I Rolling Without Slipping

[erobz]

Try answering the questions in the following problem, originally post #40 here, that features a driven wheel. I hope that it will show you how static friction works in the context of rolling without slipping. If you wish us to check your answers, please post them on a separate thread and let us know.

View attachment 358104Problem
A yo-yo of radii R1=R and R2=75R is acted upon by forces F and κF (0<κ<∞) 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=qMR22.
Given quantities are R, F, κ, 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.)

Lets imagine this wheel rolls without slipping in the direction of F in the diagram (to the left).

The displacement of force F apparently experiences the same displacement as that of force κF. This is Aristotle's Wheel Paradox in action. The inner wheel has slipped?!?



Anyhow, If I apply Work-Energy

The is no contribution of work from the force of "static friction" This leads to:

$$ -\kappa F s + Fs = \frac{1}{2}M \dot s ^2 + \frac{1}{2}I \omega^2 $$

Substitute $\dot s = R_2 \omega = \frac{7}{5}R \omega$, and $I = qMR^2$ and differentiate w.r.t. time.

$$ F ( 1- \kappa ) s = \frac{1}{2}M \dot s ^2 + \frac{1}{2} q M R^2 \frac{1}{R^2}\left( \frac{5}{7} \right)^2 \dot s^2 $$

$$F ( 1 - \kappa ) s = \frac{1}{2} M \dot s^2 \left( 1 + \frac{25}{49} q \right) $$

taking the derivative:

$$ \implies F ( 1 - \kappa ) = M \left( 1 + \frac{25}{49} q \right) \ddot s $$

Then we get the friction force $f_r$ :

$$ F - \kappa F + f = M \ddot s $$

$$ \implies f_r = F ( 1 - \kappa ) \left( \frac{1}{1+ \frac{25}{49} q } - 1 \right) $$

Now, if we let $q \to 0$ , $f_r \to 0$ . This is expected to me. As $q \to \infty$ from 0 we maintain a singular direction for the friction ( the factor involving $q$ is negative for all reasonable values of $q$. Also expected in my opinion.

Can we talk about this problem https://www.physicsforums.com/threads/rolling-without-slipping-problem.1078971/ again. I hope so, because it seems interesting to me.

 

[A.T.]

Lets imagine this wheel rolls without slipping in the direction of F in the diagram (to the left).

The displacement of force F apparently experiences the same displacement as that of force κF.

Not the displacement relevant for mechanical work done by strings under tension attached like that. The linear speed of the straight string segments would be different. So if you for example make F = κF, then the mechanical power transmitted trough the strings would be of different magnitude. If the work done by equal forces is different, then their displacement relevant for mechanical work is also different.

The inner wheel has slipped?!?

If the forces are applied by rolling-up / un-rolling strings, which neither stretch nor go slack, then the kinematics for each string (rolled vs straight part) is equivalent to clean rolling, without slippage.

 

[erobz]

Not the displacement relevant for mechanical work done by strings under tension attached like that. The linear speed of the straight string segments would be different. So if you for example make F = κF, then the mechanical power transmitted trough the strings would be of different magnitude. If the work done by equal forces is different, then their displacement relevant for mechanical work is also different.

If the forces are applied by rolling-up / un-rolling strings, which neither stretch nor go slack, then the kinematics for each string (rolled vs straight part) is equivalent to clean rolling, without slippage.

The velocity of the point of application of $F$ is $\dot s + R_1 \dot \theta$, the velocity of the point of application of $\kappa F$ is $\dot s + R_2 \dot \theta$ Correct ? Yet the points of contact manage to arrive at any location in unison here. I have trouble seeing how that isn't paradoxical.

Another thing: The traveling points of contact of $F$, and $\kappa F$ do work, but static friction, a traveling point of contact, does not. It seems an arbitrary distinction from my point of view.

 

[Lnewqban]

Let's imagine this wheel rolls without slipping in the direction of F in the diagram (to the left).

The displacement of force F apparently experiences the same displacement as that of force κF.

These two statements, if taking the flat surface as reference for slipping and displacement, seem to be contradictory to me.

Sorry, I may be missing some idea discussed in previous related threads that I have not followed.

 

[erobz]

These two statements, if taking the flat surface as reference for slipping and displacement, seem to be contradictory to me.

See the video I linked, or search Aristotle's Wheel Paradox.

 

[erobz]

I will grant that that strings are implied in the problem as a "yo-yo", but are not explicitly demanded in the OP. "frictional" forces between the string and points of contact are ultimately the motive. The strings don't slip wheel, the wheel doesn't slip. Yet something is clearly slipping.

 

[A.T.]

