Rolling without slipping problem

[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