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Rolling friction is the resistance to motion experienced by a body when it rolls upon another. It is much less than sliding friction for same pair of bodies. When one body rolls upon another, there is theoretically no sliding or slip between them. And if both are perfectly rigid, there is no surface of contact.

Then the book writes:

Different causes are present depending on the nature of the wheel and the road. For a hard wheel on a soft dirty road, as the road is slightly depressed at the point of contact and a ridge is formed in front of the wheel. Thus, the wheel is being continuously pulled up in a minute hill which causes rolling friction.

Now, my questions are: - How does the ridge provide rolling friction in the opposite direction of rolling? - There is not always soft road. What happens in the hard road? - Why is rolling friction less than sliding friction??

Please help.

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"Rolling friction" is a misnomer. Never be confused by it. Its not friction. Its ROLLING RESISTANCE-the correct name. It comes into play because objects are not perfect rigid in real life. It comes into play because of the deformation of shape of objects when in contact. When an object rolls on another WITHOUT slipping, there is a surface contact between them. enter image description here

It is the normal reaction that is responsible look at the diagram. A body rolling like this (in the diagram) is deformed and the normal forces on the front parts of the surface are always more (whether road is hard or soft) and results in a net backward force that will gradually stop the rolling body. The deformation depends on the nature of the two bodies, depending on their rigidity. Rolling resistance is usually less than static friction because the surface of contact is very very small in real life so the net normal force is also small and not enough like the static friction to prevent the body from sliding and keep it unmoved and static friction also increases with the increase of the external force.

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  • $\begingroup$ Hi. In addition to that normal forces, is there a static (before motion) / kinetic (after motion) friction force (with negative torque) too? Is its distribution parallel to the surface? Does its magnitude opposite that of the normal forces? $\endgroup$ – aayyachi Oct 28 '15 at 10:34
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    $\begingroup$ Can you add a little about sliding friction and the famous $\mu N$ too. :) Thanks. :) $\endgroup$ – H. R. Oct 4 '16 at 16:17
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There are several components to "rolling friction".

First off, there is the wheel bearing that will inevitably have some friction (though a well-designed bearing has very little).

Next, if ordinary pneumatic (or simply hard rubber) tires are used, the weight of the "load" will cause the rubber to deform. Rubber is a "viscoelastic" material, meaning that flexing it causes energy to be lost in the form of heat. This lost energy must be in turn "replaced" by draining energy from the rolling wheel (ie, friction).

Further, as the tire flexes the rubber "squirms" and rubs against the roadway. This "squirm" is worse the more the tire flexes and is also affected by the cross-sectional profile of the tire.

On the other hand, if the wheel is harder than the road, the road will flex, and virtually all paving materials (especially asphalt) absorb energy as they flex, similar to rubber.

Finally, if the road and wheel are perfectly hard, any roughness in the roadway (or the periphery of the wheel) will cause the load to bounce up and down. Most loads (especially humans) will flex as they go up and down, and energy will be lost due to imperfect elasticity.

Lost energy (and hence friction) can be minimized by using a tire with very thin walls, so that practically all flexing occurs in the air of the tire (with very little loss), picking a tire profile that minimizes "squirm", then using a tire pressure which is high enough to minimize tire flexing but low enough that roadway flex and load bounce are minimized. Of course, such a tire combo may not be ideal for traction and durability, nor will it necessarily produce a comfortable ride.

(Note that railroads essentially have a "perfectly" hard "tire" and also a "perfectly" hard "roadway". If the rails are smooth and continuously welded then rolling friction is many times lower than, say, truck tires on a roadway. Add the fact that wind resistance is reduced due to the cars being strung one after the other and you can see that rail transport is quite energy efficient.)

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  • $\begingroup$ How about adhesion between the tire and the road? (Think, fresh tarmac on a hot, sunny day.) $\endgroup$ – Solomon Slow Feb 11 '16 at 0:48
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    $\begingroup$ @jameslarge - Adhesion is a function of a number of variables. In general, on reasonably hard, dry pavements, the tire tread has little effect, and adhesion is directly dependent on the coefficient of friction between the rubber used in the tire and the paving material. But coat the surface with a viscous liquid (such as exudes from fresh tarmac/asphalt paving) and things become quite unpredictable. $\endgroup$ – Hot Licks Feb 11 '16 at 2:05
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First off, I think it is better that we draw free body diagrams and write motion’s equations related to a motorcycle (for example). Consider that motorcycle is moving with acceleration $a$ on a rigid road. Assume that motorcycle’s wheel are rigid too. Rear wheel is driver. ($T_e$ is engine torque and $R$ is radius of the wheels) enter image description here Free body diagram for rear wheel is as figure below: enter image description here $$f_r-F_1=ma\;\tag 1$$ $$N_r=mg+N_1\;\tag 2$$ $$T_e-f_rR=I\alpha\;\tag 3$$ Free body diagram for front wheel is shown below: enter image description here $$F_2-f_f=ma\;\tag 4$$ $$N_f=mg+N_2\;\tag 5$$ $$f_fR=I\alpha\;\tag 6$$ Free body diagram for chassis and rider: enter image description here $$F_1-F_2=Ma\;\tag 7$$ $$Mg=N_1+N_2\;\tag 8$$ $$N_1l_1=N_2l_2\;\tag 9$$ The condition of non-slipping: $$a=R\alpha\;\tag {10}$$

There is not always soft road. What happens in the hard road?

