Force required to start rolling a wheel ?

  • Thread starter rayallenvn
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  • #1
Hi everyone ,
I've searched the answer for this question for a while , but haven't reach an satisfying explanation yet .
Generally , from the study of friction , any force that's smaller than the static friction between the wheel and the ground would cause the wheel rolling ( Otherwise the wheel would be sliding ) .
But obviously when we push a car with a small force , it doesn't move . Somewhere on the internet , people say that's because of the internal friction between components inside the car .
Then again , when we push a wheel alone , or a tyre , the more weight the tyre/wheel has , the greater force we have to apply to move it , obviously greater than the force we push a toy wheel to cause it rolling .
So what's the exact explanation for this ? Does weight of the wheel have anything to do with the scale of the force applied to move it ? If yes , how's the calculation ?
Thank you all for reading this and waiting for your reply .
 

Answers and Replies

  • #2
sophiecentaur
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(tyre is actually spelled tire, just a note)
Like colour is supposed to be spelled 'color'
Ye Gods, they're even trying to spell sulphur "sulfur"
What are things coming to? (Or should I say 'to what are things coming?')

Once you guys have fixed your bizarre Units system then you can tell the World about spelling. And even your pints and gallons are wrong, btw.
I'll write to that nice Mr Obama when he gets back in and perhaps he'll sort you lot out. You know it makes sense.

You usually find that 'static' friction is higher than 'rolling' friction - but then, as speed increases, other losses become progressively more significant.
 
  • #3
rcgldr
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In the case of a car, even without a drivetrain, the tires tend to develop a flat spot when the car is at rest for a while, and it requires additional force because the cars center of mass is being raised a small amount initially due to rolling out of the flat spot. Once it's moving, it's a bit easier to keep it moving.

The other issue is "stiction", the static friction in the drivetrain, even if the car is in neutral. You'll have to overcome all of the static friction in the movable components of the drivetrain to start the car moving, and aftewards, it will be easier to keep the car moving because the dynamic friction in the drivetrain will be less than the static friction of the drivetrain.

Static friction between the tires and the ground isn't an issue here, unless you're comparing moving a car by sliding the tires on ice, versus rolling the tires on pavement. Rolling resistance is the term used for the opposing force to rolling:

http://en.wikipedia.org/wiki/Rolling_resistance
 
  • #4
In the case of a car, even without a drivetrain, the tires tend to develop a flat spot when the car is at rest for a while, and it requires additional force because the cars center of mass is being raised a small amount initially due to rolling out of the flat spot. Once it's moving, it's a bit easier to keep it moving.

The other issue is "stiction", the static friction in the drivetrain, even if the car is in neutral. You'll have to overcome all of the static friction in the movable components of the drivetrain to start the car moving, and aftewards, it will be easier to keep the car moving because the dynamic friction in the drivetrain will be less than the static friction of the drivetrain.

Static friction between the tires and the ground isn't an issue here, unless you're comparing moving a car by sliding the tires on ice, versus rolling the tires on pavement. Rolling resistance is the term used for the opposing force to rolling:

http://en.wikipedia.org/wiki/Rolling_resistance
Hi , thank you for you reply .
I'm talking about the force required to start rolling the wheel . Rolling resistance only matters when the wheel already roll .
Your explanation about the flat spot is helpful , but let's imagine we have 2 steel wheels ( alone ) on the rail track for example . It's not made of rubber , so I think the flat spot may not take place here . But obviously to start rolling the heavier wheel still require greater force . The theory states that any force smaller than the static friction between the wheel and the ground would start rolling the wheel . So what is the explanation for this ?
 
  • #5
rcgldr
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The theory states that any force smaller than the static friction between the wheel and the ground would start rolling the wheel.
In this case, as long as the wheel doesn't slide, the initial force would be related to static friction at the axis of the wheel. Once the wheel is rotating, then the dynamic (rotating) friction at the axis would be a bit less in most cases (some surfaces like teflon on teflon have very little difference between static and dynamic friction).
 
  • #6
So is there any formula or way to calculate the initial force here ? For example , my project require to pull a truss that have steel wheels by a winch , and on steel rail track . So how to calculate the required line pull of the winch to be sufficient for pulling the system ?
 
  • #7
rcgldr
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You would need to calculate the torque related to static friction at the axle (which is related to coefficient of friction, the load on the axle, and radius of the axle). Then the idealized minimal force would be this torque divided by the radius of the wheel. The actual required force will probably be greater due to other factors.
 
