# Forces of rolling

1. Sep 29, 2014

### Lord Anoobis

1. The problem statement, all variables and given/known data
A constant horizontal force Fapp of magnitude 10N is applied to a wheel of mass 10 kg and radius 0.30m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its centre of mass has magnitude 0.60m/s2. In unit vector notation, what is the frictional force on the wheel?

The figure in the book shows a wheel on a horizontal surface with the arrow representing the applied force pointing horizontally to the right from the centre of the wheel.

2. Relevant equations

3. The attempt at a solution
I arrived at an answer of F = (4.0N)i. The correct answer is negative. Having given the matter some thought, this is my reasoning. In order for the wheel to rotate when the force is applied, the point on the wheel in contact with the surface must be pushed to the left by the force of friction. That explains the negative sign.

However, let's assume instead that the wheel is moving under the influence of a torque applied to centre, not a horizontal force. Now I'm visualizing the bottom of the wheel pushing against the surface to the left. If the wheel does not slip, the frictional must be countering slipping by pushing to the right.

Is my thinking correct or do I have it back to front?

2. Sep 29, 2014

### BvU

Your thinking is correct. In the exercise the wheel is pulled and the friction makes it turn (no friction -> no turning).
In your 'instead' case, the wheel turns and the friction is what causes the forward acceleration (no firction -> no forward acceleration).

3. Sep 29, 2014

### Lord Anoobis

Thank you for clearing that up. Now let's consider a wheel rolling to the right with no applied force. In this case the friction must be acting to the right as well, yes? So as to provide a torque to counter the movement?

4. Sep 29, 2014

### PhanthomJay

If the wheel is rolling to the right at constant speed, with no applied force acting on it (neglecting air drag and rolling resistance), what is the net force acting on it?

5. Sep 29, 2014

### BvU

Actually, no ! No such thing as 'torque to counter the movement' exists. Once the wheel is rotating, it will keep rotating at the same angular speed until there is something to change that. Compare $F = ma$ with $\tau = I\;\alpha$. Zero $\alpha$ means zero friction needed to maintain angular speed.

6. Sep 29, 2014

### Lord Anoobis

Actually I meant in a realistic situation, ie. the wheel slows and eventually topples.

7. Sep 29, 2014

### BvU

What slows the wheel ?

And do you realize that 'constant speed' and 'the wheel slows' are contradictory ? Sorry, two different authors.

And: Do you realize that 'no applied force' and 'the wheel slows' are contradictory ?

8. Sep 29, 2014

### Lord Anoobis

The net force would be zero. So the friction which allows the wheel to move without slipping is not the one that causes it to slow down?

9. Sep 29, 2014

### Lord Anoobis

True, but I should have clarified that I was not referring to constant speed in the third case. I meant, what happens if you roll it along the ground in real terms.

10. Sep 29, 2014

### BvU

As you already hint at, a real wheel on real ground slows down. So there must be a net force in the direction opposite to its motion. You can think of various wheels and various grounds to come up with ways to get such a resultant. E.g. loose sand will slow it down faster than concrete. Why? The sand has to be pushed aside (thus leaving a track with some depth) and it doesn't push 'back up' once the weight goes off again.
A wheel with a tyre: tyre has to be compressed where coming down to the ground. It does push 'back up' when getting off again, but with a slight loss (tyres warm up).

Last edited: Sep 29, 2014
11. Sep 29, 2014

### Lord Anoobis

Okay, I believe I see the light here. It's painfully obvious now and I can't imagine where I got that "countering torque" idea from. Thank so much for the info.