Rolling without slipping problem

[cianfa72]

Homework Statement

The problem setting is given in the thread itself

Relevant Equations

$v_{cm} = R \omega$
$I\dot {\omega} = M$
$M a_{cm} = F_{ext}$

Hi, I started this thread to address the solution of this problem.

Problem

A yo-yo of radii $R_1=R$ and $R_2=\frac{7}{5}R$ is acted upon by forces $F$ and $\kappa F~~~(0<\kappa<\infty)$ as shown in the figure on the right. The yo-yo rolls without slipping on the horizontal surface. The mass of the yo-yo is $M$ and its moment of inertia about its center of mass is $I=qMR_2^2$.
Given quantities are $R$, $F$, $\kappa$, $M$ and $q$.

(a) Find the linear acceleration of the center of mass of the yo-yo in terms of the given quantities.
(b) Find the force of static friction acting on the yo-yo in terms of the given quantities.
(c) Find the value of κ such that the yo-yo rolls at constant velocity.
(d) Find the value of κ such that the force of static friction is zero.
(The ratio R2/R1 is given a numerical value to match the drawing to scale)

My attempt
Let's take the yo-yo as the "system". Vertically acting forces balance out, so they aren't part of the problem.

a) to calculate the yo-yo CoM's linear acceleration w.r.t. horizontal plane rest frame, one can proceed as follows: at any given time $t$, going in the (inertial) system's CoM momentarily comoving reference frame, conservation of angular momentum gives: $$I\dot \omega = FR_1 + R_2(f_s - \kappa F)$$ From rolling without slipping condition $v_{cm} = R_2 \omega$ follows $\dot v_{cm} = R_2 \dot \omega$.

Furthermore in the horizontal plane's rest frame: $$MR_2 \dot \omega = F(1 - \kappa) - f_s$$
Solving the above linear system for $\dot \omega$ $$\dot \omega = \frac {5F (12 - 14 \kappa)} {49MR(1 + q)}$$ hence $$\dot v_{cm} = \frac {F(12 - 14 \kappa)} {7M (1+q)}$$
b) solving the system for $f_s$ gives: $$f_s = \frac {F[7(1+q)(1- \kappa) -12 + 14 \kappa]} {7 (1+q)}$$
c) rolling without slipping at constant system's CoM velocity means $\dot v_{cm} = \dot \omega = 0$ hence $$\kappa = \frac {6} {7}$$ d) From b) setting $f_s=0$ one gets: $$\kappa = \frac {5 - 7q} {7(1-q)}$$

 

[jbriggs444]

a) to calculate the yo-yo CoM's linear acceleration w.r.t. horizontal plane rest frame, one can proceed as follows: at any given time $t$, going in the (inertial) system's CoM momentarily comoving reference frame, conservation of angular momentum gives: $$I\dot \omega = FR_1 + R_2(f_s - \kappa F)$$

You are assuming here that the sign convention on $f_s$ for ground on yoyo is positive to the right. That is fine, but it would have been good to state it explicitly. You are also adopting a sign convention that rotation is counter-clockwise positive.

You are clear on your frame of reference and I agree with the formula. We are off to a good start.

From rolling without slipping condition $v_{cm} = R_2 \omega$ follows $\dot v_{cm} = R_2 \dot \omega$.

You have a sign convention here so that $v_\text{cm}$ is positive to the left. Opposite to the convention for $f_s$. No problem as long as we keep our conventions straight.

Furthermore in the horizontal plane's rest frame: $$MR_2 \dot \omega = F(1 - \kappa) - f_s$$

And there it is. Horizontal momentum conservation. With a minus sign on $f_s$ due to the sign convention.

The set up looks good. The rest is algebra. My eyes are glazing over at this point. I am too lazy to check the algebra.

Good job.

Edit: And then the algebra turned out to be wrong. :-(

 

[erobz]

If $\kappa = 0$ can we tell which way $f_s$ is pointing?

 

[cianfa72]

I'd like to deeply check a point with you.

