Direction of friction on rolling object

[songoku]

Homework Statement

This is not really homework, I just want to understand the direction of friction for rolling object.

Please see the picture below

Relevant Equations

None

A force is given to the center of the object so the object rolls to the right without slipping. I understand that to provide clockwise rotation the static force should be directed to the left.

But if the force F is located at the very top of the object, the static friction is directed to the right. I don't understand why.

At first I thought because the rotation is clockwise then the object will exert backward force on the ground so the ground will exert forward force to the object but it seems this reasoning can not be applied if the F is at the center.

So when F is at the very top, why the static friction is directed to the right?

Thanks

 

[hmmm27]

You might consider what the rod would 'want to do' for both cases if there were no friction : either supported by a frictionless surface, or suspended in mid-air.

 

[lewando]

But if the force F is located at the very top of the object, the static friction is directed to the right.

Why do you claim that this is true?

 

[songoku]

You might consider what the rod would 'want to do' for both cases if there were no friction : either supported by a frictionless surface, or suspended in mid-air.

I am not really sure I understand your hint. I would think that the object will just slide (translational motion) to the right without rotating when F is at center and for F is at very top the object would still rotate clockwise

Why do you claim that this is true?

This is my reasoning:

At first I thought because the rotation is clockwise then the object will exert backward force on the ground so the ground will exert forward force to the object

Thanks

 

[hmmm27]

I would think that the object will just slide (translational motion) to the right without rotating when F is at center and for F is at very top the object would still rotate clockwise

Force applied to the top, does the object translate ?
[edit: talking about the object floating in mid-air]

 

[Delta2]

Yes the force if applied at the top, will have dual effect, translational and rotational. If the sphere will roll without slipping then the direction of the static friction will still be towards left.

 

[hmmm27]

Yes the force if applied at the top, will have dual effect, translational and rotational. If the sphere will roll without slipping then the direction of the static friction will still be towards left.

That particular post was regarding a non-friction environment : sorry, I forgot to include that in the response. Edited.

 

[Delta2]

That particular post was regarding a non-friction environment : sorry, I forgot to include that in the response. Edited.

ehm yes right, in order to have translational motion it must be $F-f_s>0$.

 

[songoku]

Force applied to the top, does the object translate ?
[edit: talking about the object floating in mid-air]

Yes because there is resultant force acting on the object

Yes the force if applied at the top, will have dual effect, translational and rotational. If the sphere will roll without slipping then the direction of the static friction will still be towards left.

Is there a case where the sphere rolls but the direction of static friction is to the right?

 

[Delta2]

Is there a case where the sphere rolls but the direction of static friction is to the right?

In my opinion the only way for this to happen is the ball CoM moving to the right, while the ball rotates CW with linear velocity $v=\omega R$ greater than the translational velocity.

 

[haruspex]

In my opinion the only way for this to happen is the ball CoM moving to the right, while the ball rotates CCW. But I don't know what sort of Torque will make it rotating CCW.

No, @songoku is correct. If the driving force is near the top of the disc then the static friction will be in the same direction.
The reason is that a force applied that far from the disc centre will exert such a high torque about the disc's centre that it will tend to create more angular acceleration than is needed to match the linear acceleration.
It is a standard problem to find the height of application at which the static frictional force is zero.

 

[Delta2]

No, @songoku is correct. If the driving force is near the top of the disc then the static friction will be in the same direction.
The reason is that a force applied that far from the disc centre will exert such a high torque about the disc's centre that it will tend to create more angular acceleration than is needed to match the linear acceleration.
It is a standard problem to find the height of application at which the static frictional force is zero.

Yes I think I was wrong, I edited my post, now it is more correct I hope lol.

 

[malawi_glenn]

One can also do like this. Assume $\vec{F}_f$ is directed to the left when force applied on top (assuming a solid cylinder shape, you can redo this analysis with any object having moment of inertia of the form $kmR^2$ where $k$ is a positive number)

Let $F$ and $F_f$ be the horisontal components of the respective forces.
$F-F_f = ma$ equation (1)
$FR+F_fR = I\alpha = \dfrac{mR^2\alpha}{2}$ equation (2)
Rolling without sliding: $a = \alpha R$
(2) becomes $F+F_f = \dfrac{ma}{2}$ equation (3)
Now, solve the system of equations (1) & (3) for $F$ and $F_f$. You will find something peculiar, what does it mean?
After this, consider that general moment of inertia that I gave you.
And finally, move the point of acftion of F so that its distance from the center is r/R where -R< r < R, and do the analysis once more. What will you find?

