# FeaturedI A perfectly stiff wheel cannot roll on a stiff floor?

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1. Dec 2, 2017

### PeroK

It occurs to me that if a perfectly rigid wheel cannot roll, then it cannot do anything.

The same practical arguments relating to molecular interactions etc. apply equally to its internal structure and make its existence impossible.

As soon as we say "perfectly rigid wheel" we are in the realm of an idealised mathematical model. Any talk of atoms is absurd in this model, for many reasons.

It's a bit like advancing practical reasons why a point particle, say, cannot move in a perfect circle.

2. Dec 2, 2017

### gr71cj5

Envision: High tinsel steel ball bearing, High tinsel steel plate, no deformation of either will occur, however, apply any force to the bearing either directly (even with a simple magnet to avoid physical contact) or by tilting the metal plate (imposing gravity) and it will roll quite well. No deformation of either surface is needed. In fact, any deformation of either surface simply impedes the motion even if only by absorbing the energy required to deform the surface.

3. Dec 2, 2017

### gr71cj5

Envision: High tinsel steel ball bearing, High tinsel steel plate, no deformation of either will occur, however, apply any force to the bearing either directly (even with a simple magnet to avoid physical contact) or by tilting the metal plate (imposing gravity) and it will roll quite well. No deformation of either surface is needed. In fact, any deformation of either surface simply impedes the motion even if only by absorbing the energy required to deform the surface.

4. Dec 2, 2017

### jbriggs444

It is not enough to have high tensile strength. If you want a perfectly rigid wheel with an infinitesimal contact patch, you need infinite tensile strength. That is impossible to find in nature.

[Tinsel is the stuff you put on Christmas trees]

5. Apr 9, 2018

### rcgldr

There's no reason that a perfectly rigid wheel on a perfectly rigid surface could not "roll", even with zero friction, if the wheel just happens to be rotating so that it's outer surface matches the speed that it's axis is translating with respect to the surface. Then again, I don't see an issue with the concept of having infinite friction to go along with a perfectly rigid wheel and surface. What if the wheel and the surface were geared (for perfectly smooth motion) to result in the equivalent of infinite friction?

6. Apr 9, 2018

### jmolmo

A little "offtopic", but the question is "Will it roll or not?"

7. Apr 9, 2018

### DaveC426913

Cool.

No, they will not roll.
The thread, the radii and the two touch-points of the wheels together form a trapezoid.
The first thing that rolling would theoretically result in is the lengthening of the longest side (the line between the two touch-points).
That trapezoid cannot be deformed and still maintain symmetry.

Not so sure now.

Last edited: Apr 9, 2018
8. Apr 9, 2018

### haruspex

Well, it does depend on the relationship between the ratio of the radii and the angle of the slope. What if the slope increases to 60 degrees?

9. Apr 9, 2018

### kuruman

To elaborate, static equilibrium is maintained when the torque due to gravity and the torque due to the tension in the string are in opposite directions. The former is non-zero as long as $\theta < 90^o$. The latter becomes zero when the radius of the inner wheel is large enough so that the position vector from the contact point to the point of application of the tension is horizontal. This happens when $r/R=\cos\theta$. For $r/R>\cos\theta$ the two torques are in the same direction and there can be no equilibrium. Note: Torques are calculated using the point of contact of the wheel with the incline as origin.

10. Apr 9, 2018

### Baluncore

The distance between circle centres is the length of tight string.
It is not the change in height, Sin(30°), but the change in horizontal separation, Cos(30°) that is important.

They will not roll because R/2R = 0.5 string released, is less than Cos(30°) = 0.886 string required between centres.
Roll becomes possible when Cos(slope) = R/2R = 0.5 which will be when the slope becomes > 60°.

11. Apr 10, 2018

### A.T.

It boils down to whether to string goes above or below the contact points.

12. Apr 10, 2018

### A.T.

13. Apr 10, 2018

### Baluncore

No. But that is not a genuine solvable problem. It is an advertisement for “advanced physics tutors”. It is designed to disempower and confuse students to the point where they subscribe to a study program. Now being advertised on PF.

14. Apr 10, 2018

### jmolmo

15. Apr 10, 2018

### jmolmo

You're right ... In a rigid body, you can't consider 2 simultaneous "centers of rotation" at the same time (so, a point from which all the other body points motion can be described as circles around that "point")... such body would be no rigid.

The only way you can describe the movement of the points of a wheel if the geometrical lowest point of the wheel is moving along a line, is as the composition of one translation movement that follows that "line", and one rotational movement around the center of the wheel. So, or that "center of rotation" is constantly flipping between the center of the wheel and the geometrical contact point (but never both at the same time), or there's no point in saying that the contact point has been the center of rotation at any time, even if an infinitesimal time is considered.

Last edited: Apr 10, 2018
16. Apr 10, 2018

### Baluncore

Define upwards, and what you mean by pull. Consuming string at the fixed anchor point will move the spool up the ramp. Feeding string out from the fixed anchor point will allow the spool to roll. Raising the anchor point to change the angle between the string and the slope will make a big difference at some point.

17. Apr 10, 2018

### jmolmo

If it's an string, I think it's clear what I mean by pull. If you mean how much pull, then image you do very, very little at first (0.01N) ... and then more and more.