In order to apply the conservation of angular momentum I "jumped", at any given fixed time $t$, to the momentarily comoving frame w.r.t. the system's CoM (let's call it $\mathcal A$). By definition $\mathcal A$ is an inertial frame (since it moves at constant velocity w.r.t. the horizontal plane's rest frame assumed to be inertial) so there are no pseudo/inertial forces in it.

Let's say $L$ the angular momentum about the axis of rotation $X$ through the system's CoM as measured in $\mathcal A$, $\omega$ the counter-clockwise positive angular velocity, $I$ the yo-yo's moment of inertia along $X$ and $T$ the total torque from external forces. Then one gets: $$\frac {dL} {dt} = I \frac {d\omega} {dt} = T$$ Is the above correct ? Thanks.

 

[cianfa72]

If $\kappa = 0$ can we tell which way $f_s$ is pointing?

From b) we get for $\kappa = 0$ $$f_s = \frac {F[7(1+q) -12]} {7 (1+q)}$$

 

[jbriggs444]

I'd like to deeply check a point with you.

In order to apply the conservation of angular momentum I "jumped", at any given fixed time $t$, to the momentarily comoving frame w.r.t. the system's CoM (let's call it $\mathcal A$). By definition $\mathcal A$ is an inertial frame (since it moves at constant velocity w.r.t. the horizontal plane's rest frame assumed to be inertial) so there are no pseudo/inertial forces in it.

Let's say $L$ the angular momentum about the axis of rotation $X$ through the system's CoM as measured in $\mathcal A$, $\omega$ the counter-clockwise positive angular velocity, $I$ the ball's moment of inertia along $X$ and $T$ the total torque from external forces. Then one gets: $$\frac {dL} {dt} = I \frac {d\omega} {dt} = T$$

Is the above correct ? Thanks.

Yes.

By choosing an axis of rotation at the center of mass, you have eliminated any contribution to angular momentum from the bulk linear motion of the object as a whole.

Or, looking at the same thing a bit differently, any pseudo-force arising from adopting an accelerating frame of reference will not give rise to any torque about the center of mass.

 

[erobz]

From b) we get for $\kappa = 0$ $$f_s = \frac {F[7(1+q) -12]} {7 (1+q)}$$

Which gives the expression:

$$ F \left( 1- \frac{12}{7(1+q)} \right) $$

Test some values for $q$. For instance $q = \frac{1}{2}$. Can $q$ be larger than $\frac{1}{2}$, smaller?

 

[cianfa72]

Or, looking at the same thing a bit differently, any pseudo-force arising from adopting an accelerating frame of reference will not give rise to any torque about the center of mass.

That was my point/doubt. At any fixed time $t$, taking the viewpoint of inertial momentarily comoving frame w.r.t. the system's CoM, there are no pseudo-forces involved at all in the application of the conservation of angular momentum in a such frame (i.e. only "real/physical" forces enter in the calculation).

To be more precise: if we adopt the accelerating frame in which the entire system (yo-yo) is at rest having the origin in the system's CoM, there will be two type of pseudo-forces in it.

From the point of view of rotation about the axis along the center, as you pointed out, the contribution to the torque about the CoM is null. However the system's CoM is accelerating horizontally too (w.r.t. the horizontal plane's inertial rest frame), therefore there will be a non-zero torque due to associated pseudo-forces ?

 

[erobz]

Sorry, I don't grasp your point. Do you mean it is wrong ?

I'm trying to see. Is there something fishy going on here. Physically I can see no reason why $\kappa$ can't be arbitrarily small - do you agree? Next, can we put values in for $q$ that don't appear to be an issue, but somehow manage to alter the direction of the friction, sending it from positive to negative. Is there a logical explanation for that, or is there something questionable going on here. I was hoping for you to plug in some values for $q$ and ask yourself these questions like I am asking myself.

 

[A.T.]