The velocity of the point of application of $F$ is $\dot s + R_1 \dot \theta$, the velocity of the point of application of $\kappa F$ is $\dot s + R_2 \dot \theta$ Correct ? Yet the points of contact manage to arrive at any location in unison here. I have trouble seeing how that isn't paradoxical.

Another thing: The traveling points of contact of $F$, and $\kappa F$ do work, but static friction, a traveling point of contact, does not. It seems an arbitrary distinction from my point of view.

What is relevant for work is the motion of the physical material that the force is applied to. Not the motion of a location like 'contact point', which might move differently than the physical bodies at that location.

 

[Lnewqban]

See the video I linked, or search Aristotle's Wheel Paradox.

I just did it.
Still, if taking the flat surface as reference for slipping and displacement, both statements seem to be contradictory to me.

The strings don't slip wheel, the wheel doesn't slip. Yet something is clearly slipping.

... then, the ends of one of the strings must slip respect to the center of the wheel.
Which string it would be?

 

[erobz]

What is relevant for work is the motion of the physical material that the force is applied to.

So the work is $FR_1 \theta - \kappa F R_2 \theta$. Following from the red arcs, is that what you are saying?

 

[erobz]

.. then, the ends of one of the strings must slip respect to the center of the wheel.
Which string it would be?

I don't see why it would be just one string end that would need to slip relative to center. Both have different velocities than the center of mass, and all are colinear.

 

[Lnewqban]

I don't see why it would be just one string end that would need to slip relative to center. Both have different velocities than the center of mass, and all are colinear.

Apologies, my previous statement was incorrect.

The end of the string to which force kF is applied should move at twice the velocity of the center of the wheel.

The end of the string to which force F is applied should move at a velocity with a magnitude between the two above velocities.

 

[kuruman]

I will grant that that strings are implied in the problem as a "yo-yo", but are not explicitly demanded in the OP. "frictional" forces between the string and points of contact are ultimately the motive. The strings don't slip wheel, the wheel doesn't slip. Yet something is clearly slipping.

I posted this problem originally and, yes, strings are not explicitly mentioned, but are intended to be there. Can you explain what is slipping and where?

This problem is a simplified version of a problem I put on take-home exam. Strings are attached to the yo-yo which go over pulleys to hanging masses $m$ and $\mu m$ (see figure on right.) Say the mass on the left accelerates up with acceleration $a_L$, the mass on the right accelerates down with acceleration $a_R$ and the yo-yo has angular acceleration $\alpha.$

For rolling without slipping, at any instant point of contact P on the rim is at rest relative to the surface.
Point A has linear acceleration $a_A=\alpha(R+\frac{7}{5}R)$ and so do all the points on the string to the left and hanging mass $m$.
Point B has linear acceleration $a_A=\alpha(\frac{14}{5}R)$ and so do all the points on the string to the right and hanging mass $\mu m$. One then writes four Newton's second law equations: one for each hanging mass, one for the linear acceleration of the CM of the pulley $a_{\text{cm}}$ and one for the angular acceleration about point P. The four unknowns are $a_L$, $a_R$, $a_{\text{cm}}$ and $\alpha.$ I note that the tensions are constant and have a constant ratio $\kappa$ which is a function of $\mu.$

 

[erobz]

I posted this problem originally and, yes, strings are not explicitly mentioned, but are intended to be there. Can you explain what is slipping and where?

View attachment 358310This problem is a simplified version of a problem I put on take-home exam. Strings are attached to the yo-yo which go over pulleys to hanging masses $m$ and $\mu m$ (see figure on right.) Say the mass on the left accelerates up with acceleration $a_L$, the mass on the right accelerates down with acceleration $a_R$ and the yo-yo has angular acceleration $\alpha.$

For rolling without slipping, at any instant point of contact P on the rim is at rest relative to the surface.
Point A has linear acceleration $a_A=\alpha(R+\frac{7}{5}R)$ and so do all the points on the string to the left and hanging mass $m$.
Point B has linear acceleration $a_A=\alpha(\frac{14}{5}R)$ and so do all the points on the string to the right and hanging mass $\mu m$. One then writes four Newton's second law equations: one for each hanging mass, one for the linear acceleration of the CM of the pulley $a_{\text{cm}}$ and one for the angular acceleration about point P. The four unknowns are $a_L$, $a_R$, $a_{\text{cm}}$ and $\alpha.$ I note that the tensions are constant and have a constant ratio $\kappa$ which is a function of $\mu.$

Please diagram in post 1. Do you agree with the paradox’s resolution in the video? That picture holds at every point in time as the wheel rolls in a particular direction. The force F has traveled a distance it couldn’t have if it had rolled itself through the angle, but there it has arrived nonetheless.

