How Can I Calculate the Angular Speed of a Ring Rolling Without Slipping?

[Vibhor]

Homework Statement

Homework 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

 

[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 ∅?

 

[Vibhor]

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.

Sorry . I do not understand what is meant by "lies down along the arc ".

How is that change in angle related to the change in angle of orientation of the larger circle?

What do you mean by "angle of orientation of the larger circle " ?

 

[TSny]

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 " ?

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?

 

[Vibhor]

Between the start and this moment, find the angle that the larger circle has rotated about its center.

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

 

[TSny]

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

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?

 

[Vibhor]

But what about the angle that the large ring has rotated about its center?

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

 

[TSny]

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

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

 

[Vibhor]

The angle of rotation of the large ring about its center is related to the change in orientation of the orange tangent.

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 .

Please elaborate .

 

[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.

 

[Vibhor]

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.
View attachment 204048

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 .

 

[TSny]

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 .

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?

 

[Vibhor]

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

 

[TSny]

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

Yes

 

[Vibhor]

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

 

[TSny]

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

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.

 

[Vibhor]

When the finger rotates through angle $\theta$, how much does the large ring rotate about its center?

$\theta$ - $\phi$

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

 

[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.]

 

[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) ??

 

[TSny]

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) ??

I believe that's right.

 

[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 .

 

[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

 

[Vibhor]

The point of the ring that is in contact with the finger is instantaneously at rest.

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

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).

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

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

 

[TSny]

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

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.

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

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?

 

[Vibhor]

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?

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 .

 

[Vibhor]

Is it exactly like a hoola hoop ??

 

[TSny]

In the left diagram above, you can imagine the grey circle to be a solid disk that doesn't move while the large ring rolls without slipping on the disk. The point of contact is the IAOR for the large ring.

In the right diagram I have added a blue dot that represents the finger. The finger can take the place of the solid disk as long as the finger moves along a circle (circumference of the disk). The large ring doesn't know the difference.

At least that's how I see it.

 

[TSny]

Is it exactly like a hoola hoop ??

Probably not, but I don't know what differences you are alluding to.

 

[Vibhor]

In the left diagram above, you can imagine the grey circle to be a solid disk that doesn't move while the large ring rolls without slipping on the disk. The point of contact is the IAOR for the large ring.

In the right diagram I have added a blue dot that represents the finger. The finger can take the place of the solid disk as long as the finger moves along a circle (circumference of the disk). The large ring doesn't know the difference.

At least that's how I see it.

Thank you very much

 

[zwierz]

It seems that it is supposed that the ring remains in a horizontal plane. Strange

 

[TSny]

It seems that it is supposed that the ring remains in a horizontal plane. Strange

Yes, strange. I don't see how it could move in a truly horizontal plane.

 

[zwierz]

it would be a good task to write honest equations of stationary motion that is when an angle of ring's incline is constant and the speed of ring's center is constant and ring's center moves along a horizontal cirle