# A spinning wheel

1. Feb 6, 2014

### andywelik

Let's imagine a spoke on a spinning wheel like that of a cart wheel. Now we know and can see that the outermost particles of the spoke have a higher speed than the inner particles. As we get closer and closer to the centre, we know that the speed gets slower and slower.

Here's my question. Is there a particle in its dead centre which does not spin? In my mind I am not able to think of even an atom in the dead centre that will not rotate. But that is not possible for this reason: Since speed diminishes as we get closer and closer to the dead centre, there has to be a point when the speed gets to be zero. Is that possible?

2. Feb 6, 2014

### CWatters

Funny things happen when you look at infinity or 1/infinity. Lets say you measure the surface speed at a radius of one meter, then half a meter, then a quarter of a meter, then an eighth of a meter etc. You can go on doing that indefinitely. You will get closer and closer to the axis of rotation but the speed will never get to zero. An atom in the centre will rotate.

It might be more helpful to think in terms of the angular velocity (measured in degrees per second). That remains constant all the way to the centre.

3. Feb 6, 2014

### Staff: Mentor

You are mixing up linear/translational and rotational motion. The linear speed is indeed different at different radii, but the entire wheel has the same rotational speed.

4. Feb 6, 2014

### andywelik

OK no problem. Let's convert rotational to angular. Won't there be a speed at dead centre which cannot be converted because linear zero speed has reached?

5. Feb 6, 2014

### CWatters

See my previous reply. "linear zero speed" is never reached.

6. Feb 6, 2014

### Staff: Mentor

For an arbitrary elemental size (such as a point), zero linear speed IS reached, but that doesn't make the rotational speed zero.

Not sure what conversion you had in mind...

7. Feb 6, 2014

### andywelik

Zero linear speed multiplied by any other number is still zero, no matter what the formula is. Is that not true?

8. Feb 7, 2014

### A.T.

Zero multiplied by a finite number is still zero.

9. Feb 7, 2014

### A.T.

Did you mean linear to angular?

omega = v/r = 0/0 = indeterminate, which means omega can take any finite value, including the constant value it has at r>0. There is no implication that the center doesn't spin.

10. Feb 7, 2014

### Sunfire

The problem here arises because a wheel is a finite-sized entity, for which peripheral velocity and angular velocity are well defined.

At the center, the axis of rotation crosses the plane of rotation (the plane of the flat wheel) in a single geometrical point. I mathematics, a point does not have size. Think of it as having a size of zero. Then, for this central point, linear and angular velocity are meaningless. A mathematical point is after all, an approximation to reality.

11. Feb 7, 2014

### Baluncore

Probably the best concept here is movement, or translation of a point, rather than rotation of a point.

The “Fixed Point Theorem” proves that when you stir a liquid, there will be, at every point in time, at least one point that is not moving, (but where the immediately surrounding liquid may be moving).

That can be applied to a your wheel, so long as the axle is not fixed to the chassis.

12. Feb 8, 2014

### andywelik

Thank you all for taking the trouble to explain.

13. Feb 9, 2014

### Varun Bhardwaj

there will be a point at centre, but you cannot prove it in real life

14. Feb 9, 2014

### andywelik

Let's take the centre point. Imaginarily enlarge it a million times. Take a photo of it, imagined of course. Won't that show the point you mention?

15. Feb 9, 2014

### A.T.

Zero multiplied by a finite number is still zero.

16. Feb 10, 2014

### Khashishi

This is a good question, but hard to answer because you get quantum mechanical effects.

Everything gets fuzzy when you get below a certain scale. Since the temperature is finite, particles are constantly shifting and bouncing around, so the simple answer is no, all particles are moving to some extent. The Heisenberg uncertainty principle also puts a limit on how precisely a particle can be located in phase space. If a particle is centered at the dead center, it has a wave function which extends outside the center (to infinity). In a large object like a top, the center point has overlapping contributions from many particles.

What if the top is the size of a molecule? In quantum mechanics, angular momentum is quantized. This means that a molecule is only allowed to spin at very specific frequencies, which gives rise to a rotational spectrum which is characteristic to the molecule. Depending on the symmetry of the molecule, there might not be any atom in the center.

What if the top is a single particle? In quantum mechanics, every particle has an intrinsic spin. For fundamental fermions, the spin is either +1/2hbar or -1/2hbar, (basically the smallest allowable unit in the clockwise or counterclockwise direction). Fundamental particles don't have any size, so there are no geometric features on them that we can rotate, so their spin can only be understood theoretically, through interactions. Some physicists will jump in and say spin isn't the same as orbital angular momentum, but they are both angular momentum just the same.

Back to the large macroscopic top. If there is a fermion in the dead center of the top, then it is spinning with some superposition* of clockwise and counterclockwise. I think if you spin the top rapidly, the mix ratio will change a tiny bit. You can get the center particle to spin on average equal amounts clockwise and counterclockwise, but you can't get the particle to stop spinning.

*might be more accurate to say entangled mixture of clockwise and counterclockwise

17. Feb 12, 2014

### andywelik

Take a liquid passing thru' a pipe. The molecules that adhere to the inside surface of the pipe move very slowly while those outside of the thin film on the pipe move faster. On a solid this could never be.

18. Feb 12, 2014

### Staff: Mentor

No, because there will still be a single point at the center of the enlargement, and we can ask the same question about that point. And we can repeat this process (in our imagination) forever without ever getting to an end.

There was much sense in CWatters response earlier in this thread: "Funny things happen when you look at infinity or 1/infinity"; you are basically talking about 1/infinity when you try narrowing down to a single point.

19. Feb 12, 2014

### Baluncore

Define a circle with radius r, centred at the origin, x^2 + y^2 = r^2
The centre is specified as a point, P(x,y), where x = 0 and y = 0.
Rotate that circle continuously about it's centre.

The x and y coordinates of the centre do not change with that rotation.
Therefore the point at the centre is not moving.

In mathematics, a point does not have an orientation.

20. Feb 12, 2014

### AlephZero

This is rather missing the point (groan!). If we are talking about classical mechanics of a solid, by any reasonable definition of the shear strain at the axis of rotation, the "center point" does "rotate".

The mathematical reason is that strain is not a property of an isolated point of material, it is also property of the the material close to that point.

And if we are talking about quantum mechanics, a question about "a particle in the center of the wheel" is rather meaningless.