Next, can we put values in for $q$ that don't appear to be an issue, but somehow manage to alter the direction of the friction, sending it from positive to negative. Is there a logical explanation for that,

Consider how it would slip without friction:
- With small relative moment of inertia the yo-yo will rotate too fast for its translation
- With large relative moment of inertia the yo-yo will translate too fast for its rotation
The slip would be in opposite directions, thus is the static friction which prevents it.

 

[jbriggs444]

That was my point/doubt. At any fixed time $t$, taking the viewpoint of inertial momentarily comoving frame w.r.t. the system's CoM, there are no pseudo-forces involved at all in the application of the conservation of angular momentum in a such frame (i.e. only "real/physical" forces enter in the calculation).

To be more precise: if we adopt the accelerating frame in which the entire system (yo-yo) is at rest having the origin in the system's CoM, there will be two type of pseudo-forces in it.

I am sorry. I only see one pseudo-force.

Unless you were counting gravity. That is a pseudo-force of sorts. But you'd discarded vertical forces as irrelevant.

From the point of view of rotation about the axis along the center, as you pointed out, the contribution to the torque about the CoM is null. However the system's CoM is accelerating horizontally too (w.r.t. the horizontal plane's inertial rest frame), therefore there will be a non-zero torque due to associated pseudo-forces.

Let us back up a bit.

We are adopting the frame of reference that moves with the CoM of the wheel. This is an accelerating frame. There is a pseudo-force in this frame. If the CoM has proper acceleration to the right then this pseudo-force is acting toward the left. The line of action of this pseudo-force passes through the CoM. Therefore the associated torque about the CoM is zero.

You seem to agree with this.

But you also seem to object to this.

If we continue to adopt the frame of reference that moves with the CoM of the wheel but adopt a reference axis that moves along with the point of contact of wheel on plane then yes, I agree that the leftward pseudo-force will give rise to a non-zero torque. Nonetheless, if we summed the torques, we would see that the predicted rotational acceleration matches what we would get with our original choice of reference axis.

[The two calculations are sure to match because we have the same set of four forces ($F$, $\kappa F$, $F_s$ and the pseudo-force) and we know that those forces sum to zero. If you have a set of forces that sums to zero, the net torque does not depend on the choice of reference axis]


Note that I regard the position of the reference axis to be independent of the choice of reference frame.

 

[erobz]

Consider how it would slip without friction:
- With small relative moment of inertia the yo-yo will rotate too fast for its translation
- With large relative moment of inertia the yo-yo will translate too fast for its rotation
The slip would be in opposite directions, thus is the static friction which prevents it.

κ is not friction its a factor? It scales $F$ for force acting to the right. $q$ is also not friction, it is a factor the scales the moment of inertia w.r.t. the mass having $R_2$.

The assumption it rolls without slipping in the problem statement.

Homework Statement: The problem setting is given in the thread itself
Relevant Equations: $v_{cm} = R \omega$
$I\dot {\omega} = M$
$M a_{cm} = F_{ext}$

Hi, I started this thread to address the solution of this problem.

ProblemView attachment 358160
A yo-yo of radii $R_1=R$ and $R_2=\frac{7}{5}R$ is acted upon by forces $F$ and $\kappa F~~~(0<\kappa<\infty)$ as shown in the figure on the right. The yo-yo rolls without slipping on the horizontal surface. The mass of the yo-yo is $M$ and its moment of inertia about its center of mass is $I=qMR_2^2$.
Given quantities are $R$, $F$, $\kappa$, $M$ and $q$.

If you are saying that I can't pick suitable materials such that this criterion can be met for some very reasonable $q$ here, explain?

 

[A.T.]

The assumption it rolls without slipping in the problem ststatement.

Because there is static friction preventing slipping. So that static friction is opposed to the slipping that would occur without friction.

 

[erobz]

Because there is static friction preventing slipping. So that static friction is opposed to the slipping that would occur without friction.

You're not making sense. The equations say it flips direction somewhere before $q= 1$. I ask again...explain. You are telling me "static friction" it is going to definitely flip at a specific $q$ regardless of the materials I have selected and the force $F$ that is acting...Oh and all of this is due to the very suspicious introduction of some "slipping criteria" into the problem.

