Friction on a spinning wheel when it is dropped onto the ground

[nedoX]

TL;DR Summary: I have a question about the friction on a wheel and how it works.

Since the question was in another language i will try to translate.

This wheel was spinned the way shown in the photo while it was still in air. Then the wheel was put on the ground. And it rolled without sliding without accelarating.

Regarding questions whether if they are true or false:
I- When put on ground, friction was applied in direction 1.
II- When put on ground, friction in the direction 2 is the force which rolled the wheel (to the right).

III- While rolling without accelarating (constant speed), the wheel is in equilibrium state.

Which of them are correct?

 

[berkeman]

Welcome to PF.

This wheel was spinned the way shown in the photo while it was still in air. Then the wheel was put on the ground. And it rolled without sliding without accelarating

Are you sure you translated this part correctly? The wheel cannot go from spinning in the air to rolling on the ground instantaneously without some sliding and acceleration during the initial contact phase with the ground.

 

[nedoX]

Welcome to PF.


Are you sure you translated this part correctly? The wheel cannot go from spinning in the air to rolling on the ground instantaneously without some sliding and acceleration during the initial contact phase with the ground.

I guess you gotta ignore them since they are so small in numbers. (At least thats what we do here while solving questions)

 

[berkeman]

Sorry, the question makes no sense as asked. At least not to me. Perhaps other users will be able to decode what the question is really asking.

Keep in mind that you must show your best efforts to solve your schoolwork-type problems here. Once we can clarify the scenario and understand what the problem is asking, it will be up to you to tell us what you think the answer is and why.

 

[nedoX]

Sorry, the question makes no sense as asked. At least not to me. Perhaps other users will be able to decode what the question is really asking.

Keep in mind that you must show your best efforts to solve your schoolwork-type problems here. Once we can clarify the scenario and understand what the problem is asking, it will be up to you to tell us what you think the answer is and why.

Ill leave the original question (in Turkish) you may have a look maybe.

 

[nedoX]

Translated.

 

[berkeman]

Ah, thanks, that helps. The part you left out of your original post is that "After a while it is seen that the wheel continues". So they only ask about the direction of friction during the transition phase, and then ask questions about the steady-state rolling phase.

Given all of that, what do you think the correct answers are?

 

[nedoX]

Actually i have talked this question with my friends. The actual reason why i am asking is we cannot comprehend why the friction is applied to the way it going. Also when we search on the Google there are couple other photos which tricked us a bit. Like these photos.
Why are the ways friction is applied is different? One is applied to the way its going (Its applied to slow down the wheel with torque.) Other one is being rolled down but the friction is on the other side.

 

[GeekGurl123]

I'm not sure if it's exactly the answer you're looking for but i was wondering if the concept that Friction always acts in the direction opposite to the motion would help? For example if the object is moving from left to right the frictional force would move from right to left etc.

 

[berkeman]

the frictional force would move from right to left etc.

Welcome to PF. Friction forces do not "move"; a better term to use would be "act" or something similar.

 

[berkeman]

The actual reason why i am asking is we cannot comprehend why the friction is applied to the way it going.

Perhaps a real-world photo would help you remember which way the friction force acts on a spinning wheel that is accelerating a vehicle. In this photo the dragster's rear tires are spinning counter-clockwise and the accelerating force on the tire and vehicle is pointing which way?

https://speedsport.com/drag-racing/prock-recounts-his-wild-first-run-in-a-top-fuel-dragster/

(full disclosure: I am not financially benefiting from any of the sponsors displayed on this racecar)

 

[kuruman]

I'm not sure if it's exactly the answer you're looking for but i was wondering if the concept that Friction always acts in the direction opposite to the motion would help?

That would be the case if the friction is kinetic and the two surfaces in contact have non-zero relative velocity. Here, the wheel is slipping for a while and then rolls without slipping which is the stage of the wheel's motion that is of interest. The friction here is static which means that the point of the wheel that is in contact with the ground is instantaneously at rest relative to the ground. Static friction has whatever magnitude and direction are necessary to provide the observed acceleration.

Example
You are standing still and at some point you start walking forward on firm ground. Your acceleration is forward and horizontal to the ground. The only horizontal force is static friction acting on the soles of your feet. It is in the forward direction, the same direction as the velocity, as required by Newton's second law.