 

[songoku]

The reason is that a force applied that far from the disc centre will exert such a high torque about the disc's centre that it will tend to create more angular acceleration than is needed to match the linear acceleration.

You mean if the friction is to the left then $a=\alpha R$ would be violated? It would become $a<\alpha R$ so rolling without slipping is not possible?

One can also do like this. Assume $\vec{F}_f$ is directed to the left when force applied on top (assuming a solid cylinder shape, you can redo this analysis with any object having moment of inertia of the form $kmR^2$ where $k$ is a positive number)
View attachment 304801
Let $F$ and $F_f$ be the horisontal components of the respective forces.
$F-F_f = ma$ equation (1)
$FR+F_fR = I\alpha = \dfrac{mR^2\alpha}{2}$ equation (2)
Rolling without sliding: $a = \alpha R$
(2) becomes $F+F_f = \dfrac{ma}{2}$ equation (3)
Now, solve the system of equations (1) & (3) for $F$ and $F_f$. You will find something peculiar, what does it mean?

The $F_f$ would be negative so it means that the direction should be to the right.

 

[malawi_glenn]

The Ff would be negative so it means that the direction should be to the right.

Exactly.

Consider that general moment of inertia that I gave you, so you are conviced it does not depend on the moment of inertia I.
And finally, move the point of acftion of F so that its distance from the center is r/R where -R< r < R, and do the analysis once more. What will you find?

 

[songoku]

Exactly.

Consider that general moment of inertia that I gave you, so you are conviced it does not depend on the moment of inertia I.
And finally, move the point of acftion of F so that its distance from the center is r/R where -R< r < R, and do the analysis once more. What will you find?

The direction of friction depends on the location where the force acts and also depends on the moment of inertia of the object

 

[malawi_glenn]

The direction of friction depends on the location where the force acts and also depends on the moment of inertia of the object

as long as F is applied on top, does it depends on the moment of inertia?

 

[songoku]

as long as F is applied on top, does it depends on the moment of inertia?

No, the static friction will always be to the rightI also want to ask about the type of friction acting on rolling object. When doing the exercises, the friction is always about static friction. In what case the friction will be kinetic? Is it for case rolling with slipping?

And is there an example in daily life to show the motion of rolling without slipping and rolling with slipping?

Thanks

 

[malawi_glenn]

In what case the friction will be kinetic? Is it for case rolling with slipping?

yes

And is there an example in daily life to show the motion of rolling without slipping and rolling with slipping?

Drive a bike on asphalt vs on ice :)
You can also drop a marble on an incline, at some angle the marble will both rotate and slide i.e. roll with slipping.

 

[songoku]

yes

Drive a bike on asphalt vs on ice :)
You can also drop a marble on an incline, at some angle the marble will both rotate and slide i.e. roll with slipping.

What is the condition to say that the motion is rolling with slipping?

If rolling without slipping, linear velocity = angular velocity x radius

If $v> \omega r$ or $v< \omega r$, both are rolling with slipping?

Thanks

 

[malawi_glenn]

Rolling without slipping is that the point of contact with the surface have 0 momentary speed, i.e. is instantaneously at rest.

This results in ω = v/R.

All the other cases are considered rolling with slipping. But I prefer not to call it "rolling" here, I prefer to call it "sliding and rotating".

 

[songoku]

Rolling without slipping is that the point of contact with the surface have 0 momentary speed, i.e. is instantaneously at rest.

This results in ω = v/R.

All the other cases are considered rolling with slipping. But I prefer not to call it "rolling" here, I prefer to call it "sliding and rotating".

Does the term rolling refer to rolling without slipping?

And what would be the difference between sliding and rotating in case of $v> \omega r$ and $v< \omega r$?

Thanks

 

[malawi_glenn]

Does the term rolling refer to rolling without slipping?

I think it is better to just say rolling but it is not all who does that

And what would be the difference between sliding and rotating in case of v>ωr and v<ωr?

What are your own thought? Can you draw pictures of those two?

 

[songoku]

What are your own thought? Can you draw pictures of those two?

