Wooden sphere rolling on a double metal track

[erobz]

Would this result fit?

If you simplified correctly, the supporting theory looks ok to me.

 

[Hak]

If you simplified correctly, the supporting theory looks ok to me.

So, the other two equations I had set up are correct?

 

[erobz]

Sum of forces along track: $$mg sin \alpha - 2 F_{friction} = ma_{C}$$

Looks correct

Sum of forces perpendicular to track: $$2N - mg cos \alpha = 0$$

Was not correct, but you understand that now.

Sum of torques about the center of mass: $$2 F_{friction}h = I_C \frac{a_{C}}{h}$$, or, alternatively, about the axis of rotation: $$mghsin \alpha = (I_C + mh^2) \frac{a_{C}}{h}$$, where $\frac{a_{C}}{h}$ is the angular acceleration, obtained by deriving $\omega$.

Both seem ok to me.

 

[erobz]

Latex tip: the trig functions are:

##\sin \alpha , \cos \alpha ##

$\sin \alpha, \cos \alpha$

 

[Hak]

According to first equation, doubt had occurred to me that I should advance the same reasoning as for the normal force $N$, but it turns out I had actually conceived the equation correctly.

 

[erobz]

According to first equation, doubt had occurred to me that I should advance the same reasoning as for the normal force $N$, but it turns out I had actually conceived the equation correctly.

I don't know if this is what you mean, but when you are looking from the side ( summing forces in the $x$ direction ) you are seeing the "true length" of the frictional force vector.

 

[Hak]

I don't know if this is what you mean, but when you are looking from the side ( summing forces in the $x$ direction ) you are seeing the "true length" of the frictional force vector.

Yes, I understood all that. It was only a passing doubt....

 

[erobz]

Yes, I understood all that. It was only a passing doubt....

Ok, all good then?

 

[Hak]

Ok, all good then?

Yes, now I am trying to understand more about the possibility of "rotational skid." I don't really know where to start...

 

[kuruman]

There must be a rotational skid, like a ball spinning on the spot on a horizontal surface.

Isn't it true, though, that if the contact is truly at a single point, i.e. the surface is completely flat, the contact force of friction can exert no torque that will affect the skid?

I think in this case one can see what's going on if one considers the torques about the center of the sphere point O. The weight generates no torque. By symmetry, the frictional force at point $A$ has the same magnitude and direction as at point $B$. This means that friction generates no net torque about the center of the sphere either. The only non-zero torque is generated by gravity and that is about line segment $AB$. Note that the ends of $AB$ are prevented from rotating about a vertical axis by the static friction $A$ and $B$. If that "prevention" can no longer hold, the sphere will slide.

 

[haruspex]

if the contact is truly at a single point, i.e. the surface is completely flat, the contact force of friction can exert no torque that will affect the skid?

The principle I apply to idealisations is that they are the limiting cases of realistic scenarios. It would be necessary to solve the latter to a first approximation then take the limit.

A possible model is to suppose the sphere "rolls" on two narrow conical sections, so that each makes instantaneous contact along a line length w. Because of the different radii across the sections, friction on them acts down the slope where the radius is large and up the slope where it is small. Assuming the normal force is evenly distributed along the line, the net frictional force depends on the relative lengths of those two portions of w. If these lengths are x at the large radius end, w-x at the small radius end, the net frictional force on each is $\mu_k N(1-2x/w)$ up the slope.
We could define "rolling contact" in this model as x>0.

the frictional force at point A has the same magnitude and direction as at point B. This means that friction generates no net torque about the center of the sphere either.

The two frictional forces have the same magnitude and direction but O does not lie in the plane containing their lines of action. Consequently they do have a net torque about O.

 

[Hak]

The principle I apply to idealisations is that they are the limiting cases of realistic scenarios. It would be necessary to solve the latter to a first approximation then take the limit.