The center displaced by $s$ and so have points A and B. Miraculously they have arrived at the final position a distance $s$ away from the start positions, while the points of force contact and center of mass all travel at different velocities over the same timespan.

 

[A.T.]

So the work is $FR_1 \theta - \kappa F R_2 \theta$.

Not in the rest-frame of the ground. Note that work is frame dependent, and the strings move differently in the ground-frame, than in the center-of-wheel-frame.

Do you agree with the paradox’s resolution in the video?

In that video, the inner wheel slips on a rail that is a rest relative to the ground on which the outer wheel rolls. But here, there is no reason to assume the unrolled string applying F is at rest relative to the ground. If the unrolled string stays taut, the relative kinematics of the wheel & unrolled string is that of clean rolling without slippage.

 

[jbriggs444]

Another thing: The traveling points of contact of $F$, and $\kappa F$ do work, but static friction, a traveling point of contact, does not. It seems an arbitrary distinction from my point of view.

For mechanical work, the relevant motion is not that of the point of contact. It is the motion of the contacted material.

 

[kuruman]

Do you agree with the paradox’s resolution in the video?

No, because I have a more transparent resolution. Consider the setup on the right. The small hanging masses keep the strings taut at all times. Initially I hold the yo-yo at rest as shown. Points A and B are at the 12 o'clock position. The two hanging masses are initially at the same height.

I now allow the yo-yo to roll without slipping to the right and stop it when points A and B are again at the 12 o'clock position.

Question:
What is the difference in height between the two masses?

Analysis:
The center of the yo-yo moves distance $S=2\pi r_A.$ The amount of top string that unwinds, assumed without slipping, is $L_A=2\pi r_A.$ So the overall vertical drop of the mass attached to the top string is $h_A=S+2\pi r_A.$ Similarly for the bottom string, $h_B=S+2\pi r_B.$ Thus $$\Delta h =h_A-h_B=2\pi(r_A-r_B)$$ Note that the answer would be the same if the yo-yo were just rotating without rolling about a fixed (relative to the ground) axis through its center.

The video misleads the viewer to think that just because points A and B start and end at the 12 o'clock position and have the same horizontal displacement as the CM of the wheel, they have traveled the same distance as the CM of the wheel. That is simply not true. The distance traveled by each point is equal to the height by which its corresponding mass has descended.

 

[erobz]

So we need to consider the point of contact for a force in the displacement for the work...the end of the rope. Ignoring $\kappa F$, I find that for $F$ in this case:

the displacement of $F$ is $x= ( R_2 + R_1 ) \theta$?

 

[kuruman]

There are two kinds of work that increase the kinetic energy of the rolling yo-yo.
Translational work done by the external force on the CM
$W_{\text{tran}}=F~\Delta s _{\text{cm}}=F R_2\Delta \theta.$
Rotational work done by the external torque about the CM on the wheel
$W_{\text{rot}}=F R_1~\Delta \theta.$
The total work that is equal to $\Delta K$ is
$W_{\text{total}}=W_{\text{tran}}+ W_{\text{rot}}=F( R_1+R_2)~\Delta \theta.$
Note that this is also the work done by the external torque about the point of contact with the surface.

If you are asked to find the speed of the CM assuming that the yo-yo then it is easiest to calculate the rotational energy about the point of contact. If the moment of inertia about the CM is $I_{\text{cm}}=qMR_2^2$, then about the point of contact it is $I_{\text{P}}=(q+1)MR_2^2.$
Assuming that the yo-yo starts from rest, $$W_{\text{total}}=\Delta K=K=\frac{1}{2}(q+1)MR_2^2\omega^2=F( R_1+R_2)~\Delta \theta.$$ For rolling without slipping $V_{\text{cm}}=\omega R_2$ in which case the expression above gives $$V_{\text{cm}}^2=\frac{2F( R_1+R_2)~\Delta \theta}{(q+1)M}.\tag 1$$Also note that the angular acceleration about the point of contact is the torque divided by the moment of inertia, $$\alpha=\frac{F( R_1+R_2)}{(q+1)MR_2^2}\implies a_{\text{cm}}=\alpha R_2=\frac{F( R_1+R_2)}{(q+1)MR_2}.$$ We put this result in equation (1) to get $$ V_{\text{cm}}^2-0^2=2a_{\text{cm}}R_2\Delta \theta=2a_{\text{cm}}\Delta s_{\text{cm}}$$ which is exactly what one would expect from the kinematics. There are no surprises anywhere.

