# Homework Help: Ring rolling without slipping

1. May 22, 2017

### Vibhor

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

2. Relevant equations

3. The attempt at a solution

These are my two observations for this problem .

1) Center of mass of the ring moves in a circle of radius (R-r) about point O . O is the center of the smaller circle in figure 2 .

2) From the geometry , the angular speed of the CM of the ring is equal to the angular speed with which the finger is rotating i.e ω0 .

Now I am not sure how do I calculate the angular speed of the ring about its CM ( which coincides with its center ) .

I would appreciate if someone could help me with the problem .

Thanks

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2. May 22, 2017

### TSny

Your observations look good. One way to get to the angular speed of the ring is to consider the geometry shown below

The green arcs have the same length. The red and orange lines are tangents. As the arc of the larger circle lies down along the arc of the smaller circle, think about the change in the angle of the orange tangent. How is that change in angle related to the change in angle of orientation of the larger circle? Can you express this change in terms of the angles θ and ∅?

3. May 22, 2017

### Vibhor

Sorry . I do not understand what is meant by "lies down along the arc ".
What do you mean by "angle of orientation of the larger circle " ?

4. May 22, 2017

### TSny

Sorry, I wasn't clear.

Let the larger ring roll without slipping on the smaller circle. During the rolling, each point of the green arc on the larger circle will make momentary contact with a point of the green arc on the smaller circle until the orange and red tangent lines coincide. Between the start and this moment, find the angle that the larger circle has rotated about its center. Hint: How does this angle compare to the initial angle between the orange and red tangent lines?

5. May 22, 2017

### Vibhor

$\phi = \theta \frac{r}{R}$

6. May 22, 2017

### TSny

That's the correct relation between the angles $\phi$ and $\theta$. But what about the angle that the large ring has rotated about its center?

7. May 22, 2017

### Vibhor

Isn't $\phi$ the angle that the larger ring has rotated about its center ?

8. May 22, 2017

### TSny

No. The angle of rotation of the large ring about its center is related to the change in orientation of the orange tangent.

9. May 22, 2017

### Vibhor

Sorry , but I think I am not understanding this part .

When the finger is rotated by angle $\theta$ , doesn't the ring rotate about its center by angle $\phi$ ?

How is angle rotated by ring related to change in orientation of orange tangent ?

I am quite confused and finding it hard to visualize .

10. May 22, 2017

### TSny

OK, forget the specific problem for a moment. The figure below shows some circle with a tangent line attached at the blue dot. The circle starts in the initial position and then it is translated and rotated to the final position shown. How much has the circle rotated about its center? Assume that the amount of rotation is less than a full rotation.

11. May 22, 2017

### Vibhor

This is how I think .

The angle initially made by tangent with the vertical is 30 deg , so angle made by normal is 30 deg with the horizontal . Finally normal makes angle 70 deg with horizontal .Angle rotated by normal is 40 deg which is equal to the angle rotated by the circle . Hence angle rotated by circle is 40 deg .

12. May 22, 2017

### TSny

Yes. So going back to the original problem.

The left figure is as before except I added the blue dots. Point $p$ is the initial point of contact. Later, point $A$ of the large ring will make contact with point $a$ of the small circle. The right figure shows the orientation of the tangent line of the large ring for the initial and final configurations. I have drawn them side by side rather than overlapping them. Can you identify the angles represented by question marks in terms of $\theta$ and $\phi$ of the left figure?

13. May 22, 2017

### Vibhor

Left ? = $\phi$ and Right ? = $\theta$

14. May 22, 2017

### TSny

Yes

15. May 22, 2017

### Vibhor

So angular speed of ring about its center is equal to the angular speed of the finger I.e ω0 ??

16. May 22, 2017

### TSny

No. When the finger rotates through angle $\theta$, how much does the large ring rotate about its center? Use the same reasoning that led to the 40 degree answer before.

17. May 22, 2017

### Vibhor

$\theta$ - $\phi$

So when finger makes one full rotation $\theta$ = 360 deg , but what is $\phi$ ?

18. May 22, 2017

### TSny

Yes, the amount of rotation of the large ring is $\theta - \phi$.

Can you express this solely in terms of $\theta$? (See your post #5)

To get the rate of rotation of the large ring, divide the angle of rotation of the large ring by time.
[Or, once you express the angle of rotation of the large ring solely in terms of the angle of rotation of the finger, $\theta$, you can then see how much the large ring rotates when $\theta$ = 360 degrees.]

19. May 22, 2017

### Vibhor

OK . So the angular speed of the larger ring about its center is $\omega _0\frac{R-r}{R}$ ??

And the linear speed of CM of ring is $\omega _0(R-r)$ ( considering the angular speed of the CM about the center of small circle ) ??

Option A) ??

20. May 22, 2017

### TSny

I believe that's right.

21. May 22, 2017

### Vibhor

Ok .

Do you think the point of contact of ring with the finger could be considered the instantaneous AOR of the ring ?

I am asking this because , this question came in a prestigious national exam and some of the best minds have been assuming the Contact point as the IAOR and getting option A) .

I don't think it is correct , as the contact point itself is moving . IAOR should be at rest momentarily .

22. May 22, 2017

### TSny

I think they are correct that the instantaneous point of contact of the ring with the finger is the location of the IAOR of the ring. The point of the ring that is in contact with the finger is instantaneously at rest.

When a wheel rolls in a straight line along the ground with no slipping, then the IAOR of the wheel passes through the point of contact of the wheel with the ground (even though the point of contact is moving).

Using the IAOR makes the problem much easier. I didn't think of that approach

23. May 22, 2017

### Vibhor

I don't think so . Isn't the finger rotating ?

There is a difference in the above example and the current problem .

In our problem , the contact point remains same and is in motion

24. May 22, 2017

### TSny

The finger moves in a circle, but it does not rotate. That is, the fingernail will always point in the same direction. It's like a "hoola-hoop" where your waist takes the place of the finger.

If the same point of the ring is always in contact with the finger, wouldn't that imply that the ring is slipping on the finger?

25. May 23, 2017

### Vibhor

So the Contact point continuously change just like how a wheel rolls without slipping .But in that case the underlying surface is at rest ??

Somehow I am finding it hard to visualize how the Contact point of the ring is at rest momentarily when both the finger and the ring are in motion .