Will the coin spin forever or will it be de-accelerated by the gravitational force?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ The gravitational force on earth doesn't slow the coin down, friction does. $\endgroup$– user121330Commented Mar 10, 2021 at 15:37
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
Assume there is no other force (such as friction), and say the coin is near the Earth and experiences the Earth's gravitational field, then with classical mechanics, the gravitational force only acts on the center of mass of the coin, and produce no torque, so the coin, if it is spinning initially, will remain spinning forever.
This is the rotational analogy of Newton's 1st law, which states that any object that's moving in constant velocity, without any external force, will remain moving in constant velocity.
$\begingroup$
$\endgroup$
2
Strictly speaking, the spinning coin loses kinetic energy because it emits gravitational waves. This loss is so small though that there will be no perceptible change during a lifetime.Aa
-
1$\begingroup$ Another tiny effect which will slow the coin down is that the photons it emits/reflects from the parts facing the direction of rotation will on average have a slightly shorter wavelength than the ones leaving in the opposite direction due to the Doppler effect. Since a photon of shorter wavelength has higher momentum, this results in a weak slowing effect on the rotation of the coin. Effectively a "friction" against the electromagnetic field. $\endgroup$ Commented Mar 10, 2021 at 15:32
-
$\begingroup$ Assuming that this is the case (I don't see why it shouldn't though), then this effect will be even larger, I think. $\endgroup$ Commented Mar 10, 2021 at 15:38
$\begingroup$
$\endgroup$
If would like to elaborate on the answer give by Leo L.
Tidal forces have to be considered, too. Those will eventually lead to a rotation of the coin locked to its orbit, like moon. In this specific set up, however, I reckon that tidal forces are so small that they can be ignored for all practical purposes. The term „practical purpose“ here includes the moon which had enough time to synchronise its rotation.
The crucial aspect is the relation of the radius of an extended object to its orbit. For the coin it is much smaller than for moon and the timescale of synchronisation increases with a very high power of that ratio.
Edit:
Elaborating on my own answer: tidal forces only apply when the body is not completely rigid but is deformable by the gravitational forces.
If the coin is considered rigid then you can do the maths and integrate all forces over all of the volume of the coin and the result is as described in the answer by Leo L.
$\begingroup$
$\endgroup$
0
The coin is spinning because you put energy in it, Which is force in this case and according to the second law of thermodynamics energy is escaping, It will run out of energy at some point. I don't think coin would spin forever.
and i'm not proper english speaker but i hope you get the point