 

[cianfa72]

I am sorry. I only see one pseudo-force.

Unless you were counting gravity. That is a pseudo-force of sorts. But you'd discarded vertical forces as irrelevant.

No of course, gravity as pseudo-force can be discarded as irrelevant.

We are adopting the frame of reference that moves with the CoM of the wheel. This is an accelerating frame. There is a pseudo-force in this frame. If the CoM has proper acceleration to the right then this pseudo-force is acting toward the left. The line of action of this pseudo-force passes through the CoM. Therefore the associated torque about the CoM is zero.

Suppose to adopt the frame of reference $\mathcal B$ that moves with the yo-yo's CoM but does not rotate with it. As you pointed out, there exists a pseudo-force field in this frame acting horizontally towards the right (assuming yo-yo's CoM has proper acceleration to the left as in the OP diagram). Therefore such vector field evaluated at the origin $O$ of the frame $\mathcal B$ has zero torque about the CoM.

Regarding the pseudo-force vector field evaluated at different points in $\mathcal B$ they are all parallel to the line of action of the pseudo-force in $O$ (i.e. to the pseudo-force vector field evaluated at $O$).

Do they contribute to the total torque about the system'CoM ? Ah yes, thinking about it turns out that they actually do not contribute at all to the total torque about CoM.

As a result, we can safetly adopt the accelerating frame $\mathcal B$ (accelerating in the sense of proper/newtonian acceleration) writing down the conservation of angular momentum without adding any torque from pseudo-forces in it.

 

[A.T.]

The equations say it flips direction somewhere before $q= 1$.

Yes, and R₁/R₂ < 1. So where is the problem?

 

[erobz]

Yes, and R₁/R₂ < 1. So where is the problem?

I ask again...explain. You are telling me "static friction" it is going to definitely flip at a specific $q$ regardless of the materials I have selected and the force $F$ that is acting...Oh and all of this is due to the very suspicious introduction of some "slipping criteria" into the problem.

 

[erobz]

I'm pulling it with a little tiny force $F$ - we aren't jerking it or anything like that in the very precise region around the special $q$, that I will calculate if you desire. The friction goes one way. Then I change out the wheel with a tiny incremental change in size...suddenly, I'm to believe the friction acts the other way...How ever could we explain this clearly fundamental constant $q_{switch}$ of nature.

I can't see it...should I just say, there is nothing suspect about that at all! Move along, "the Radi have a ratio less than 1...nothing to see here!"

Just follow the yellow brick road! Follow the yellow brick road?

 

[cianfa72]

Just to be clear: in the OP solution I took as frame $\mathcal A$ to write down the conservation of angular momentum the inertial comoving frame momentarily at rest with the system's CoM at a given fixed (newtonian) time $t$. By definition this frame $\mathcal A$ is inertial and I used it to write down the conservation of angular momentum just for time $t$.

I'm not sure whether it makes sense or makes not though

 

[jbriggs444]

Furthermore in the horizontal plane's rest frame: $$MR_2 \dot \omega = F(1 - \kappa) - f_s$$
Solving the above linear system for $\dot \omega$ $$\dot \omega = \frac {5F (12 - 14 \kappa)} {49MR(1 + q)}$$

This is clearly erroneous since $R$ is never defined.

 

[A.T.]

I ask again...explain.

Explain what exactly your issue with the solution is. All you have stated are vague suspicions.

 

[cianfa72]

This is clearly erroneous since $R$ is never defined.

Why not ? $R=R_1$ is a given quantity in the problem statement.

 

[erobz]

Explain what exactly your issue with the solution is. All you have stated are vague suspicions.

Read the thread. I'm not going to keep repeating myself.

 

[A.T.]

...suddenly, I'm to believe the friction acts the other way...

Why suddenly? If you vary q smoothly, then fs will smoothly pass through 0.