To @nedoX :
Read the above the example and then decide in what direction the static friction must be in the example you posted.

 

[haruspex]

I'm not sure if it's exactly the answer you're looking for but i was wondering if the concept that Friction always acts in the direction opposite to the motion would help? For example if the object is moving from left to right the frictional force would move from right to left etc.

Unfortunately that will not help because it is wrong. Friction acts to oppose relative motion of surfaces in contact.

Consider a wheel spinning clockwise to the observer, placed on the ground, initially with zero velocity relative to the ground. Where the wheel contacts the ground, it is moving left relative to the ground. The frictional force on it therefore acts to the right, accelerating the wheel to the right but exerting an anticlockwise torque, slowing the rotation.
The wheel now moves to the right, but at first there is still a mismatch between the linear velocity and the angular velocity, so the wheel continues to skid. So now the wheel is moving to the right, but the frictional force on it continues to act to the right. Why? Because the bit of the wheel on the ground is still moving to the left.
Eventually, the increasing linear speed and reducing angular speed bring the two to the point where $v=\omega r$, i.e. the bit of the wheel in contact with the ground is instantaneously at rest. At this point, we have rolling contact and frictional forces vanish.

Interesting trick: we can find the final velocity v by considering angular momentum relative to a point on the ground, P. Initially it is $I\omega_0$; since the frictional force exerts no torque about P, angular momentum is conserved, so $I\omega_0=mvr+Iv/r$.

 

[erobz]

Interesting trick: we can find the final velocity v by considering angular momentum relative to a point on the ground, P. Initially it is $I\omega_0$; since the frictional force exerts no torque about P, angular momentum is conserved, so $I\omega_0=mvr+Iv/r$.

More interestingly it doesn't depend on $\mu_k$. I got the same result with forces/torques (assuming uniform cylindrical disk), $v = \frac{R \omega_o}{3}$

I find it a bit suspicious. It's certainly incorrect for $\mu_k = 0$, but arguably that scenario is not real ... What about close to it?

 

[kuruman]

More interestingly it doesn't depend on $\mu_k$.

And it shouldn't. The angular speed of starts at $\omega_0$ and decreases with time as the center of the wheel acquires increasing velocity $v$ relative to the ground. When the decreasing $\omega$ and the increasing $v$ match the condition $v=\omega R$ ($R$ is the contact radius), we have the onset of rolling without slipping.

What depends on $\mu_k$ is the time interval required for this onset to occur. This interval becomes infinite if the contact is frictionless.

 

[erobz]

And it shouldn't. The angular speed of starts at $\omega_0$ and decreases with time as the center of the wheel acquires increasing velocity $v$ relative to the ground. When the decreasing $\omega$ and the increasing $v$ match the condition $v=\omega R$ ($R$ is the contact radius), we have the onset of rolling without slipping.

What depends on $\mu_k$ is the time interval required for this onset to occur. This interval becomes infinite if the contact is frictionless.

Well, I feel like there is something still off with the result...as a result. Its taught ( I know its been said here repeatedly) to check what happens in the limit. If you were shown just this about this wheel problem, you would be forced to draw a false conclusion about what happens when $\mu_k \to 0$. If you wrote Newtons Second it becomes apparent...yes, that if $\mu_k \to 0, \dot v \to 0$, and $\omega \to \omega_o$. I'd prefer that reflected in the result.

 

[haruspex]

Well, I feel like there is something still off with the result...as a result. Its taught ( I know its been said here repeatedly) to check what happens in the limit. If you were shown just this about this wheel problem, you would be forced to draw a false conclusion about what happens when $\mu_k \to 0$. If you wrote Newtons Second it becomes apparent...yes, that if $\mu_k \to 0, \dot v \to 0$, and $\omega \to \omega_o$. I'd prefer that reflected in the result.

The trick I described in post #13 is for finding the velocity when rolling contact is achieved. As @kuruman pointed out, if there is no friction that state is never achieved.
This is how it would be analysed in symbolic logic:
p is the condition that rolling contact has been achieved
q is the condition that the trick works
claim: $p\implies q$
In the limiting case of zero friction, p is always false.
According to logic theory, a false statement implies all statements, so $p\implies q$ still applies.

 

[kuruman]

If you wrote Newtons Second it becomes apparent...yes, that if $\mu_k \to 0, \dot v \to 0$, and $\omega \to \omega_o$. I'd prefer that reflected in the result.

