# Homework Help: Can someone explain this quote about centripetal motion?

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1. Aug 3, 2016

### ooohffff

1. The problem statement, all variables and given/known data
In describing a wheel's circular motion: "Circular motion arises purely from the resistance of the ground upon which it is applied…this resistance is equal to the force which draws the wheel in the right line"

2. Relevant equations
F=m(v^2/r)

3. The attempt at a solution
Is this talking about how the only forces acting on the wheel is the normal force and gravity, thus making the centripetal force? How is this equal to the force which draws the wheel forward in linear motion?

2. Aug 3, 2016

### haruspex

It is a very odd statement. Is it a translation?
I don't think it is discussing centripetal force. Sounds more like a wheel rolling along. But in that case it is not true that the propulsive force equals the static friction from the ground, unless both are zero.

3. Aug 3, 2016

### ooohffff

It is from a very old reference book which I'm using to write a paper. The wording is confusing so I think I might not use it.

4. Aug 3, 2016

### haruspex

That sounds wise.

5. Aug 3, 2016

### TSny

The quoted passage is from a 1795 book. See pages attached below.

It deals with a paradox about rolling without slipping that apparently goes back to at least Aristotle. Consider a wheel rolling without slipping and the wheel has a hub (or nave). After one revolution, points a and b go to a' and b', respectively. The lengths of lines aa' and bb' are equal. Yet the circumference of the nave is much smaller than the circumference of the rim of the wheel.

The discussion is interesting.

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6. Aug 3, 2016

### TSny

I'm guessing that the words "force" and "resistance" are not used in the strict sense of Newton's laws. I think this passage is just a way of "arguing" that the point of contact of the wheel (point a) with the ground is instantaneously at rest. For purely translational motion of the wheel, point a would have the same speed as the center of the wheel. But the "resistance" of the ground on point a exactly cancels this speed so that point a is instantaneously at rest.

7. Aug 3, 2016

### rcgldr

My interpretation of the article in the old book: First, it's not about centripetal motion. The books point: consider 'a' as the point of contact between wheel and surface, and 'b' as the point of contact between the nave and an imaginary horizontal plane. At any moment in time, 'a' does not move with respect to the surface, but 'b' has a component of velocity with respect to the surface (to the right in the example from post #5), so 'b' can be considered to be sliding along the imaginary plane, while 'a' is not sliding along the surface.

No frictional force from the ground is required. The wheel could be moving in space, free from any forces, rotating and translating in a manner that would be the equivalent of rolling.

As an analogy, consider a yo-yo moving at constant speed and an infintely thin string kept taught that wraps under and around the nave (axle), also at constant speed; the string moves to the right with respect to the surface that the yo-yo rolls along.

Last edited: Aug 3, 2016
8. Aug 3, 2016

### TSny

Hi rcgldr. I certainly agree with everything you wrote.

Regarding the passage from the text,

I get the feeling that it is partly concerned with why the wheel starts rotating as the coach is drawn into translational motion. As the coach starts to move, the wheel starts to rotate in such a way that the wheel rotates without slipping. It refers to how the rotational motion of the wheels “arises”. It is not necessarily assuming rolling at constant speed. Just my opinion.

Whether or not the argument can be considered “correct” is debatable. But it's fun to read these old writings.

9. Aug 3, 2016

### TomHart

Haruspex, could you please explain the last sentence in your quote (above) to me. My degree of being stumped by your statement has exceeded my fear of looking stupid by asking you to explain it.

10. Aug 3, 2016

### haruspex

Suppose there is a horizontal force applied at the axle. If there is an equal and opposite frictional force at the ground then there is no net force, so no acceleration. But there is a net torque, so there is angular acceleration. That is not consistent with rolling contact.

11. Aug 4, 2016

### CWatters

I think the book might be confusing rolling resistance with static friction. At constant velocity the rolling resistance is equal to the pulling force.