A possible model is to suppose the sphere "rolls" on two narrow conical sections, so that each makes instantaneous contact along a line length w. Because of the different radii across the sections, friction on them acts down the slope where the radius is large and up the slope where it is small. Assuming the normal force is evenly distributed along the line, the net frictional force depends on the relative lengths of those two portions of w. If these lengths are x at the large radius end, w-x at the small radius end, the net frictional force on each is $\mu_k N(1-2x/w)$ up the slope.
We could define "rolling contact" in this model as x>0.

The two frictional forces have the same magnitude and direction but O does not lie in the plane containing their lines of action. Consequently they do have a net torque about O.

Really interesting...

 

[kuruman]

The two frictional forces have the same magnitude and direction but O does not lie in the plane containing their lines of action. Consequently they do have a net torque about O.

Yes, of course. That torque is along $AB$ as it should be. I don't know what I was thinking.

 

[Hak]

A friend of mine solved point 2 this way, but I cannot understand where the $\sqrt{3}$ factor comes from.

"According to the condition of the problem, the ball rolls without slipping, therefore, the speeds of those points of the ball that at a given moment of time touch the metal at AB are equal to zero. Considering the ball to be an absolutely rigid body (that is, the distance between any two points of the ball is unchanged), we conclude that at a given moment in time all points of the ball lying on the segment AB are motionless. And this means that at each moment of time the movement of the ball is a rotation about the axis AB. (It is clear that points A and B - the points where the ball touches the metal - are moving with the speed $v$.)

The instantaneous velocity of any point of the ball is $\omega \rho$, where $\omega$ is the angular velocity of rotation, $\rho$ is the distance from the point to the axis AB. The speed of the center of the ball (point O in the figure) is equal to $v$ ; the distance from point O to axis AB is $$\rho_0 = |OC| = \frac{\sqrt{3}}{2} R.$$ Hence,
$$\omega = \frac{v}{\rho_0} = \frac{2v}{\sqrt{3}R}.$$

It is clear that the points of the ball most distant from the axis AB have the maximum speed. From geometric considerations it is clear that at any moment there is only one point that is maximally distant from the axis - in the figure this is point Q. The distance from point Q to the axis of rotation is $$\rho' = \rho_0 + R = R \bigg(1 + \frac{\sqrt{3}}{2}\bigg)$$, and speed of Q is:

$$v_{max} = \omega \rho' = v \bigg(1 + \frac{2}{\sqrt{3}} \bigg)$$".

Can anyone help me understand? Thanks.

 

[Hak]

A friend of mine solved point 2 this way, but I cannot understand where the $\sqrt{3}$ factor comes from.

"According to the condition of the problem, the ball rolls without slipping, therefore, the speeds of those points of the ball that at a given moment of time touch the metal at AB are equal to zero. Considering the ball to be an absolutely rigid body (that is, the distance between any two points of the ball is unchanged), we conclude that at a given moment in time all points of the ball lying on the segment AB are motionless. And this means that at each moment of time the movement of the ball is a rotation about the axis AB. (It is clear that points A and B - the points where the ball touches the metal - are moving with the speed $v$.)

The instantaneous velocity of any point of the ball is $\omega \rho$, where $\omega$ is the angular velocity of rotation, $\rho$ is the distance from the point to the axis AB. The speed of the center of the ball (point O in the figure) is equal to $v$ ; the distance from point O to axis AB is $$\rho_0 = |OC| = \frac{\sqrt{3}}{2} R.$$ Hence,
$$\omega = \frac{v}{\rho_0} = \frac{2v}{\sqrt{3}R}.$$

It is clear that the points of the ball most distant from the axis AB have the maximum speed. From geometric considerations it is clear that at any moment there is only one point that is maximally distant from the axis - in the figure this is point Q. The distance from point Q to the axis of rotation is $$\rho' = \rho_0 + R = R \bigg(1 + \frac{\sqrt{3}}{2}\bigg)$$, and speed of Q is:

$$v_{max} = \omega \rho' = v \bigg(1 + \frac{2}{\sqrt{3}} \bigg)$$".

Can anyone help me understand? Thanks.

?

 

[haruspex]

$\sqrt 3$ is wrong, as you suspected. Your friend seems to think AB has length R.

