Calculating the average linear speed of all points in a rotating sphere

AI Thread Summary
Calculating the average linear speed of all points in a rotating sphere involves understanding the distribution of speeds, as points closer to the axis rotate slower than those at the equator. The average speed cannot be simply derived from r/2 and pi due to the complexities of rotational dynamics. Kinetic energy and momentum are key concepts, with the average velocity of a stationary spinning disc being zero, despite its rotational energy. The discussion highlights the distinction between kinetic energy and momentum, emphasizing that while a sphere has rotational energy, its linear momentum can be zero. The conversation also touches on the implications of these dynamics at both micro and macro scales, particularly in relation to theoretical physics.
Darius Kaufmann
Messages
2
Reaction score
0
Is there a way to calculate the average linear speed of all points in the volume of a sphere rotating on a single axis? Since points closer to the axis of rotation and the poles move slower than points further out, would the average speed be a simple function of r/2 and pi? It would seem that there should be a significant relationship between that average and energy dynamics on both micro and macro scales. For example, points on the equator of a spinning proton are presumed to be moving at c, according to current theories – What about the gradient of speeds approaching the axis and poles?
 
Physics news on Phys.org
Hello Darius, ##\qquad## :welcome: ##\qquad## !

easier variation: how about a disk ? Well, kinetic energy from rotation is ##{1\over 2} I\omega^2## with ##I= {1\over 2} M R^2## for a disk.
Kinetic energy of a point mass M moving along at speed ##v## is ##{1\over 2} M v^2##

Equate the energies $${1\over 2} \left ( {1\over 2} M R^2\right) \omega^2 = {1\over 2} M v^2\ \Rightarrow \ v^2 = {1\over 2} \omega^2 R^2\ \Rightarrow \ v = {1\over 2} \sqrt 2\ \omega R$$

Can you now do the 3D sphere by yourself ?
 
Thank you so much -- However, I'm a musician, not a physicist, so I don't even know what all the symbols mean in the above equations - Can you clarify a little? Thanks again for taking the time to respond!
 
Darius Kaufmann said:
For example, points on the equator of a spinning proton are presumed to be moving at c, according to current theories

This is nonsense. Why do you think that?
 
Darius Kaufmann said:
Summary: Is there a way to calculate the average linear speed of all points in the volume of a sphere rotating on a single axis?

Is there a way to calculate the average linear speed of all points in the volume of a sphere rotating on a single axis?
I'm wondering what it is, exactly that you want to know. We need some basic things about motion. The average velocity of a stationary spinning disc (in the reference frame of the Lab) is zero - that is if you use the word "average" to signify the "Mean" - add em all up and divide by the number of points' - and it is going nowhere so the mean is zero. But a spinning object has angular velocity.
There are two quantities that indicate the Motion and 'stopability' of a body. In the past they were confused with one another but eventually they were acknowledged to be two different things. One is the Kinetic Energy and the other is the Momentum. KE involves the square of the velocity and that is always positive. Momentum involves Velocity Vectors which have direction as well as speed.
A stationary, spinning sphere clearly has Energy, despite a mean velocity of zero. @BvU discusses the KE idea, above. But the linear momentum is zero so what about the rotation?
Linear Momentum is just mv but angular momentum depends on the distribution of the mass as well as the total mass. Its Moment of Inertia is the result of adding all the individual particle masses times the (distance from the centre of rotation) squared. So the outer parts have a bigger contribution than the inner parts (flywheels usually have the mass in an outer ring for this reason).
 
  • Like
Likes BvU
PeroK said:
This is nonsense. Why do you think that?
Wholeheartedly agree.

Must admit I only read the first half of #1 and (re)acted on that.

Reading the second half feels like hearing a chainsaw go through a violin...
 
  • Haha
Likes sophiecentaur and berkeman
Back
Top