I am not sure what kind of pictures you mean. I can imagine the two velocities at the bottom of the object pointing to opposite direction and have different values but I can't use the information to analyse the motion

 

[malawi_glenn]

I am not sure what kind of pictures you mean. I can imagine the two velocities at the bottom of the object pointing to opposite direction and have different values but I can't use the information to analyse the motion

A picture of an object that rotates and slides, with arrows indicating direction of translational movement and rotation respectively.

Consider this set-up

The black part of this "skateboard-ramp" has very high friction and the grey part has very low friction (you can assume it is very close to zero.

What would the motion be of a small ball that is released at the top of the black part when it enters the grey part? Assume that the ball can roll (without sliding) on that black part at all times.

What would the motion of that same ball be, if you release it at the top of the grey part when it enters the black part?

 

[haruspex]

You mean if the friction is to the left then $a=\alpha R$ would be violated? It would become $a<\alpha R$ so rolling without slipping is not possible?

I mean this:
Start with a frictionless surface. If the force F is applied sufficiently high up it will have such a high torque that $a<\alpha R$. Now introduce friction. Friction opposes the relative motion of two surfaces in contact, so the frictional force on the wheel is in the same direction as F.

Btw, I see some authors reserve "rolling" to mean without slipping, while others distinguish that case as "pure rolling" and allow mere "rolling" to include rotating while slipping. I'm not sure whether that is even if the rotation is 'backwards'. If so, it seems it might as well include the pure slipping case!

 

[Delta2]

mean this:
Start with a frictionless surface. If the force F is applied sufficiently high up it will have such a high torque that a<αR. Now introduce friction. Friction opposes the relative motion of two surfaces in contact, so the frictional force on the wheel is in the same direction as F.

Can we say that the friction force is always such as to lessen the gap between $a$ and $\alpha R$?

 

[haruspex]

Can we say that the friction force is always such as to lessen the gap between $a$ and $\alpha R$?

Yes, a good way of putting it.

 

[songoku]

A picture of an object that rotates and slides, with arrows indicating direction of translational movement and rotation respectively.

You mean like this?

Consider this set-up
View attachment 304856
The black part of this "skateboard-ramp" has very high friction and the grey part has very low friction (you can assume it is very close to zero.

What would the motion be of a small ball that is released at the top of the black part when it enters the grey part? Assume that the ball can roll (without sliding) on that black part at all times.

What would the motion of that same ball be, if you release it at the top of the grey part when it enters the black part?

When a ball is released from the leftmost part of the ramp, it will roll in clockwise direction (the direction of friction will be in opposite direction to the motion and tangent to the ramp) but I don't know whether it would be rolling without slipping or rolling with slipping. The speed will increase but the acceleration decreases. When it enters the grey part, the speed will decrease and the ball will only slide without rotating

When the ball is released from the rightmost part of the ramp, it will slide down the ramp (the speed will increase but the acceleration decreases). When it enters the black part, it will rotate clockwise (I also don't know if it would be rolling without slipping or rolling with slipping) and decelerate.

Thanks

 

[malawi_glenn]

What do you think "very high friction" vs. "very low friction" means? ...

Remember, you asked

And what would be the difference between sliding and rotating in case of v>ωr and v<ωr?

Now since you have done some effort, I will give you the answer.

When the marble is released from the top of the black side, where the friction is very high. This means that we can assume it will roll (without slipping) at all times. This means at any instant we have v = ωr.
Then it enters the grey part, where the friction is very low. This means that we can assume that the angular velocity does not change due to zero net external torque wrt the CM of the marble. Let the marble enter the grey region with speed v₁ and angular velocity ω₁ = v₁r (clockwise). Then its angular velocity will always be ω₁ = v₁r (clockwise) in the grey region. However, the linear speed of the marble will decrease. So, here we have the situation of "v<ωr"

In the situation where the marble was released at the top of the grey part, the marble will enter the black part with ω=0 but with some velocity v₂. Here the force of friction will exert a net torque on the marbles CM, which will increase its angular velocity (from zero to something that is not zero) in the counterclockwise direction. The speed of the marble will decrease both due to friction and from gravitational acceleration. At some point, we will reach ω=v/r and the marble starts to roll (without slipping). But in between that transition, you would have a situation of "v>ωr".

 

[songoku]

I am sorry for late reply

Thank you very much for the explanation hmmm27, lewando, Delta2, haruspex, malawi_glenn