 

[Hak]

$\sqrt 3$ is wrong, as you suspected. Your friend seems to think AB has length R.

I thought so too, but he claims that the $\angle AOB$ angle measures 60° since $A$ and $B$ are not on the same level. I struggle to understand...

 

[erobz]

I struggle to understand...

Is there something you are not understanding about the trig that allows you to have doubt about it?

 

[haruspex]

I thought so too, but he claims that the $\angle AOB$ angle measures 60° since $A$ and $B$ are not on the same level. I struggle to understand...

They are on the same level as each other, and obviously not on the same level as O. Maybe he means AB is not in the same vertical plane as O, so the height of O above AB is less than $R/\sqrt 2$. But that height is not relevant here, and depends on $\alpha$.

 

[Hak]

They are on the same level as each other, and obviously not on the same level as O. Maybe he means AB is not in the same vertical plane as O, so the height of O above AB is less than $R/\sqrt 2$. But that height is not relevant here, and depends on $\alpha$.

At this point, I am no longer sure about this. Interestingly, this document from an American competition, in problem 2, reports the result with $\sqrt{3}$, without proof. I can't understand...

 

[Hak]

Is there something you are not understanding about the trig that allows you to have doubt about it?

No, I cannot understand why A and B should be considered not on the same level and why the $\angle OAB$ angle should measure 60° instead of 90°...

 

[haruspex]

At this point, I am no longer sure about this. Interestingly, this document from an American competition, in problem 2, reports the result with $\sqrt{3}$, without proof. I can't understand...

In the diagrams you attached, it is not entirely clear where points P and Q are. Are they (1) respectively the lowest and highest points of the sphere, or (2) the point nearest where the plates meet and the point furthest from it?

The right-hand diagram appears to show the plates end-on, so is looking up the slope. That means (2) is the correct interpretation. This makes the answer 'Q' correct. The text is wrong to refer to them as (1).
Under interpretation (2), the speed of Q is v multiplied by the ratio of distances from Q to AB and O to AB, namely, $(1+1/\sqrt 2)/(1/\sqrt 2)=1+\sqrt 2$.

Under interpretation (1), Q is not the fastest point, and its speed is indeterminate because we do not know the angle of the slope.

My guess is that your friend has taken the answer at the aapt site on trust and has fabricated an argument to justify it.
As to how the aapt answer came to be wrong, perhaps it was a rehash of an old question and they forgot to change everything consistently. It happens.

It sometimes helps to consider an extreme case. Suppose the plates are vertical, which they could be, as far as we know. The point at the top of the sphere is $R\sqrt{3/2}$ from AB, so moves at speed $v\sqrt{3}$. Meanwhile, the point furthest from the plate join still moves at $v(1+\sqrt 2)$.

 

[Hak]

In the diagrams you attached, it is not entirely clear where points P and Q are. Are they (1) respectively the lowest and highest points of the sphere, or (2) the point nearest where the plates meet and the point furthest from it?

The right-hand diagram appears to show the plates end-on, so is looking up the slope. That means (2) is the correct interpretation. This makes the answer 'Q' correct. The text is wrong to refer to them as (1).
Under interpretation (2), the speed of Q is v multiplied by the ratio of distances from Q to AB and O to AB, namely, $(1+1/\sqrt 2)/(1/\sqrt 2)=1+\sqrt 2$.

Under interpretation (1), Q is not the fastest point, and its speed is indeterminate because we do not know the angle of the slope.

My guess is that your friend has taken the answer at the aapt site on trust and has fabricated an argument to justify it.
As to how the aapt answer came to be wrong, perhaps it was a rehash of an old question and they forgot to change everything consistently. It happens.

It sometimes helps to consider an extreme case. Suppose the plates are vertical, which they could be, as far as we know. The point at the top of the sphere is $R\sqrt{3/2}$ from AB, so moves at speed $v\sqrt{3}$. Meanwhile, the point furthest from the plate join still moves at $v(1+\sqrt 2)$.

