Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work done by friction by wheels

Tags:
  1. Nov 7, 2015 #1
    Hello all.

    I have a doubt about the work done by friction force on a wheels , in two diferent situations:

    a) In Morin's mechanics book, chapter 5 (page 146) he considers the situation in which a car is braking without skidding. He claims that the friction from the ground on the tires causes the car to slow down. But then, he says that this force is not doing work on the car, because the force acts over zero distance. I do not really undertand why there is no work here.

    b) Imagine the same situation, but now, the car slows down while skidding. In this situation, the wikipedia says that there is work done by friction (see https://en.wikipedia.org/wiki/Work_(physics), section "Moving in a straight line (skid to a stop)". I guess that friction does work only if the wheels of the car are skidding, but it does no work if the wheels are just rolling. I tink it is related to the point of application of the force, but I can't see the point obout it

    Can anybody explain me why?


    Best regards
     
  2. jcsd
  3. Nov 7, 2015 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First, take a simple example. If you push a heavy object, but can't get it moving because of static friction, then no work is done. But, if you get it moving then work is done by kinetic friction.

    In case b), you have an example of kinetic friction at work. The tyres will heat up etc.

    In case a) the work is being done in the braking mechanism. The brakes will get hot. But, braking does rely on static friction on the ground. If the car were moving through the air, then applying the brakes would do nothing. To explain this, you can see that the tyre is not actually moving along the ground - it's rolling. One of the things about rolling (without slipping) is that the part of the tyre/wheel that touches the ground is instantaneously at rest. It grips using static friction, rather than kinetic friction. This is what allows rolling to be so efficient: each small area of the wheel touches the ground without slipping, grips momentarily using static friction, then moves off the ground. And the next small area of the wheel takes over. So, there is effectively no heat or energy loss to friction: either in accelerating, braking or moving with constant speed.
     
  4. Nov 7, 2015 #3
    Thanks a lot.

    Trata helps clarify.
     
  5. Dec 13, 2015 #4
    Sorry, but after reading Kleppner - Kolenkow (2nd ediction) example 7.16, I have another doubt about work done by friction.

    In this example, a uniform drum of radius b, mass M, weight W = Mg, and moment of inertia I = Mb2/2 is on a plane of angle β. The drum starts from rest and rolls without slipping, and we have to find the speed V of its center of mass after it has descended a height h.

    In the solution, they calculate the total work done by all the forces while the certer of mass moves along the plane. For them, the forces doing work are the weight W (that is acting at the center of mass), and the friction f between the drum and the plane as well. So they consider that work done by friction is not zero, but -fL (being L the distance covered) (!!).

    After that, they calculate the work done by the total torque on the drum. The only force providing a torque is friction (because weight is acting on the center of mass, and normal force is paralell to position vector). So this torque is responsible for the rolling of the drum, and accounts for the change of rotational kinetic energy of the drum. In adition, they get that work done by torque is fL, so it equals the work done by friction.

    Finally, they conclude that work done by friction (-fL) is decreasing the the center of mass kinetic energy in exactly the same amount that torque exerted by friction (fL) is increasing the rotational energy. So here friction simply transforms mechanical energy from one mode to another.

    Now, if it is true, the friction must be doing work, but previously you told me that friction is not doing work. I am really confused. What is really happening here??

    Thanks for your replies.
     
  6. Dec 13, 2015 #5

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    K & K sum up that example very well at the end by saying that in rolling without slipping friction transforms linear to rotational energy with no energy being dissipated as heat. Although, you could also say that friction transforms gravitational PE to rotational KE.

    "Doing work" is putting energy in or taking energy out of a system. Transforming energy from one form of KE to another isn't "work".

    In fact, if you look at the definition of work in K & K: secion 5.3.3. it gives ##W_{ba} = K_b - K_a##. In this case, all the change in KE is due to gravity. Friction, by definition, does no work in this example.
     
  7. Dec 13, 2015 #6

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Just as they say: The linear and rotational work done by friction cancel each other, so friction is doing no work in total.
     
  8. Dec 14, 2015 #7
    But then, or Morin is wrong, or Kleppner and Kolenkow are wrong. Let me explain:

    1) As I mentioned in the first contribution to this thread, in Morin's mechanics book, chapter 5 (page 146) he considers the situation in which a car is braking without skidding. He claims that the friction from the ground on the tires causes the car to slow down. But then, he says that this force is not doing work on the car, because the force acts over zero distance. So, for Morin, friction is nor doing work.

