# Rolling Friction experiment

1. Jul 18, 2009

### Meggett

Hi. I am doing an experiment on golf balls. I know that the dimples are supposed to reduce the resistance when travelling through the air. I am testing the effects that dimples have on rolling resistance when on the ground.
I am using 3 golf balls of the same type/brand.
One, I am keeping it how it came (new, with dimples). The second I have filed HALF of the dimples off (then sanded that half to make it smooth). The third I have filed all the dimples off and sanded it to make it smooth. I have constructed a pendulum with a sinker on the end so that I can pull it back a set distance so it should hit the ball with the same force every time. This will keep the experiment controlled (every single time the ball should have the same initial velocity - I think...). I am going to conduct the experiment on a smooth concrete surface.

The only things I can think of to measure is:
1. The time to taken to stop rolling
2. The distance travelled.

I am stuck on how to prove the friction part of things. Like, what formulas should I use??

i know the formula:
Friction force = coefficient of friction x normal force

except i dont know how to find the coefficient of friction.

Last edited: Jul 18, 2009
2. Jul 18, 2009

### tiny-tim

Welcome to PF!

Hi Meggett! Welcome to PF!
But your 3 balls have different masses … so they won't have the same initial speed.

You'll need to compare the accelerations … why not use a sloping surface, or maybe the inside of a bowl?

3. Jul 18, 2009

### Meggett

oh... thanks!
so if i use an inclined plane i can place the ball at the same spot each time, and release without using any force...
is: [a = g x sin theta] the right equation to use to find acceleration?

but then how would i find the friction?

4. Jul 18, 2009

### queenofbabes

With friction, a = g x sin theta - F/m.

You're probably going to do something along the lines of finding the time taken to roll down a certain distance, then using kinematics equations to work out acceleration, and hence friction.

5. Jul 18, 2009

### Meggett

No wait... if i use that equation, won't it be the same for each ball..?
can i use [s = ut + 1/2 at^2] for inclined planes, or is that just for flat surfaces?

6. Jul 19, 2009

### RoyalCat

That would apply to any movement under constant acceleration.
But it's irrelevant since your acceleration will be the sum of two other accelerations. The acceleration brought on by the gravitational force $$g\sin{\theta}$$ and a second one, in the direction opposite the velocity, $$\frac{F_{friction}}{m}$$

You'll have to make a few key assumptions and a few key observations in order for the experiment to be useful.

For instance, you'll have to neglect air resistance (Which isn't that big of a crime, considering the velocities you'll be dealing with) and you'll have to take note of rolling versus sliding for each of the balls, since the models you'll be using for the consuming forces would have to be different (Kinetic friction as opposed to http://en.wikipedia.org/wiki/Rolling_resistance" [Broken]).

Last edited by a moderator: May 4, 2017
7. Jul 19, 2009

### Meggett

cool thanks for your help so far...

ok so I've got:
a = (g*sin theta) - (Ff / m)
= (9.8*sin9) - (Ff / 0.05)

How do i find Ff, frictional force?

8. Jul 19, 2009

### RoyalCat

You measure a and then get the frictional force from the equation you wrote above. You'll get a time-averaged result, since you're making the assumption that the frictional force is constant.

9. Jul 19, 2009

### Meggett

ok i'm stuck again...
Í was looking over in a text book and the 'F' in the equation doesn't stand for frictional force, but the net force. Is that the same as friction? i'm so confused...
And i have no way of measuring 'a' so there are two unknowns in the equation and i cannot solve it...

sorry if i'm annoying

10. Jul 19, 2009

### RoyalCat

Don't sweat it, we're glad to help. :)

F is just the filler letter for force. Don't get too mixed up with memorizing the exact names certain equations put on things. Problems will rarely be consistent with the way the book names things. Look at the meaning rather than the names, and try not to mix the two up.

Putting an index on things, or using different notation should always clear things up for you.
For the net force, for instance, I usually stick to the notation $$\Sigma \vec F$$ rather than indexing it as $$F_{net}$$, or worse still, just leaving it as $$F$$
That's just a recipe for utter confusion.

What exactly is your experiment going to be? What will you be measuring?

If you know the time it takes the balls to roll down an incline (The distance is known), starting from rest, then you can use kinematic equations to extract the equivalent acceleration.

$$S(t) = \frac{\bar{\vec a} t^2}{2}$$

The net force in the direction parallel to the inclined plane is:
$$\Sigma \vec F_{incline}=mg\sin{\alpha} - F_{friction}$$

If we let ourselves use an average, constant value for the force of the friction, or the rolling resistance, then we can extract the acceleration using the second law.
$$\Sigma \vec F_{incline}=m\vec a_{incline}$$
$$\vec a_{incline}=g\sin{\alpha} - \frac{F_{friction}}{m}$$

Do you understand now?

Oh, and one more major thing. A rolling ball is not slowed down by friction, as the point of contact of the ball with the surface its rolling on is at rest (No kinetic friction to slow it down).
What does slow a rolling ball down is what's called rolling resistance. I linked to the wiki article on it in my earlier post. It too depends on the normal, but it doesn't work in the same way as the friction you're used to working with, so you should start thinking of things in a different way.

Last edited: Jul 19, 2009
11. Jul 19, 2009

### Meggett

Ok I think I’ve got it… is this right?

TO FIND a:
S = 3m, t= 3.05sec

s = ut + ½ at2
3 = 0*3.05 + ½ * a*(3.05)2
3 = ½ * a* 9.303
0.322 = a / 2
2*0.322 = a
a = 0.644m/s2

TO FIND Ff:
Mass = 50g

a = (g*sin theta) - F/m
0.644 = (9.8*sin9) - F/0.05
0.644 = 1.533 - F/0.05
-0.889= - F/0.05
-0.044 = - F

Therefore frictional force = 0.044

Is that right??

12. Jul 19, 2009

### RoyalCat

Yeah, that looks right. :) Just don't forget that it's in Newtons, and that you need to repeat the experiment a couple of times (And then use statistical tools to choose the best kind of average) for it to have any value.

Last edited: Jul 19, 2009
13. Jul 20, 2009

### Meggett

wicked :rofl: thanks so much for all your help... you're awesome!!