 

[erobz]

Why suddenly? If you vary q smoothly, then fs will smoothly pass through 0.

You cant vary $q$ smoothly. It is a fixed property of the wheel.

 

[A.T.]

You cant vary $q$ smoothly. It is a fixed property of the wheel.

Then what do you mean by fs suddenly changing direction? If q is fixed then so is the fs direction.

 

[erobz]

I have a wheel its $q$ value is just less than $\frac{5}{7}$. I'm applying $F$. The wheel rolls without slipping. The calculations indicate the friction point to the left( or right), then I make a small incremental change to $q$ such that its now greater that $\frac{5}{7}$ and repeat the experiment, the calculation tells me the friction force now points to the right (or left). I've changed nothing else. The wheel still goes in the direction I pull it in both cases. One case has a friction force pointing right, the other left!

 

[A.T.]

I have a wheel its $q$ value is just less than $\frac{5}{7}$. I'm applying $F$. The wheel rolls without slipping. The calculations indicate the friction point to the left( or right), then I make a small incremental change to $q$ such that its now greater that $\frac{5}{7}$ and repeat the experiment, the calculation tells me the friction force now points to the right (or left). I've changed nothing else. The wheel still goes in the direction I pull it in both cases. One case has a friction force pointing right, the other left!

Yes, and?

 

[erobz]

Yes, and?

and its absurd.

 

[A.T.]

and its absurd.

Why?

 

[erobz]

Why?

follow the yellow brick road, tin man.

 

[A.T.]

follow the yellow brick road, tin man.

Your issues seem to go beyond physics.

 

[erobz]

Your issues seem to go beyond physics.

I may have no brain, but you got no heart, tin man.

 

[jbriggs444]

Why not ? $R=R_1$ is a given quantity in the problem statement.

A result that is independent of $R_2$ would be surprising. So let us try to solve the linear system to see where the error lies. I hate doing algebra.

We start with two equations in two unknowns. The unknowns being $\dot \omega$ and $f_s$.

$$I\dot \omega = FR_1 + R_2(f_s - \kappa F)$$

$$MR_2 \dot \omega = F(1 - \kappa) - f_s$$

You say that you solved this for $\dot \omega$. So let us begin by solving both of the above for $f_s$
and then setting the two results equal to each other. That will eliminate $f_s$ and allow us to solve for $\dot \omega$.

Starting with the first equation:$$I\dot \omega = FR_1 + R_2(f_s - \kappa F)$$
$$I\dot \omega - FR_1 + \kappa F R_2 = R_2 f_s$$
$$\frac{I \dot \omega - FR_1 + \kappa F R_2}{R_2} = f_s$$
Now with the second equation: $$MR_2 \dot \omega = F(1 - \kappa) - f_s$$
$$f_s = F - F \kappa - MR_2 \dot \omega$$
One moment for a sanity check with dimensional analysis. All terms in the numerator in the first result have units of energy. The denominator has units of distance. The result has units of force. Check. In the second result, all terms have units of force. Check.

I ran with this and got a result which fits with what I expected. It depends on both $R_1$ and $R_2$. The dependence on $q$ is also what I expected. It is just part of a multiplier for mass. It affects magnitudes but not directions. Any result which claims that the zero point for $f_s$ depends on $q$ is simply wrong.

The bulk of the algebra in #1 above is meaningless since it follows from an incorrect result.

I am mistaken. Since the ratio of $R_1$ to $R_2$ is fixed, the dependence is on $R = R_1$ only.[/S][/S]

 

[erobz]

It affects magnitudes but not directions

I thank you for making this clear. There are not many here like here like you.

 

[vela]

and its absurd.

Having read through your exchange with @A.T., I have to admit I'm confused as well about your objection. I thought @A.T. explained it pretty clearly in post #10.

 

[cianfa72]

Sorry, that's not the point of the problem as stated. It explicitly states that $R_1=R$ and $R_2 = \frac {7} {5}R$.

Therefore I plugged those expressions into formulas to get the results.