  • #8
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rayallenvn I am also having a lot of difficulty finding a definitive answer for this question...

Similarly to an example of the force required to move a stationary block (a fairly simple problem) I am looking to work out the force required to move a stationary wheel. However in my case there is no resisting forces created anywhere (in axles, bearings etc) as the wheel is independent of any components. It is simply just a wheel at rest on a surface. How would I go about calculating the force required to move this wheel?

It must just depend on a factor of its mass somehow, but how?
 
  • #9
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The simple way to understand the static and rolling "friction" of the tires is to understand that the main frictional losses in the tires is flexing friction, both in the sidewalls, and the tire tread squirming on the roadway. These frictional losses occur once per revolution, so the power loss is proportional to wheel RPM. Tires have a RRC (or rolling resistance coefficient), typically about 0.01 (or 1%). This simply means that if you have a 1600 Kg mass vehicle with 4 tires, each tire is pressing on the roadway with a downward force of 400 x 9.81 = 3900 Newtons. Thus the minimum horizontal force to push the car is
[tex] F_{horiz}=RRC \cdot F_{vert} = RRC \cdot mg = RRC \cdot 1600 \cdot 9.81 = 157\text{ Newtons (= 16 Kg force)} [/tex]The car will not move unless you push with at least this force.
 
  • #10
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Thanks Bob. To clarify- even when a wheel is stationary, the minimum force required to initiate motion if RRC * Normal Force?
 
  • #11
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Thanks Bob. To clarify- even when a wheel is stationary, the minimum force required to initiate motion is RRC * Normal Force?
Yes, in theory. May be more if tires develop flat spots. For a vehicle with velocity v, the power required to overcome tire rolling resistance is [itex] P=RRC \cdot mg \cdot v \space \space [/itex] watts.
 
  • #12
Exactly the same confusion as Mart7x . When we consider a wheel alone , with no friction of bearings or axles , like a homogeneous steel wheel , still don't know how to calculate the force needed to make the wheel roll from rest .
The rolling resistance , in theory , only takes place when the wheel is already rolling . I have searched quite a lot on the internet but still couldn't find satisfying explanation for this .
 
  • #13
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rayallenvn I think I have found a suitable answer on a separate thread (https://www.physicsforums.com/showthread.php?p=3861678#post3861678). Basically what I got confused at is modelling the sphere either in an real or ideal world:

In an ideal world (the surface and wheel are both rigid), the only property of the sphere resisting motion is its mass. Even the smallest of forces will start a stationary ball rolling. The acceleration it has is simply the force applied over its "rolling mass".

However, in the real there is an resistive forces (rolling resistance) determined by the the coefficient of rolling resistance multiplied by the normal force. This force shouldn't be confused with a frictional force. This is where a lot of people complicate things by factoring in fricition in bearings/axles etc that are irrelevant for our examples.

I have come to believe that rolling resistance is applicable to a stationary object too and not just a rolling object (I can't confirm this but after much searching its the only plausible explanation I can find!). My reasoning is that a stationary wheel will still have a deformation at the point of contact, as it does when rolling.

Therefore to find the force to move a stationary wheel (in the real world):

Force Applied > Fresistive = Crr * N
 
Last edited:
  • #14
Therefore to find the force to move a stationary wheel (in the real world):

Force Applied > Fresistive = Crr * N
I thought this is the force required to keep the wheel rolling ( when it is already rolling ) . Once it stops , I believe we need additional force to make it roll again ( it may be because of the flat spots that Bob mentioned ) . But how to calculate that additional force is still vague .
 
  • #15
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A fairly comprehensive discussion and table of the rollng resistance coefficient for various wheels may be found at http://en.wikipedia.org/wiki/Rolling_resistance. RRC (or Crr) ranges from about 0.001 (steel railroad car wheels on steel track) to 0.3 (ordinary car tires on sand). Most efficient car tires on highways are about 0.01. Rolling resistance refers specifically to the coefficient of force that occurs at constant velocity, and is (nearly) independent of wheel RPM. In this case, the power P needed is equal to Crr×Fload×velocity). The additional force required to start a wheel rolling (e.g., "nylon thump") is not included, nor is inertial force, meaning the non-frictional rotational inertia of a wheel (moment of inertia Iwheel ≈ 0.8 mr2, where m = mass of wheel, and r=radius of wheel).
 

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