 

[erobz]

There are two kinds of work that increase the kinetic energy of the rolling yo-yo.
Translational work done by the external force on the CM
$W_{\text{tran}}=F~\Delta s _{\text{cm}}=F R_2\Delta \theta.$
Rotational work done by the external torque about the CM on the wheel
$W_{\text{rot}}=F R_1~\Delta \theta.$
The total work that is equal to $\Delta K$ is
$W_{\text{total}}=W_{\text{tran}}+ W_{\text{rot}}=F( R_1+R_2)~\Delta \theta.$
Note that this is also the work done by the external torque about the point of contact with the surface.

If you are asked to find the speed of the CM assuming that the yo-yo then it is easiest to calculate the rotational energy about the point of contact. If the moment of inertia about the CM is $I_{\text{cm}}=qMR_2^2$, then about the point of contact it is $I_{\text{P}}=(q+1)MR_2^2.$
Assuming that the yo-yo starts from rest, $$W_{\text{total}}=\Delta K=K=\frac{1}{2}(q+1)MR_2^2\omega^2=F( R_1+R_2)~\Delta \theta.$$ For rolling without slipping $V_{\text{cm}}=\omega R_2$ in which case the expression above gives $$V_{\text{cm}}^2=\frac{2F( R_1+R_2)~\Delta \theta}{(q+1)M}.\tag 1$$Also note that the angular acceleration about the point of contact is the torque divided by the moment of inertia, $$\alpha=\frac{F( R_1+R_2)}{(q+1)MR_2^2}\implies a_{\text{cm}}=\alpha R_2=\frac{F( R_1+R_2)}{(q+1)MR_2}.$$ We put this result in equation (1) to get $$ V_{\text{cm}}^2-0^2=2a_{\text{cm}}R_2\Delta \theta=2a_{\text{cm}}\Delta s_{\text{cm}}$$ which is exactly what one would expect from the kinematics. There are no surprises anywhere.

I was heading there... For the original problem:

$$ F( R_1 + R_2 ) \theta - \kappa F R_2 \theta = \frac{1}{2}M R_2^2 \dot \theta^2+ \frac{1}{2} q MR_1^2 \dot \theta^2 $$

With $R_2 = \frac{7}{5}R_1$, after substitution and differentiating w.r.t. I end up with:

$$ MR_1 \ddot \theta = \frac{F ( 12-7\kappa ) }{5 \left( \frac{49}{25} + q \right)} $$

Do you agree with that?

I then go to Newtons Second:

$$ F - \kappa F + f_r = M R_2 \ddot \theta $$

So I get after subbing in $R_2 = \frac{7}{5}R_1$ for the force of friction $f_r$:

$$ f_r = \frac{7}{25} \left( \frac{F( 12-7 \kappa)}{ \left( \frac{49}{25} + q \right) } \right) - F( 1-\kappa) $$

So the friction force does not change direction as a function of $q$ ( I checked $\kappa = 0$ as a special case - same as the other thread)? This is the point of contention in which this whole shebang -initiated my grave concerns - closing a thread- about the validity of the solution accepted as correct had began. See: https://www.physicsforums.com/threads/rolling-without-slipping-problem.1078971/

So, is there an issue with the solution proposed in that thread, this post, maybe they agree?


EDIT: I cleaned mine up and they disagree already at $\dot \omega$. I'm getting:

$$\dot \omega = \frac{5F( 12-7\kappa)}{MR_1 ( 49 + 25 q )} $$

 

[TSny]

I was heading there... For the original problem:

$$ F( R_1 + R_2 ) \theta - \kappa F R_2 \theta = \frac{1}{2}M R_2^2 \dot \theta^2+ \frac{1}{2} q MR_1^2 \dot \theta^2 $$

For the second term on the left, note that the distance from the bottom of the wheel to the top of the wheel is $2R_2$ instead of $R_2$.

For the second term on the right, the subscript on $R$ should be 2 instead of 1 according to the problem statement.

 

[erobz]

View attachment 358349
For the second term on the left, note that the distance from the bottom of the wheel to the top of the wheel is $2R_2$ instead of $R_2$.

For the second term on the right, the subscript on $R$ should be 2 instead of 1 according to the problem statement.

If the big wheel rolls without the rope slipping, the force $\kappa F$ on has displacement $R_2 \theta$. This is the same idea I applied to conclude the displacement of $F$ in post 17.

And $R_1 = R$ in the original problem. I stand corrected on that inertia term it apparently says $I = q MR_2^2$. I'll re compute.

 

[erobz]

When I recompute I get:

$$\dot \omega = \frac{5 F ( 12- 7\kappa)}{49 M R(1+q)}$$

Against the other analysis there is still a discrepancy.