I wrote Newton's second law equations and I got expressions
$V_{cm}(t)=\mu_k~g~t$ for the velocity of the wheel's CM and
$\omega(t)=\omega_0-\dfrac{\mu_k~g~t}{q~R}~$ for the angular velocity about the center. Here $q$ is the moment of inertia factor defined by $I \equiv qmR^2$ ($q=\frac{1}{2}$ for a solid disk.)

At critical time $t_c$ $$V_{cm}(t_c)=\omega(t_c)R\implies t_c=\frac{q}{q+1}\frac{\omega_0}{\mu_kg}.$$ The result reflects that as $\mu_k \rightarrow 0$ the critical time becomes very large as expected.

 

[jbriggs444]

Well, I feel like there is something still off with the result...as a result. Its taught ( I know its been said here repeatedly) to check what happens in the limit. If you were shown just this about this wheel problem, you would be forced to draw a false conclusion about what happens when $\mu_k \to 0$. If you wrote Newtons Second it becomes apparent...yes, that if $\mu_k \to 0, \dot v \to 0$, and $\omega \to \omega_o$. I'd prefer that reflected in the result.

Are you trying to draw a conclusion about what happens after something that never happens?

Edit...

Oh, or is this rather a matter of the ambiguity of whether the final velocity is

1. The velocity after rolling without slipping is achieved
2. The velocity after a final equilibrium is achieved.

 

[erobz]

I wrote Newton's second law equations and I got expressions
$V_{cm}(t)=\mu_k~g~t$ for the velocity of the wheel's CM and
$\omega(t)=\omega_0-\dfrac{\mu_k~g~t}{q~R}~$ for the angular velocity about the center. Here $q$ is the moment of inertia factor defined by $I \equiv qmR^2$ ($q=\frac{1}{2}$ for a solid disk.)

At critical time $t_c$ $$V_{cm}(t_c)=\omega(t_c)R\implies t_c=\frac{q}{q+1}\frac{\omega_0}{\mu_kg}.$$ The result reflects that as $\mu_k \rightarrow 0$ the critical time becomes very large as expected.

This is what I did as well (less the business with $q$).

It implies that when this happens the velocity is $v = \mu_k g t_c$, which gives:

$$v = \cancel{\mu_k g} \frac{q}{q+1} \frac{\omega_o}{\cancel{\mu_k g} } $$

My point is if you saw this, without inspecting Newtons Laws, one might be tempted to believe that even if $\mu_k = 0$ this would be the case...because that is what it "on its face" is saying.

 

[erobz]

Are you trying to draw a conclusion about what happens after something that never happens?

I'm saying that I would trust an expression for $v$ that went to zero as $\mu_k$ went to zero. This expression at face value says if $\mu_k = 0.0000...01$ then $v = \frac{R \omega_o}{3}$, if $\mu_k = 0$ then $v = 0$.

 

[kuruman]

I'm saying that I would trust an expression for $v$ that went to zero as $\mu_k$ went to zero. This expression at face value says if $\mu_k = 0.0000...01$ then $v = \cdots$, if $\mu_k = 0$ then$v = 0$

Sorry, whether there is or is not a horizontal force that accelerates the center of the wheel is a binary decision. Either there is such a force or there isn't. One cannot say that $1$ is $0$ for small values of $1$.

 

[erobz]

Sorry, whether there is or is not a horizontal force that accelerates the center of the wheel is a binary decision. Either there is such a force or there isn't. One cannot say that $1$ is $0$ for small values of $1$.

I get it $0.000 ... 01 \neq 0$. My point was also that the result for $v$ should be piecewise defined then. If you circumvent a time-based analysis, one might inadvertently neglect this distinction.

$$ v= \begin{cases} \frac{q}{q+1} \omega_o , & \mu_k > 0 \\ 0 , & \mu_k = 0 \end{cases} $$

 

[kuruman]

I get it $0.000 ... 01 \neq 0$. My point was also that the result for $v$ should be piecewise defined then. If you circumvent a time-based analysis, one might inadvertently neglect this distinction.

$$ v= \begin{cases} \frac{q}{q+1} \omega_o , & \mu_k > 0 \\ 0 , & \mu_k = 0 \end{cases} $$