Thank you so much, but... could you explain me the last paragraph? I don't think I understand this correctly, including the calculations...

 

[Hak]

My friend claims to refer to the following link: https://earthz.ru/solves/Zadacha-po-fizike-6534.

 

[erobz]

My friend claims to refer to the following link: https://earthz.ru/solves/Zadacha-po-fizike-6534.

That is a different problem... The left rail is not tangent to the sphere at the point of contact.

 

[haruspex]

Thank you so much, but... could you explain me the last paragraph? I don't think I understand this correctly, including the calculations...

The point O is $R/\sqrt 2$ from AB.
In the vertical plates model, that is a horizontal distance. The top of the sphere is R above O, so distance $R\sqrt{(1/\sqrt 2)^2+1^2}=R\sqrt{3/2}$ from AB.
Being $\frac{R\sqrt{3/2}}{R/\sqrt 2}=\sqrt 3$ times further away from AB, it moves that much times faster than O.

 

[haruspex]

That is a different problem... The left rail is not tangent to the sphere at the point of contact.

Looks suspiciously like a case of sloppy plagiarism by an aapt contributor.
I have emailed aapt.

 

[Hak]

Thank you very much to everyone. You have all been very kind and professional. A wonderful disquisition.

 

[Hak]

A friend of mine, regarding point 3, set up a different procedure.

"The weight force acting on the center of mass $O$ forms the angle $\alpha$ with its projection on the plane $AOB$, $AB$ being the instantaneous axis of rotation. Therefore the component of the weight force acting along the inclined plane, rib of the guide, results $F_P = Mg \sin \alpha$. The other component $F_H = Mg \cos \alpha$ has two components, one on $AO$ and one on $BO$, perpendicular to the tracks of $A$ and $B$ on the guides. Their value is $N = Mg \frac{\sqrt{2}}{2} \cos \alpha$. We determine the minimum value of $\mu$, i.e., $\mu_m$, by imposing constancy of $v$, i.e., that twice the friction force $F_{friction}$ (because it develops in $A$ and $B$) is equal to the force $F_P$. Meanwhile, it is $F_{friction} \le \mu N$, whence $\mu \ge \frac{F_{friction}}{N}$. Since it must be, for the constancy of $v$, $2F_{friction} = 2 Mg \sin \alpha$, it follows that $\mu$ must be at least equal to $\mu = \sqrt{2} \tan \alpha$".

All this is different from the procedure carried out together a few days ago. What is wrong with this?

 

[erobz]

A friend of mine, regarding point 3, set up a different procedure.

"The weight force acting on the center of mass $O$ forms the angle $\alpha$ with its projection on the plane $AOB$, $AB$ being the instantaneous axis of rotation. Therefore the component of the weight force acting along the inclined plane, rib of the guide, results $F_P = Mg \sin \alpha$. The other component $F_H = Mg \cos \alpha$ has two components, one on $AO$ and one on $BO$, perpendicular to the tracks of $A$ and $B$ on the guides. Their value is $N = Mg \frac{\sqrt{2}}{2} \cos \alpha$. We determine the minimum value of $\mu$, i.e., $\mu_m$, by imposing constancy of $v$, i.e., that twice the friction force $F_{friction}$ (because it develops in $A$ and $B$) is equal to the force $F_P$. Meanwhile, it is $F_{friction} \le \mu N$, whence $\mu \ge \frac{F_{friction}}{N}$. Since it must be, for the constancy of $v$, $2F_{friction} = 2 Mg \sin \alpha$, it follows that $\mu$ must be at least equal to $\mu = \sqrt{2} \tan \alpha$".

All this is different from the procedure carried out together a few days ago. What is wrong with this?

How many friends do you have? Do any of them agree with the method found in this thread?

 

[Hak]

How many friends do you have? Do any of them agree with the method found in this thread?

The "friends" you speak of are only two, with the term "friends" not so appropriate. You are absolutely right, but could you tell me what is wrong with my "friend" 's reasoning? As I understand it, he assumed $v$ constant, while we considered some acceleration of the center of mass other than 0. Perhaps he considered "rolling without sliding" simply as "pure rolling"? Could you provide some guidance on this?

