friction and rolling resistance, and work done queriesby sgstudent Tags: rolling, work energy theorem 

#1
Mar2912, 09:54 AM

P: 637

1) when a wheel turns there is a forward acting friction but the 'frictional' force that opposes it is rolling resistance right? So when a wheel successfully turns and move does it mean the friction is greater than the rolling resistance?
Then in a car the resisting force will be this rolling friction only? Because at the back wheels where there is no forfeit turning it, then which force acts on the back wheel to turn it, if so then won't the force be moving in the same direction as the wheel's motion and the rolling resistance? So how will the free body diagram of the back wheel look like? 2) I learned that work done to bring an object up to eg 1m is the same for different steepness because the source will increase despite a smaller incline distance. But if I do that, when I minus the work done against gravity, I will get the same work done against friction (taking that there is no net work done). But if I take that work dine against friction and divide it by the displace, then the fictional force will vary from the different angle of inclines. So how can this be so? 3) I'm not sure if I can use the formula net Work done=final KEinitial KE for all scenarios like in an incline situation. Can I do that? Thanks for the help everyone! 



#2
Mar2912, 03:56 PM

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hi sgstudent!
the static friction from the road makes the car move forward opposing this are the friction in the bearings, and the rolling resistance, which isn't friction but is basically the energy lost in the continual deforming of the tyre (we don't normally put the rolling resistance or the friction in the bearings onto the free body diagram … the assumption is that they are already subtracted from the engine torque to give a net torque, which does go on the diagram) thw work done against friction is different 



#3
Mar2912, 05:42 PM

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#4
Mar2912, 05:58 PM

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friction and rolling resistance, and work done queries 



#5
Mar2912, 06:07 PM

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Oh okay! I read the rolling resistance already but I'm unsure what's the purpose of having high friction tires. I'm also not very sure about why the frictional force on the nonbrake wheels is opposite to the frictional force. Because the wheel doesn't have any of its own force to turn. So I'm confused about why the fictional force will be in that direction. Thanks for the help!




#6
Mar3012, 10:58 AM

P: 637

um.. so could anyone give me a clearer explanation on the nondriving wheel's frictional force? I don't really get why the force can produce the turning motion of the wheel. Also, if the wheel turns faster does it mean that the friction will also be faster? So, if i double the speed of a car will my friction be doubled too? Thanks for the help!




#7
Mar3012, 11:02 AM

P: 642

The second question, static friction just stops something from moving up to a point. Kinetic friction, if I remember correctly, which is what we're talking about, is dependent on normal force and nothing else. So, if you push on the axle twice as hard with something, friction will be twice as strong. It's constant otherwise. To be very exact, a flux integral would be ideal for calculating normal force on "odd" surfaces, or it's what I'd use.




#8
Mar3012, 01:29 PM

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once the nondriving wheels have started turning (at a particular speed), they need no force to continue at that speed, do they? 



#9
Mar3012, 01:31 PM

P: 642

Correct. But, I think, if I'm getting the problem correct, there's torque being applied, whether we like it or not, because we have friction. This is a force/torque, so it will slow down. We need another force/torque to counter it.




#10
Mar3012, 01:35 PM

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from the pf library …On the wheels of a nonskidding car: The wheels are rolling, and so the point of contact of each tyre with the road is instantaneously stationary, and static friction applies. There is also the "rolling friction", or rolling resistance, an additional small force, caused by deformation of the tyre, which in exam questions can usually be ignored. to find the direction of static friction, always ask "what would happen if there was no friction?" (for example, if one surface was ice) … If a brake is being applied to a particular pair of tyres (on the same axle), imagine that the road under those tyres suddenly becomes ice (but the road under the nonbraked tyres remains normal): the braked tyres will go slower, but the car will stay the same speed (because there are no external braking forces on it): looking down from the window, you see the road going backward faster than the braked tyres are going backwards, so the road will try to drag those tyres backward with it In other words: the tendency is for the road to move backward relative to the braked tyres, so the friction on the road is forward, and the friction on the braked tyres is backward However, the nonbraked tyres (usually the front, steering, tyres) will have a forward friction force from the road at the same time …Similarly, for a car being accelerated by its engine, the friction from the road on the driving tyres is forward, but on the nondriving tyres is backward. 



#11
Mar3012, 10:22 PM

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#12
Mar3112, 01:39 AM

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hi sgstudent!
(just got up ) (btw, i'm tinytim, i was named after a famous character from dickens' "a christmas carol" ) there's nothing else to make it rotate forward!!(the forward rotation of the driving wheels comes from the engine) adapting a paragraph from above, if the car is accelerating forward … To see this, imagine instead that the nondriving tyres (only) are on ice: then they will go at the same speed, but the car will go faster: looking down from the window, you see the road going backward faster than the nondriving tyres are going backward, so the road will try to drag those tyres backward. 



#13
Mar3112, 08:47 AM

P: 637

So the nondriving wheels have an opposite friction to turn the wheels, but they don't have any forward friction? Also, in this link: http://www.physicsclassroom.com/mmedia/energy/ie.cfm I don't quite understand what direction the friction will be since its just rolling down. In this case would the wheels friction be like the barked ones or the non brakes ones? And why can't I include them into my work done against/by friction? I read somewhere that its because its static friction so it works only at each individual points so there is no distance like in the work done=force x distance formula. But then again, why would this be static friction and not kinetic friction/sliding friction? Thanks tinyTim! 



#14
Mar3112, 10:17 AM

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The force of friction does not do work upon the cart because it acts upon the wheels of the cart and actually does not serve to displace either the cart nor the wheels.… i've read something like this many times, but it's wrong! friction (from the road) does do work on a rolling wheel … in the obvious case of an accelerating car on a horizontal road, the only external horizontal force is the friction (from the road) … if it's not doing the work, what is??!!work done = force "dot" displacement of the point of application of the force, the point of application is the point of contact with the road, and that's moving! suppose the car has mass M, and the wheels have total mass m radius r and total moment of inertia I … if a horizontal force P pushes the car (on a horizontal road), and if the total friction force is F, and if the reaction force between the car and the wheels is R, then … P  F = (M+m)ain other words: either we must take F (the friction) into account, or we must add the "rolling mass" to the actual mass of the wheels and the car 



#15
Apr112, 02:50 AM

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#16
Apr112, 02:57 AM

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(and you know the drill here … you tell us what you think, and why, first ) 



#17
Apr112, 03:16 AM

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#18
Apr112, 04:13 AM

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it's simply torque = moment of inertia times angular acceleration …
the torque has to be in the same direction as the angular accelerationif you push the cart, then looking from the righthand side, the wheels will be accelerating clockwise, and the only external forces on the wheel are the weight (irrelevant), some friction from the axle (negligible), and the friction from the road so the friction from the road has to be clockwise, which means it's backwards 


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