# Pure Rolling

1. Feb 5, 2009

### ritwik06

1. The problem statement, all variables and given/known data

There are two spheres of equal radii. The bottom sphere is fixed in space. The other sphere purely(i.e. without slipping) rolls over the fixed sphere with constant angular velocity $$\omega 1$$ about its axis of rotation. Find the angular velocity of the axis of the rotating sphere about the center of the fixed sphere $$\omega 2$$ . (Assume a gravity free environment)
http://img201.imageshack.us/img201/6729/sphrert9.jpg [Broken]
3. The attempt at a solution

The problem is that I get two different and inconsistent answers from two different methods.

Method 1:
I follow a basic approach:
The sphere has a constant angular velocity about its own axis. The linear speed of the center of mass will be R$$\omega1$$. The center of mass will move in a circle of 2R. Its velocity = R$$\omega 1$$
The angular velocity of the center of mass of the rotating sphere(wrt to centre of fixed sphere) is =(R$$\omega1$$)/2R
$$\omega 2$$=$$\omega 1$$*0.5

Method 2:
As the radii of the two spheres is the same. When the rolling sphere rotates through an angle of 2$$\pi$$ along its axis, it would have reached its original position. It implies that the centre of mass of the rotating sphere has rotated through an angle 2$$\pi$$ about the centre of the fixed sphere (in the same time interval)

2$$\frac{\pi}{\omega 1}$$=2$$\frac{\pi}{\omega 2}$$

$$\omega2$$=$$\omega1$$

Now which of the results is plausible. Both seem right to me. Please help!!!

Last edited by a moderator: May 4, 2017
2. Feb 5, 2009

### tiny-tim

Hi ritwik06!
No, it will only have gone half-way round the fixed sphere.

(do it with two bicycle wheels instead of spheres, and then gradually tilt the movable wheel until it lies almost flat on top of the fixed wheel, and you see how to deform a 720º rotation into a 0º rotation … something you can't do with a 360º rotation! )

3. Feb 6, 2009

### ritwik06

I dont see through it. I used a pair of chalk pieces instead. I marked the initial contact point of the two chalks. Then I rolled one of them over the fixed one. I ended up with only one rotation when the marked point met each other.

Please explain this to me. I will explain all my problems through different diagrams.
I am assuming that the angular velocity of the moving circle(about its centre) is same in both cases.
First I consider a flat surface over which 2D circle is rolling without slipping.

http://img8.imageshack.us/img8/1194/flatsurfacedz0.png [Broken]
Suppose the circle rotates through an angle $$\theta$$.
The contact point as well as the centre suffer a linear displacement $$R\theta$$.

Now I consider a circular surface over which another 2D circle rotates without slipping(Radius of both is same)
http://img87.imageshack.us/img87/1227/curvedsurfacezg2.jpg [Broken]

Now the rotating circle rotates through an angle $$\theta$$ about its axis. Then the point of contact will move along the circumference of the fixed sphere a distance of $$R\theta$$. I hope you agree with me till here.

Now if the contact point travels a distance of $$R\theta$$, then the line joining the centres of the two circles will sweep an angle $$\theta$$. Wont it????
I am confused. Please help me (if possible with paint diagrams as in my case). I shall be very grateful!!

Last edited by a moderator: May 4, 2017
4. Feb 6, 2009

### Staff: Mentor

Better double check that. (Use two quarters, slowly rotating one about the other.)

If the disk merely rolled along a straight path, then when it rolled 1/4 of its circumference it would have made 1/4 of a revolution. But here it's rolling along a curved path--when it rolls 1/4 of its circumference, it's actually made 1/2 of a revolution.

(Your "Method 1" in your first post was correct.)

5. Feb 6, 2009

### ritwik06

Ok, But I am not convinced totally. Can you please give me a mathematical proof? Or tell me if I can get it from somewhere? I shall be very grateful.

One more question: We can consider each infinitesimally small part of the circumference as a straight line. So what difference does a curved path make?
Thanks a lot for the help!

Last edited: Feb 6, 2009
6. Feb 6, 2009

### Staff: Mentor

Rather than get hung up on a proof, try this: Instead of the outer disk rolling around the inner disk, have it just slide around. (In other words, have the same point of the moving disk making contact.) Would you not agree that in traversing the circumference, that the disk makes a complete rotation even without the rolling?

Now add in the rolling motion and the amount that the disk rotates is even greater.
Sure, each small part of the circumference can be considered a different straight line segment--but that's not the same as being on a single straight line path.

7. Feb 6, 2009

### tiny-tim

ok …

roll the moving sphere halfway round the fixed sphere …

the point of contact has moved the same distance on both spheres …

let's label the moving sphere like a clock, with 12 o'clock at the top …

if it starts at the top, with its 6 o'clock mark in contact with the fixed sphere, then the new point of contact will be halfway round, at 12 o'clock on the moving sphere …

but that means that 12 o'clock must be at the top …

so the moving sphere has gone halfway round the fixed sphere, but has rotated a whole revolution.