 

[TSny]

When I recompute I get:

$$\dot \omega = \frac{5 F ( 12- 7\kappa)}{49 M R(1+q)}$$

This looks ok with $R = R_1$. The result simplifies a bit if expressed in terms of $R_2$.

 

[erobz]

I still get something that friction direction flips because of a change in $q$. A change from positive to negative falls out of the result.

at $\kappa = 0$ as a special case, somewhere between $0 \leq q < \infty$, there is a flipping point of friction.

I suppose I'm at an impasse that these results are flawed unless someone can explain how there can be a $q$, effectively changing the mass of the yo yo that can make the wheel roll without any static fiction, on any selected mating surfaces (frictionless surfaces included), at any acceleration. Choose a mass such that it executes pure rolling without friction. Thats a paradox to me.

 

[Lnewqban]

at $\kappa = 0$ as a special case, somewhere between $0 \leq q < \infty$, there is a flipping point of friction.

What q stands for, if I may ask?

Would you mind explaining that "flipping point of friction" a little more detailed, please?

 

[vela]

If the big wheel rolls without the rope slipping, the force $\kappa F$ on has displacement $R_2 \theta$. This is the same idea I applied to conclude the displacement of $F$ in post 17.

For $F$, you concluded $x = (R_1+R_2)\theta$. The $R_1\theta$ term corresponds to the length of string winding onto the spool of radius $R_1$, and the $R_2\theta$ term corresponds to how far the center of mass moves. For $\kappa F$, where the string unspools from a spool of radius $R_2$ and the center of mass still moves $R_2\theta$, you should get $x = (R_2 + R_2)\theta = 2R_2\theta$.

 

[jbriggs444]

What q stands for, if I may ask?

Pulled from the other thread where the problem was originally posted.

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 $I=qM{R_2}^2.$
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 $\kappa$ such that the yo-yo rolls at constant velocity.
(d) Find the value of $\kappa$ such that the force of static friction is zero.
(The ratio $R_2/R_1$ is given a numerical value to match the drawing to scale.)

 

[erobz]

What q stands for, if I may ask?

Would you mind explaining that "flipping point of friction" a little more detailed, please?

$q$ is that parameter for scaling $I$.

For $F$, you concluded $x = (R_1+R_2)\theta$. The $R_1\theta$ term corresponds to the length of string winding onto the spool of radius $R_1$, and the $R_2\theta$ term corresponds to how far the center of mass moves. For $\kappa F$, where the string unspools from a spool of radius $R_2$ and the center of mass still moves $R_2\theta$, you should get $x = (R_2 + R_2)\theta = 2R_2\theta$.

The wheel rolls without slipping on the ground, and teh string does not slip on the wheel. In the case I have described in the diagram the big wheel must take up $R_2 \theta$ of string.

For the force $F$ string the red dot marking the end of the rope has travelled $R_2 \theta + \ell + R_1 \theta - \ell = x$, the little wheel pays off $R_1 \theta$ - no slip, but the rope end has translated too because of what the outer wheel is doing.

The big wheel will take up $R_2 \theta$

I haven't drawn post 17 diagram to scale, but the diagram seems clear enough.

 

[kuruman]

I still get something that friction direction flips because of a change in $q$. A change from positive to negative falls out of the result.

at $\kappa = 0$ as a special case, somewhere between $0 \leq q < \infty$, there is a flipping point of friction.

I suppose I'm at an impasse that these results are flawed unless someone can explain how there can be a $q$, effectively changing the mass of the yo yo that can make the wheel roll without any static fiction, on any selected mating surfaces, at any acceleration. Thats a paradox to me.

You cannot get a $q>1$ if the yo-yo is rolling about its largest radius $R$. That's because if all the mass is at radius $R$ from the center, you have a ring of moment of inertia about its center $I_{\text{ring}}=mR^2$ which implies $q=1.$ You cannot put more mass farther out without increasing the rolling radius. I suppose you can increase $q$ beyond $1$ if you make the larger cylinder out of styrofoam and the inner cylinder out of depleted uranium, but let's keep things simple and assume that the yo-yo has uniform density in which case $0<q<1.$

 

[jbriggs444]

You cannot get a $q>1$ if the yo-yo is rolling about its largest radius $R$.

You are right, of course. A ring has the maximum moment of inertia ($mR^2$) for any shape bounded by radius $R$.

However, there is nothing preventing us from contemplating a wheel rolling on an elevated track with a larger coaxial wheel attached to the same axle and extending below the tracks. We can get arbitrarily large $q$ by such measures. This sort of thing is almost the defining property of a yoyo.