Friction always prevents relative motion of surfaces in contact. Driver wheel (rear wheel in case above) tends to rotate due to torque exerted by engine ($T_e$). So, friction force ($f_r$) acts in a direction to oppose with wheel’s rotation. Driven wheel (front wheel in case above) tends to translate due to force transmitted to it by the chassis ($F_2$). So, friction force ($f_f$) acts in a direction to oppose with wheel’s translation. We can see all of these in above diagrams without assumption that road is soft.

How does the ridge provide rolling friction in the opposite direction of rolling?

As you can see in free body diagrams of the wheels, there is no need to have a ridge for existence of friction force. It is enough that two bodies are in contact and tend to move relative each other.

Why is rolling friction less than sliding friction?

I don’t agree. Rolling friction isn’t less than sliding friction always. Because rolling friction is static friction while sliding friction is kinetic friction and we know $\left(f_s\right)_{\textrm{max}}\gt f_k$. Probably, the book means “usually rolling friction is less than maximum static friction ($\left(f_s\right)_{\textrm{max}}$) and sliding friction ($f_k$)”. But, it isn’t so always. For checking this matter we need numerical data. Unfortunately, I couldn’t find much information, but by getting help from some data that I found from sites below, we can do a simple comparison.

$\mu_s$ and $\mu_k$ from: https://en.wikibooks.org/wiki/Physics_Study_Guide/Frictional_coefficients

Maximum torque, motorcycle mass and wheel radius from: http://www.yamaha-motor.eu/eu/products/motorcycles/hyper-naked/mt-10.aspx?view=featurestechspecs

First, consider to uniform motion on straight line (I guess the book has mentioned this case). Assume that a Yamaha MT-10 motorcycle is moving with its maximum torque in a constant speed (maximum speed). Also, assume that the rider mass is $60\;\mathrm{kg}$. From equation $4$, as $\alpha=0$, we have $$f_r=\large{\frac{\left(T_e\right)_{max}}R}$$ Numerical data is as below:

$\left(T_e\right)_{max}=111\;\mathrm{Nm}$

$R=0.22\;\mathrm m$

$M=270\;\mathrm{kg}$ ($M$ is summation of motorcycle and rider masses)

$\mu_s=0.85$

$\mu_k=0.67$

So, we obtain $$f_r= 505\;\mathrm N$$ On the other hand, If we assume that approximately $l_1=l_2$, then we have

$N_r=\large{\frac{Mg}2}=1350\;\mathrm N$

and thus

$\left(f_s\right)_{\textrm {max}}=\mu_sN_r=1148\;\mathrm N$

And

$f_k=\mu_kN_r=905\mathrm N$

So, obviously $f_r$ is much less than $f_k$ and $\left(f_s\right)_{\textrm {max}}$

There is no need to calculations for comparing in a case that motorcycle moves with acceleration. Because, if rider uses maximum possible acceleration for starting motion from rest without sliding; then certainly $f_r$ will become $\mu_sN_r$ that is obviously greater than $\mu_kN_r$

It may you claim that “it is possible that if we calculate that maximum acceleration by assuming $f_r=\mu_sN_r$, it isn’t matchable with maximum engine torque”. I.e. the possible maximum acceleration doesn’t warrantee that $f_r=\mu_sN_r$. This claim is valid and you can check it by numerical data (if you have), but we experimentally know that if we turn gas lever too much and suddenly, then rear wheel certainly will slide (note that we have a Yamaha MT-10!)

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  • $\begingroup$ Hey, @lucas, it's you ;P Situations were different two years earlier ;-)) $\endgroup$ – user36790 Jun 17 '16 at 11:24
  • $\begingroup$ Sorry @MAFIA36790 But I couldn't get your comment meaning. Do you mean that you don't need this answer now? $\endgroup$ – lucas Jun 17 '16 at 11:27
  • $\begingroup$ I didn't read your answer; my problem was solved quite long ago.... but that doesn't avert you to post a good answer. No, I've no problem with your answer :) $\endgroup$ – user36790 Jun 17 '16 at 11:29
  • $\begingroup$ @MAFIA36790 But, you haven't accepted an answer yet. This was the reason that I posted my answer. I think if none of the answers aren't correct and you know the correct answer, you should post your answer and accept that. Am I right? $\endgroup$ – lucas Jun 17 '16 at 11:33
  • $\begingroup$ Just because I've not accepted any answer doesn't mean all answers are wrong. It's simply that I've not accepted any answer - that's it. You have full permission to write answer to any question irrespective of whether it has any accepted answer or not. $\endgroup$ – user36790 Jun 17 '16 at 11:41
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Imagine a weel rotating clockwise (the car moving to the right). At the point of contact with the floor, that point on the weel will be moving to the left, so the friction will be towards the right (it is what actually makes the car move forward).

Regarding why rolling friction is usually larger that the static one. It is because the static friction (rolling friction is a form of static friction) can have a maximum value of up to $\mu_{static} N$. It up to, but usually less. Imagine a mass at rest on a table with no horizontal forces applied to it. In such a case the static friction is zero, and will start increasing up to a maximum if you start pushing the mass horizontally. Once you reach the maximum static friction, the mass will start to move. Now the friction is the dynamic one, and it is constant $F_{sliding}=\mu_{dynamic}N$. Usually $\mu_{dynamic}<\mu_{static}$, thus the sliding friction is usually less than the maximum possible static friction.

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Rolling friction has a complex origin unlike static and sliding friction. During rolling, the surfaces on contact get momentarily a little, and this results in a Donnie are (not a point) of the body being on contact with the surface. The net effect is that the component of the contract force parallel to the surface opposes motion.

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