Sure. For completeness then, one should include the case $\mu_k =\infty~$ (the wheel falls in a puddle of Gorilla glue) and write $$v= \begin{cases} \frac{q}{q+1} \omega_o , & \mu_k > 0 \\ 0 , & \mu_k = 0 \\ 0 , & \mu_k = \infty \end{cases}$$

 

[haruspex]

I get it $0.000 ... 01 \neq 0$. My point was also that the result for $v$ should be piecewise defined then. If you circumvent a time-based analysis, one might inadvertently neglect this distinction.

$$ v= \begin{cases} \frac{q}{q+1} \omega_o , & \mu_k > 0 \\ 0 , & \mu_k = 0 \end{cases} $$

No, this is wrong. As I wrote in post #17, $v$ is defined here as the velocity of the wheel when skidding ceases. If there is zero friction it does not cease, therefore $v$ is undefined, not 0.

 

[erobz]

No, this is wrong. As I wrote in post #17, $v$ is defined here as the velocity of the wheel when skidding ceases. If there is zero friction it does not cease, therefore $v$ is undefined, not 0.

You say: "We can find the final velocity $v$ etc..." my emphasis added.

To me this is what velocity the wheel will have for the remainder of "eternity", which is 0 at $\mu_k = 0$

 

[haruspex]

You say: "We can find the final velocity $v$ etc..." my emphasis added.

To me this is what velocity the wheel will have for the remainder of "eternity", which is 0 at $\mu_k = 0$

I accept that describing it as the final velocity was not sufficiently general. But if you look at the equations you can see that the actual assumption being made is that skidding has ceased. Therefore that is the definition on which the calculation is based.

 

[jbriggs444]

Sure. For completeness then, one should include the case $\mu_k =\infty~$ (the wheel falls in a puddle of Gorilla glue) and write $$v= \begin{cases} \frac{q}{q+1} \omega_o , & \mu_k > 0 \\ 0 , & \mu_k = 0 \\ 0 , & \mu_k = \infty \end{cases}$$

I disagree with the result for the case there $\mu_k = \infty$. The analogy with a puddle of Gorilla glue fails because Gorilla glue does not permit rolling without slipping while $\mu_k = \infty$ does permit rolling without slipping.

@haruspex has often pointed out that the appropriate way of reasoning about infinities like this is to take the limit as the finite parameter ($\mu_k$ in this case) increases without bound.

 

[kuruman]

I disagree with the result for the case there $\mu_k = \infty$. The analogy with a puddle of Gorilla glue fails because Gorilla glue does not permit rolling without slipping while $\mu_k = \infty$ does permit rolling without slipping.

Yes, I was being a bit facetious with the gorilla glue.

@haruspex has often pointed out that the reliable way of reasoning about infinities like this is to take the limit as the finite parameter ($\mu_k$ in this case) increases without bound.

OK, let's look at limits à la @haruspex. We have seen in post #18 that the critical time required for the onset of rolling without slipping is given by $$t_c=\frac{q}{q+1}\frac{\omega_0}{\mu_kg}.$$ In the limit $\mu_k \rightarrow \infty,~~t_c \rightarrow 0.$ Thus, in this limit, the onset of rolling without slipping is instantaneous. However $\dots$

This is problematic because it seems to imply that there is no energy loss to friction in which case one can use energy conservation to find the angular speed when the wheel is rolling without slipping. The result is $$\omega^2=\frac{q}{q+1}\omega_0^2$$ whereas the momentum conservation approach correctly yields $$\omega=\frac{q}{q+1}\omega_0$$ which is independent of $\mu_k.$

So we conclude that to reconcile the results, we must look into instantaneous energy dissipation and infinite power. It seems that we have shifted one problematic infinity to another. I am not sure how to handle this one.

 

[jbriggs444]

So we conclude that to reconcile the results, we must look into instantaneous energy dissipation and infinite power. It seems that we have shifted one problematic infinity to another. I am not sure how to handle this one.

An infinite force over an infinitesimal distance. Yes. That is problematic. A disallowed multiplication. I am most comfortable handling it as the limit as $\mu_k$ increases without bound.

 

[haruspex]

This is problematic because it seems to imply that there is no energy loss to friction

Umm, how?
Instead of glue, consider dropping a gear wheel onto a toothed rack. Big sudden impulse, lots of opportunity to lose energy.