be applied. But if (e.g.) the string goes
into a hole drilled into the body A and the string is actually only attached to A at the center of the body A, then the centripetal force would be applied to A there (the center of the body A). And if the centripetal force is supplied by the Newtonian gravity field of a spherical mass located at O, then the point of application is approximately the center of A.
-- jt]]

Eventually, if there is no wobbling, the resultant centripetal force
must be simply attached to the center of mass: in fact, always
equal and opposite to the centrifugal one. (No?)

Then I'd have a question: to the centripetal force, in accordance
with Newton's third law, corresponds an equal and contrary force
that the string exerts on the central pivot. But how does the
centrifugal force satisfy that same law, if it does? (I understand
the two forces really exist in the two distinct frames, but all the
more so the question remains).

Thanks in advance for any insights,

Julio

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Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

In my animation
https://www.geogebra.org/m/uucnkfhy
there is the chord OD which exerts a blue centripetal force on the
rotating body A and there is an opposite red centrifugal force which is present only in the rotating reference (and not in the inertial one).

Having specified this, is it correct to say that the point of
application of the two opposing forces is point D?

[[Mod. note -- No, not really.

The *centrifugal* "force" acts on A's entire mass, so if we're going to
model it as a force acting at a single point, it (the centrifugal "force") must have A's center of mass as its point of application.

Actually calculating it for an extended body can be simplified
by noting that the centrifugal force can be derived from a potential.
(equal to - 1/2 \Omega^2 r^2,
with r the distance from the axis of rotation)

Where the *centripetal* force is applied depends on how A is teathered to
the central point O. E.g. if A is attached to O by a string which connects to A at point D (the innermost point of A), then that point (D) is where
the centripetal force would be applied. But if (e.g.) the string goes
into a hole drilled into the body A and the string is actually only attached to A at the center of the body A, then the centripetal force would be applied to A there (the center of the body A). And if the centripetal force is supplied by the Newtonian gravity field of a spherical mass located at O, then the point of application is approximately the center of A.

For the Earth/Moon system it isn't.
The simplest way to derive the tides (Earth/Moon only)
is to calculate the combined potential in co-rotating coordinates.
(gravity from Earth and Moon, and centrifugal potential
from the rotation around their common centre of mass)

It is the shortest way to seeing that there must be two tidal bulges,

Jan

[[Mod. note -- You're right. I was thinking of the case where the
central mass O is fixed in position in an inertial reference frame.
That's a bit unrealistic... :)
-- jt]]

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Luigi Fortunati alle ore 18:05:54 del giorno domenica ha scritto:

In my animation
https://www.geogebra.org/m/uucnkfhy
there is the chord OD which exerts a blue centripetal force on the
rotating body A and there is an opposite red centrifugal force which is present only in the rotating reference (and not in the inertial one).

Having specified this, is it correct to say that the point of
application of the two opposing forces is point D?

[[Mod. note -- No, not really.

Where the *centripetal* force is applied depends on how A is teathered to
the central point O. E.g. if A is attached to O by a string which connects to A at point D (the innermost point of A), then that point (D) is where
the centripetal force would be applied.
-- jt]]

This is the case of my animation and, therefore, the point of
application of the centripetal force (of the string on body A) is,
without a doubt, point D.

But at point D, in addition to the centripetal action of the string on
body A, there must also be an opposite (and, therefore, centrifugal)
reaction force of body A on the string (third principle).

I would like to know if there is any other point (of the string or of
the body A) where the third principle is not valid, that is where the
action of the centripetal force is not opposed by any type of opposite
reaction (ie directed in the centrifugal direction).

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J. J. Lodder alle ore 18:18:20 del giorno mercoled=EC ha scritto:

The simplest way to derive the tides (Earth/Moon only)
is to calculate the combined potential in co-rotating coordinates.
(gravity from Earth and Moon, and centrifugal potential
from the rotation around their common centre of mass)

It is the shortest way to seeing that there must be two tidal bulges,

Why does centrifugal force that "appear" ONLY in accelerated references=20 generate bulges that ALSO appear in inertial references?

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Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

J. J. Lodder alle ore 18:18:20 del giorno mercoledì ha scritto:

The simplest way to derive the tides (Earth/Moon only)
is to calculate the combined potential in co-rotating coordinates.
(gravity from Earth and Moon, and centrifugal potential
from the rotation around their common centre of mass)

It is the shortest way to seeing that there must be two tidal bulges,

Why does centrifugal force that "appear" ONLY in accelerated references generate bulges that ALSO appear in inertial references?

It is a physical effect,
so it cannot depend on any particular reference frame.
It is just that the derivation is simpler in one frame than in another.

In non-rotating coordinates you have a dynamic problem
rather than a static one, so you have to derive the motions.

The bulge then comes about because the solid bulk of the Earth
falls faster towards the Moon than the water on the far side,

Jan

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