Wooden sphere rolling on a double metal track

[Hak]

Thank you so much!

 

[haruspex]

Postscript.

I notified AAPT of the issue …

A student has contacted me regarding this document (
https://www.aapt.org/Common/upload/2020-Fma-Exam-A_solutions.pdf). He is baffled by a formula quoted in the solution. I share his bafflement.

First, there seems to be some confusion regarding point Q in the diagram. The text defines it as the "highest point" of the sphere. The diagram implies it is not that; rather, it is the point furthest from the junction of the plates. Indeed, the highest point is not the fastest moving, whereas the point furthest from the junction of the plates is.

But what bothers the student is the formula quoted, v (2+sqrt(3))/sqrt(3). It seems clear it should be v(1+sqrt(2)).

… and got an appreciative response:

Thank you for the correction, we completely agree! 2020 was an unusual year, and the problems and solutions that year were much less polished due to the pandemic -- we didn't even bother grading the exams. We'll update the file on the website later this year, and we'd appreciate hearing about any other issues as well.

 

[Hak]

Postscript.

I notified AAPT of the issue …

A student has contacted me regarding this document (
https://www.aapt.org/Common/upload/2020-Fma-Exam-A_solutions.pdf). He is baffled by a formula quoted in the solution. I share his bafflement.

First, there seems to be some confusion regarding point Q in the diagram. The text defines it as the "highest point" of the sphere. The diagram implies it is not that; rather, it is the point furthest from the junction of the plates. Indeed, the highest point is not the fastest moving, whereas the point furthest from the junction of the plates is.

But what bothers the student is the formula quoted, v (2+sqrt(3))/sqrt(3). It seems clear it should be v(1+sqrt(2)).

… and got an appreciative response:

Thank you for the correction, we completely agree! 2020 was an unusual year, and the problems and solutions that year were much less polished due to the pandemic -- we didn't even bother grading the exams. We'll update the file on the website later this year, and we'd appreciate hearing about any other issues as well.

Thank you so much, @haruspex!

 

[etrnlccl]

Look at the figure below. The axis of rotation is the red dotted line AB. All points on the sphere rotate about that axis with a common angular velocity $\mathbf{\omega}$. Their linear velocity relative to AB, which is at rest relative to the track, is $\mathbf{v}=\mathbf{\omega}\times \mathbf{r}.$ Here, $\mathbf{r}$ is the position vector of a point on the sphere perpendicular to the axis AB. How do you find the point at which $\mathbf{v}$ has the largest magnitude?

View attachment 331652

Hi, this problem is pretty interesting and it does not seem simple to understand. Particularly, I think I disagree with the notion that the line joining the points A and B is the rotation axis. I think the rotation axis is the line passing through the center O of the sphere and parallel to the AB segment. This is because otherwise the sphere would not simply roll (without slipping) through the plates. In order to see it, remove the plates from your picture and imagine the sphere rotating about the AB axis: its movement would not correspond to a rolling (without slipping) through the plates! In fact, the center O of the sphere would go below the AB axis during part of such a movement. Therefore, I think either one of the following options would be true: the rotation axis is still the diameter parallel to the AB axis, or the movement is not simply rolling without slipping and we should have a more complicated dynamics (maybe this is what haruspex is also finding difficult to figure out regarding this problem, see this post: https://www.physicsforums.com/threa...-on-a-double-metal-track.1055489/post-6930364 ).

 

[haruspex]

Hi, this problem is pretty interesting and it does not seem simple to understand. Particularly, I think I disagree with the notion that the line joining the points A and B is the rotation axis. I think the rotation axis is the line passing through the center O of the sphere and parallel to the AB segment. This is because otherwise the sphere would not simply roll (without slipping) through the plates. In order to see it, remove the plates from your picture and imagine the sphere rotating about the AB axis: its movement would not correspond to a rolling (without slipping) through the plates! In fact, the center O of the sphere would go below the AB axis during part of such a movement. Therefore, I think either one of the following options would be true: the rotation axis is still the diameter parallel to the AB axis, or the movement is not simply rolling without slipping and we should have a more complicated dynamics (maybe this is what haruspex is also finding difficult to figure out regarding this problem, see this post: https://www.physicsforums.com/threa...-on-a-double-metal-track.1055489/post-6930364 ).

You are confusing instantaneous centre of rotation with a constant centre of rotation.
By your argument, a wheel rolling along the road can't be rotating about the point of contact because it would penetrate the road. But the point of contact must be the instantaneous centre of rotation because that is the only part of the wheel that is instantaneously stationary.

 

[etrnlccl]

I am not familiar with this concept of instantaneous center of rotation. I think a wheel rolling along the road is clearly rotating about its diameter. In fact, if it was rotating about the point of contact it would not simply roll along the road.

 

[kuruman]

I am sure you have watched a wheel spin in place about a fixed axis. You should know that each point on the rim has velocity directed tangent to the rim and linear speed $v=\omega~R$ where $\omega$ is the angular speed. That's the background.

You should also be familiar with the following situation. You are riding a bicycle moving at 15 mph on a road. You look over the handle bar down at the front wheel. What do you see?

You see the axle of the wheel at rest relative to you while the wheel is spinning about it. Based on what was said in the first paragraph, you also see the top of the wheel moving forward with speed $v=\omega~R=15~$mph and the bottom of the wheel where it touches the ground moving backward with speed $v=\omega~R=15~$mph. Furthermore, you see the road surface moving backwards at 15 mph.

This is a relative velocity question. If the road is moving backwards at 15 mph and the point of contact is also moving backwards at 15 mph, what is the velocity of the road relative to point on the rim that is (Instantaneously) in contact with the road?

 

[haruspex]

I am not familiar with this concept of instantaneous center of rotation. I think a wheel rolling along the road is clearly rotating about its diameter. In fact, if it was rotating about the point of contact it would not simply roll along the road.

If the wheel is not skidding then the point of contact is instantaneously still. A point distance y above it is moving forward at speed $\omega y$, a point distance y below it is moving backward at speed $\omega y$, a point distance y in front is moving down at speed $\omega y$, a point distance y behind it is moving up at speed $\omega y$.
It is the instantaneous centre of rotation.

If that bothers you, you will find it easier to think of the motion as the sum of the linear motion of the ball's centre and a rotation about that. The algebra works either way.