Wooden sphere rolling on a double metal track

[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.