 

[erobz]

Having read through your exchange with @A.T., I have to admit I'm confused as well about your objection. I thought @A.T. explained it pretty clearly in post #10.

So you too are saying the result that the OP posted solution is the correct one. Care to share your analysis?

It clear that A.T. was trying to argue with me for the sake of argument as spill over from the other thread of the OP - from which this problem exploration stems. I decided instead of jumping to the standard math to put my convictions to the test try some intuition about the microscopic view of the rolling rough wheels on rough ground. Thats what that exchange is really about in my purview.

Maybe I got lucky. I think the random "friction" flipping directions is wrong with the parameter $q$, and in the special case we looked at ( $\kappa = 0$) the force of "friction" would be in the same direction as the force $F$.

This is your chance to help your friend crush me.

 

[A.T.]

I decided instead ... to try some intuition

Here is your problem.

I think the random "friction" flipping directions is wrong with the parameter q,

Why random? It's a simple formula, which you have correctly interpreted. You just reject the correct conclusion based on some misguided intuition.

 

[jbriggs444]

Sorry, that is not the point of the problem as stated. It explicitly states that $R_1=R$ and $R_2 = \frac {7} {5}R$.

Thank you. I missed that. With that in mind, let me re-examine the formula you arrived at:

Solving the above linear system for $\dot \omega$ $$\dot \omega = \frac {5F (12 - 14 \kappa)} {49MR(1 + q)}$$

OK. The formula has become plausible. We have the $1+q$ modifier on mass where it belongs. The angular acceleration scales inversely with $R$ as it should. The linear velocity will not scale at all with $R$ as it should not. We have a formula involving $\kappa$ to determine whether the acceleration is clockwise or counterclockwise. That also matches expectation.

So that equation is likely correct.

hence $$\dot v_{cm} = \frac {F(12 - 14 \kappa)} {7M (1+q)}$$

And with the outer radius of $\frac{7}{5}$ of the inner radius $R$, that one becomes correct as well

b) solving the system for $f_s$ gives: $$f_s = \frac {F[7(1+q)(1- \kappa) -12 + 14 \kappa]} {7 (1+q)}$$

Is this result plausible? Yes, perhaps.

c) rolling without slipping at constant system's CoM velocity means $\dot v_{cm} = \dot \omega = 0$ hence $$\kappa = \frac {6} {7}$$ d) From b) setting $f_s=0$ one gets: $$\kappa = \frac {5 - 7q} {7(1-q)}$$

Let us give this one the sniff test. If we increase $q$ without bound so that the system is extremely hard to rotate, we get $\kappa \approx \frac{-7q}{-7q}$. This makes sense. To keep the system from translating, we need $F \approx \kappa F$. On the other hand, we can reduce $q$ to zero. In this case, rotation is easy. But you have to balance torques to avoid runaway rotation. And the result is a $\kappa$ equal to $\frac{5}{7}$.

Yes, this passes the sniff test.

 

[erobz]

So we are back to the point where these impulses from non-sliding microscopic hills must be from the front side of the hill(s) - net effect (as opposed to the back side) if a very certain moment of inertia met, irrespective of the force $F$ and material selections, etc...?

To me this "sniffs" like the artificial result of a questionable model. I guess I will die alone on my little tiny hills.

 

[A.T.]

The dependence on $q$ is also what I expected. It is just part of a multiplier for mass. It affects magnitudes but not directions. Any result which claims that the zero point for $f_s$ depends on $q$ is simply wrong.

@jbriggs444 Has already crossed this, but it might be helpful to state the correct version:

The parameter q affects only angular, but not linear inertia. So it should be obvious, there can be a value for q, such that angular and linear accelerations have the right relationship to generate rolling kinematics without the need for the constraint force by static friction. For this value of q the static friction is zero, while below/above that q value, it has opposite directions.

 

[jedishrfu]

I think this thread has run its course as the OP has gotten an answer that he is satisfied with and so its a good time to close the thread. Thank you all for contributing here.

Jedi