Wooden sphere rolling on a double metal track

[Hak]

I don't know how to do it. You said that the vertical components of the normal forces must balance the perpendicular weight $mg cos \alpha$. So, what is wrong in my equation should be the expression of $N$, right? Would you have more in-depth hints on how to write it?

 

[erobz]

I don't know how to do it. You said that the vertical components of the normal reaction must balance the perpendicular weight $mg cos \alpha$. So, what is wrong in my equation should be the expression of $N$, right? Would you have more in-depth hints on how to write it?

The component of weight is in the plane of the page. The normal force is pointing into the page at a 45 degree angle. It is only the component of $N$ parallel to the plane of the page which balances the component of weight...not the entire magnitude $N$.

 

[Hak]

Yes, you're right. $N \sqrt{2} = mg cos \alpha$ is correct actually?

 

[Hak]

So, is the correct solution $\mu_{min} = \frac {2 \sqrt{2}}{9} tan \alpha?$

 

[kuruman]

I cannot prove that ##Q is the point with maximum velocity.

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?

 

[kuruman]

Yes, you're right. $N \sqrt{2} = mg cos \alpha$ is correct actually?

No. Where does $\cos\alpha$ come from? Look at the picture in post #26. Points $A$, $B$ and $O$ are all in the same plane. In that plane are also the normal force vectors and the weight.

 

[erobz]

No. Where does $\cos\alpha$ come from? Look at the picture in post #26. Points $A$, $B$ and $O$ are all in the same plane. In that plane are also the normal force vectors and the weight.

That comes from the balance of forces along the incline.

 

[Hak]

No. Where does $\cos\alpha$ come from? Look at the picture in post #26. Points $A$, $B$ and $O$ are all in the same plane. In that plane are also the normal force vectors and the weight.

So, I would say $N \sqrt {2} = mg$. Correct?

 

[Hak]

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

From the looks of it, I would say that the point with maximum linear velocity is the point with maximum relative velocity $v_{rot}$. Since it is given by the vector product $v_{rot} = \omega \times r$, the vector product under exam is maximum when the sine of the angle between $\omega$ and $r$ is maximum, that is, at 90°. This means that the point with maximum velocity is the one having rotational velocity perpendicular to the position vector $\textbf{r}$, that is, parallel to the axis of rotation $AB$, which is the point $Q$ at the top of the sphere. Correct?

 

[Hak]

That comes from the balance of forces along the incline.

It seems to me that there are two different versions regarding this point. What should we do?

 

[erobz]

It seems to me that there are two different versions regarding this point. What should we do?

 

[Hak]

View attachment 331654

I thank you very much, but could you show me (only if you want) how, through triangles and angles, we arrive at the $N \frac{\sqrt{2}}{2}$ component of each normal force? Just to confirm.

 

[erobz]

I thank you very much, but could you show me (only if you want) how, through triangles and angles, we arrive at the $N \frac{\sqrt{2}}{2}$ component of each normal force? Just to confirm.

On the left we are looking up the incline. I'm not drawing to scale, and I didn't draw the other normal force.

 

[Hak]

Thank you so much, it seems correct. I did not understand, therefore, @kuruman's answer...

 

[erobz]

Thank you so much, it seems correct. I did not understand, therefore, @kuruman's answer....

Well there has been much back and forth about this now about your own confusion on this point, and you never put in a FBD showing your coordinates. I completely understand the confusion.

 

[haruspex]

From the looks of it, I would say that the point with maximum linear velocity is the point with maximum relative velocity $v_{rot}$. Since it is given by the vector product $v_{rot} = \omega \times r$, the vector product under exam is maximum when the sine of the angle between $\omega$ and $r$ is maximum, that is, at 90°. This means that the point with maximum velocity is the one having rotational velocity perpendicular to the position vector $\textbf{r}$, that is, parallel to the axis of rotation $AB$, which is the point $Q$ at the top of the sphere. Correct?

I cannot follow your reasoning. How are you defining "rotational velocity"?
All points of the sphere have the same angular velocity, about any axis you choose. Do you mean the instantaneous linear velocity relative to the sphere's centre?
Where exactly is Q? Is it "at the top of the sphere" or the point furthest from the intersection of the two plates?

One thing troubles me about this problem. The axis of rotation of the sphere has a component normal to each plate, so it is not really rolling contact. There must be a rotational skid, like a ball spinning on the spot on a horizontal surface. I cannot work out what that does to the frictional forces.

 

[Hak]

I cannot follow your reasoning. How are you defining "rotational velocity"?
All points of the sphere have the same angular velocity, about any axis you choose. Do you mean the instantaneous linear velocity relative to the sphere's centre?
Where exactly is Q? Is it "at the top of the sphere" or the point furthest from the intersection of the two plates?

One thing troubles me about this problem. The axis of rotation of the sphere has a component normal to each plate, so it is not really rolling contact. There must be a rotational skid, like a ball spinning on the spot on a horizontal surface. I cannot work out what that does to the frictional forces.

Yes, I mean

the instantaneous linear velocity relative to the sphere's centre.

So, is my reasoning correct? How can it be further adapted to your assertions?

 

[Hak]

So, is the correct solution $\mu_{min} = \frac {2 \sqrt{2}}{9} tan \alpha?$

Would this result fit?

 

[erobz]

One thing troubles me about this problem. The axis of rotation of the sphere has a component normal to each plate, so it is not really rolling contact. There must be a rotational skid, like a ball spinning on the spot on a horizontal surface. I cannot work out what that does to the frictional forces.

To me it seems like it would be kinetic friction, but there would be zero displacement. So indistinguishable in effect from rolling without slipping?

 

[Hak]

If I may be allowed to say so, such situations would be really complex to analyze.

 

[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.