[simplified (OP)]

Axe of rotating sphere is motionless relatively of me.I know mass of the sphere and speed of rotating of the sphere.The density of the sphere is homogeneous.How to define impulse of  the rotating sphere relatively of me?

[Soul Surfer]

Please restate your question in a different way simplified  I appreciate that your first language is probably not english but what you have stated is not clear as a question.

If you want to understand about the angular momentum energy of a rotating sphere.   This also depends on the distribution of the mass throughout the sphere(although you have already said it is uniform) and its rigidity because the angular momentum of each particle is defined by its velocity and its distance from the axis.

if you wish to know what impulse you would receive if you landed on the sphere  this depends where you land on it.  you would be rapidly accelerated to the rotational velocity of the sphere at that point assuming that you stayed in contact with the surface.  For the equator of the earth this is a rapid acceleration from 0 to 1000mph
or  1600 Km per hour.

[Bored chemist]

If it has uniform density and it's spinning about its axis then the centre of mass is also on that axis.
If the axis is not moving relative to you then the centre of mass is also not moving relative to you. In that case, the momentum it has is zero and so the impulse is also zero.

[simplified (OP)]

Thanks.
Then I would like to know average speed of all particles of the sphere(relatively of me).

[lightarrow]

Thanks.
Then I would like to know average speed of all particles of the sphere(relatively of me).

If
ω = angular speed of rotation
R = sphere's radius

then, if I computed correctly the integral, it should be (3/16)πωR.

[Soul Surfer]

If you are stationary with respect to a uniform rotating sphere in a vacuum the average velocity of all the particles in that rotating sphere is zero because there are always as many moving away from you as there are moving towards you. 

If the sphere is moving through a medium like the air there is a force on it  across the direction it is moving because of the interaction with the air like a swinging tennis ball.

[simplified (OP)]

Thanks.
Then I would like to know average speed of all particles of the sphere(relatively of me).

If
ω = angular speed of rotation
R = sphere's radius

then, if I computed correctly the integral, it should be (3/16)πωR.

Thank you very much.

[lightarrow]

If you are stationary with respect to a uniform rotating sphere in a vacuum the average velocity of all the particles in that rotating sphere is zero because there are always as many moving away from you as there are moving towards you. 

Of course, but he talked of "speed" not of "velocity"  []

[Soul Surfer]

I agree but people often confuse the two and I can se absolutely no point in wanting to calculate the average speed.

Als I think that the average speed you quote will also depend on how close you are to the sphere.

[Bored chemist]

"Als I think that the average speed you quote will also depend on how close you are to the sphere."
I don't.
As long as I'm not moving towards or away from it it has some constant value of rotational kinetic  energy no matter where I am.
That's a sum of the KE of each particle it's made from.
So the sum of all those 1/2 MV^2 terms must be constant.
The M can't vary so the sum of the V^2 terms must also be constant.
I'm fairly sure that if the sum of the v^2 terms (for each atom the sphere is made from) is constant then the mean speed is constant.

I also can't see much point calculating the mean speed, but it's perfectly well defined and it's not trivial (the mean velocity is, as you say, always zero which is trivial and dull)

[Soul Surfer]

I agree that the rotational KE does not change and that it is quite possible that this is what our questioner wanted to know.  I was talking about  the mean speed of all the particles in the sphere with respect to someone standing outside of the sphere which is very different.

The moment of inertia of a sphere   I  (kg m^2)  =  2/5 MR^2     M=mass in Kg  R=radius in meters.

[simplified (OP)]

Als I think that the average speed you quote will also depend on how close you are to the sphere.

If you are right,then show  your calculatings of the average speed relatively of different observers. []

[Geezer]

Axe of rotating sphere is motionless relatively of me.I know mass of the sphere and speed of rotating of the sphere.The density of the sphere is homogeneous.How to define impulse of  the rotating sphere relatively of me?

What does "impulse" mean?

[Bored chemist]

http://en.wikipedia.org/wiki/Impulse_(physics)

I showed that the average of the square of the velocity is constant for any observer.
Since that is constant, the mean square of the speed is also independent of the position of the observer.
I'm pretty sure that the symmetry of the system means that the mean speed is also independent of the location of the observer.

The mean speed is an odd thing to calculate, but it makes sense in the context of trying to find an impulse.

[Geezer]

http://en.wikipedia.org/wiki/Impulse_(physics)

Then it's a silly question. The rotation of the sphere makes no difference. It might as well be a stationary lump of coal.

[simplified (OP)]

http://en.wikipedia.org/wiki/Impulse_(physics)

Then it's a silly question. The rotation of the sphere makes no difference. It might as well be a stationary lump of coal.

Yes, and now we at all do not discuss my first question. []

[simplified (OP)]

I study motion of mass. Is it useless or useful, but it exists. []

[Soul Surfer]

The "impulse" of one object on another only makes sense if the objects come into physical contact but "me" and the sphere are motionless with respect to one another but the sphere is rotating so there is no effect unless thew speeds involved are relativistic and frame dragging is involved.

Re the effect of distance on average velocity  let us assume first that you are some distance from the sphere and it subtends an angle of a few degrees looking at the equator and through the object integrating the relative speeds of all the particles that make up the sphere many are moving at right angles and show little velocity.  no consider you are much closer and the sphere subtends say 160 degrees in your field of view there are vastly more of the particle travelling significantly towards and away from you and so the total integral of speed is much greater.

[simplified (OP)]

The "impulse" of one object on another only makes sense if the objects come into physical contact but "me" and the sphere are motionless with respect to one another but the sphere is rotating so there is no effect unless thew speeds involved are relativistic and frame dragging is involved.

Re the effect of distance on average velocity  let us assume first that you are some distance from the sphere and it subtends an angle of a few degrees looking at the equator and through the object integrating the relative speeds of all the particles that make up the sphere many are moving at right angles and show little velocity.  no consider you are much closer and the sphere subtends say 160 degrees in your field of view there are vastly more of the particle travelling significantly towards and away from you and so the total integral of speed is much greater.

I do not need impulse contact. I suspect gravitational-kinematic contact creates something. But it is a secret here. []

[Bored chemist]

http://en.wikipedia.org/wiki/Impulse_(physics)

Then it's a silly question. The rotation of the sphere makes no difference. It might as well be a stationary lump of coal.

You would have thought someone would have pointed that out earlier.
Oh, hang on, I did.
However far I am from the sphere, only the particles that form a circle exactly on my line of sight are moving perpendicular to me. Their number is constant.
So there's no evidence that the sum of the speeds is zero.
More importantly, they may have zero velocity from my point of view, but they still have exactly the same speed as the others.
Apart from particles on the axis, every bit of the sphere traces out a circle. The radius of that circle is the distance from the particle to the axis. The circumference is 2 pi times that and it travels that distance each time the sphere revolves.
That fixes the distance it travels, and the time it takes to do so. So that sets its speed.
Averaging that over all particles will give the mean speed.
There's nothing there about how far away from it I  (or you) might be.