 

[erobz]

There is no reason why you shouldn't be able to reconcile these errors on your own at this point if you insist on checking everyone else's work. There are obvious errors, but I want you to find them with confidence, and report back your findings if it is that important to you.

 

[Hak]

There is no reason why you shouldn't be able to reconcile these errors on your own at this point if you insist on checking everyone else's work. There are obvious errors, but I want you to find them with confidence, and report back your findings if it is that important to you.

I found two errors in the resolution I posted previously.

1. He says that "twice the friction force (because it develops in $A$ and $B$) is equal to the force $F_P$", but then inexplicably concludes that $2F_{friction} = 2Mg \sin \alpha$, considering twice the tangential component of the weight force. Therefore, his result, according to his reasoning, should be half of what he obtained, i.e., $\mu = \frac{\sqrt{2}}{2} \tan \alpha$.

2. The one just quoted is the result that would be obtained by setting the acceleration of the center of mass equal to 0. I do not understand, however, why should $v$ be considered constant . There is a net torque produced by the forces (obviously, different depending on the axis chosen), which produces angular acceleration, so the center of mass possesses its own acceleration .

Are these objections correct? In what do I mistake?

 

[erobz]

I found two errors in the resolution I posted previously.

1. He says that "twice the friction force (because it develops in $A$ and $B$) is equal to the force $F_P$", but then inexplicably concludes that $2F_{friction} = 2Mg \sin \alpha$, considering twice the tangential component of the weight force. Therefore, his result, according to his reasoning, should be half of what he obtained, i.e., $\mu = \frac{\sqrt{2}}{2} \tan \alpha$.

2. The one just quoted is the result that would be obtained by setting the acceleration of the center of mass equal to 0. I do not understand, however, why should $v$ be considered constant . There is a net torque produced by the forces (obviously, different depending on the axis chosen), which produces angular acceleration, so the center of mass possesses its own acceleration .

Are these objections correct? In what do I mistake?

Your objections are correct, however you can immediately ignore explaining the silly math in you first objection, because your second objection outweighs it. The center of mass is accelerating as you point out.

There is another error yet.

 

[Hak]

The other component $F_H = Mg \cos \alpha$ has two components, one on $AO$ and one on $BO$, perpendicular to the tracks of $A$ and $B$ on the guides.

Would that be the other mistake? I could not understand the meaning of his statement, could you explain it to me?

 

[erobz]

Would that be the other mistake? I could not understand the meaning of his statement, could you explain it to me?

they appear to be talking about the normal forces being components of the component of weight resolved perpendicular to the incline. It’s not something one should try to understand… they are saying ∑ forces perpendicular to plane…not well.

 

[Hak]

they appear to be talking about the normal forces being components of the component of weight resolved perpendicular to the incline. It’s not something one should try to understand… they are saying ∑ forces perpendicular to plane…not well.

Maybe I expressed myself wrongly. I didn't mean that "I can't understand it," I meant that "I couldn't understand the meaning." In other words, I did not understand why he conceived such an argument, which is why I asked you... So, are these the main errors?

 

[Hak]

they are saying ∑ forces perpendicular to plane…not well.

Could you explain more about this statement?

 

[erobz]

Could you explain more about this statement?

When they say there are two components of the resolved weight force, they mean there are two forces that oppose the resolved weight force.

They are not correctly stating what they mean( and appear to understand )based on the result they got for $N$. That error is forgivable in comparison to the others, but still should be addressed.

 

[haruspex]

Maybe I expressed myself wrongly. I didn't mean that "I can't understand it," I meant that "I couldn't understand the meaning." In other words, I did not understand why he conceived such an argument, which is why I asked you... So, are these the main errors?

It's inaccurate to say $F_H$ has two components. Rather, it can be resolved into two components, each normal to a plate, and normal to each other. Since there is no acceleration normal to a plate, these must equal the normal forces from the plates.

As discussed, the constant velocity argument is nonsense.

 

[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.