Coefficient of rolling friction question


by Draggu
Tags: coefficient, friction, rolling
Draggu
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#1
Mar14-09, 03:44 PM
P: 92
1. The problem statement, all variables and given/known data

Describe how you could use the procedure from this experiment to measure the coefficient of rolling friction between the tires of a car and the road surface for different pressures of air in the tires?

Procedure =
1. Slide a mass along the surface of a lab desk and release the mass at a preselected "starting point"
2. Use a stopwatch and record time/dist taken for the mass to come to a complete stop.


2. Relevant equations



3. The attempt at a solution

I really have no idea. Any hints would be nice :(
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LowlyPion
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Mar14-09, 04:18 PM
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Quote Quote by Draggu View Post
1. The problem statement, all variables and given/known data

Describe how you could use the procedure from this experiment to measure the coefficient of rolling friction between the tires of a car and the road surface for different pressures of air in the tires?

Procedure =
1. Slide a mass along the surface of a lab desk and release the mass at a preselected "starting point"
2. Use a stopwatch and record time/dist taken for the mass to come to a complete stop.

I really have no idea. Any hints would be nice :(
All things being equal, if different pressures produce different stopping distances what can you say about acceleration? And if there are different accelerations - that would be due to what?
Draggu
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#3
Mar14-09, 04:29 PM
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Err, the car tires with more pressure will accelerate faster than the one's with less pressure.

tiny-tim
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Mar14-09, 05:07 PM
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Coefficient of rolling friction question


Hi Draggu!

Assume it's the same as ordinary friction

what equation relates the stopping distance (from a particular speed) to the coefficient of friction?
Draggu
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#5
Mar14-09, 05:12 PM
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Well, I read the question over and over and I guess it's trying to say that if we put more/less pressure in the car, it can help measure the coefficient of friction?? If the tires have more pressure they weigh more (more air) but they also would roll faster than tires with less pressure, and less weight. But! newtons law states that Force= mass* acceleration

Or, Force of Friction = coefficient of friction * mass * gravity

I just don't get it! and I have no idea what equation relates to that, tiny-tim. I'm just crappy at physics I guess
tiny-tim
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Mar14-09, 05:19 PM
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Quote Quote by Draggu View Post
I guess it's trying to say that if we put more/less pressure in the car, it can help measure the coefficient of friction??
Perhaps I'm missing the point ,

but I don't think they're asking you to theorise about how the pressure affects it

just to measure the effect
LowlyPion
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#7
Mar14-09, 06:08 PM
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Quote Quote by Draggu View Post
Err, the car tires with more pressure will accelerate faster than the one's with less pressure.
So long as you are keeping in mind greater acceleration here means less deceleration. What you are measuring of course is negative acceleration - i.e the slowing down kind of acceleration. Firmer tires will roll longer yes? (Consider the extreme case of flat tires for instance.)

From kinematics then you can determine what each distance means wrt the slowing deceleration and isn't that then a result of your friction?
Draggu
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Mar14-09, 06:51 PM
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Quote Quote by LowlyPion View Post
So long as you are keeping in mind greater acceleration here means less deceleration. What you are measuring of course is negative acceleration - i.e the slowing down kind of acceleration. Firmer tires will roll longer yes? (Consider the extreme case of flat tires for instance.)

From kinematics then you can determine what each distance means wrt the slowing deceleration and isn't that then a result of your friction?
I understand in theory how it works, but I do not know how to calculate it... I just don't get it.
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Mar14-09, 07:06 PM
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Quote Quote by Draggu View Post
I understand in theory how it works, but I do not know how to calculate it... I just don't get it.
You're looking for a kinematic equation then:
http://www.physicsforums.com/showpos...63&postcount=2

One that yields a from distance and velocity perhaps?
Draggu
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#10
Mar14-09, 07:23 PM
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Quote Quote by LowlyPion View Post
You're looking for a kinematic equation then:
http://www.physicsforums.com/showpos...63&postcount=2

One that yields a from distance and velocity perhaps?
Ok, so I would roll tires with different pressures, measure the distance they each covered and the time it took, then use d=v2t-(1/2)at^2 to calculate a. The force of friction (Ff) will be the MASS of the tire times the acceleration. Fn will be the force of gravity times the mass (9.8)(x)

then the coefficient of friction will be Ff/Fn , and it will be negative, or positive?
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#11
Mar14-09, 07:26 PM
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Quote Quote by Draggu View Post
Ok, so I would roll tires with different pressures, measure the distance they each covered and the time it took, then use d=v2t-(1/2)at^2 to calculate a. The force of friction (Ff) will be the MASS of the tire times the acceleration. Fn will be the force of gravity times the mass (9.8)(x)

then the coefficient of friction will be Ff/Fn , and it will be negative, or positive?
Personally I might want to choose something that relates initial velocity and distance to stopping to try to eliminate time as something that has to be measured with precision.
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#12
Mar14-09, 07:28 PM
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Quote Quote by LowlyPion View Post
Personally I might want to choose something that relates initial velocity and distance to stopping to try to eliminate time as something that has to be measured with precision.
In the original experiment we used time anyway. But that being said, is my formulae correct?
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#13
Mar14-09, 07:31 PM
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Quote Quote by Draggu View Post
In the original experiment we used time anyway. But that being said, is my formulae correct?
Sure. If you have the time and distance and know the initial velocity then you can figure the a.
LowlyPion
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#14
Mar14-09, 07:34 PM
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As to whether μ is positive or negative ... it will always be positive (and usually less than 1 ... usually). It's basically a ratio of 2 weights right?


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