    2) On the other hand, in Kleppner and Kolenkow mechanics book (2nd ediction) example 7.16, there is a drum rolling without slipping on a plane. The situation is the same than in the example in Morin's book. Nevertheless, they do not claim that work is zero, and compute the work that is being done by friction as the integral of f (the friction force) between x1 and x2, obtaining that this work equals to -fL (where L is the distance travelled between x1 and x2). So, for Kleppner and Kolenkow, work done by friction is not zero.

    Evidently, one of them must be wrong, because the reasoning they follow is opposite, indepentdently that the torque due to the friction force is doing work or not.

    What do you think about it?

    PS: By the way, if friction does no work, how can Morin claim that the friction from the ground on the tires causes the car to slow down? By the W-E theorem, if friction does no work, the friction would not contributte to the loss of kinetic energy of the car...

    As you can see, I am reallu puzzled about this issue...
     
  9. Dec 14, 2015 #8

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    They both agree that friction is doing zero work. They just decompose that work into linear and rational parts differently, by choosing a different reference point.
     
  10. Dec 14, 2015 #9
    I see...

    So in Morin, the reference point is the contact point between the wheel and the road. But in Kleppner- Kolenkow, the referece point is the center of mass of the drum. Am I right?

    Nevertheless, when we change the reference point to the center of mass, is the friction force is still acting over zero distance, or not? This issue is not still clear to me...

    Thank you so much.
     
  11. Dec 14, 2015 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're missing a subtlety in K&K's solution. They integrate ##f## with respect to the angle around the drum. From the drum's reference frame ##f## works it's way round the circumference and does work equal to ##fb\theta##. And that's what causes the rotational motion.

    Then, they take the force down the slope as ##F-f## where ##f## is doing negative work equal to the positive work it does above. Overalll, therefore, the work done is:

    ##(Fl - fl) + fl##

    You can now interpret this two ways:

    a) Friction is doing no overall work.
    b) Friction is transforming linear KE into rotational KE by doing both positive and negative work at the same time.

    The problem is perhaps clearer if you look at the overall energy. ##E_{initial} = GPE = E_{final} = LKE + RKE##. With no energy lost to friction/heat. So that's clear. If you now reread K&K's final note, I think this is what they are trying to explain: friction in this case was not a dissipative force. They used it (rather cleverly) as a transformative force.
     
  12. Dec 14, 2015 #11

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The final point, which perhaps you may get from my post above, is that from the drum's perspective friction works its way round the circumference as a motive force.
     
  13. Dec 14, 2015 #12
    Mmmmm.

    I think now I understand it all.

    The key point is that Morin chooses as his reference point the point of contact between the wheel and the ground (which is instantaneously at rest); so for him, friction does no "linear" work nor "rotational" work at all. On the other hand, Kleppner and Kolenkow choose as their reference point the center of mass of the rolling drum (which, in the center of mass reference frame is at rest); so for them, friction does both "linear" and "rotational" work, which happen to be identical (but with opposite signs), and cancel each other, so the total work done by frction is zero.

    I find this argumentation plausible, but I need to work it out to check if it is right.

    Thanks A.T, and PeroK for your clever indications. If I had any other doubts, I would post them here as well.

    Regards.
     
  14. Dec 14, 2015 #13
    OK, I have tried to solve Kleppner - Kolenkow example taking the point of contact between the drum and the ground as the origin of the reference frame.
    That implies the new moment of inertia of the drum is I=ICM+Mb2=(3/2)Mb2. Now the problem is easier, because the W-E theorem for the rotational motion equation is enough to get the answer. Now the torque is only due to the weight, and not to the friction, so we get:

    ∫τ dθ=(1/2)Iω2


    bWsenβ θ=(1/2)(3/2)Mb2ω2

    , so:
    V=√(4gh/3)

    which is the same answer that Kleppner - Kolenkow obtain by using the W-E ecuations both for the lineal and rotation motion.

    Now, I have a doubt. Altough it is not necesary for the problem, how should I use the W-E theorem for the traslational motion with this new reference point (the point of contact between the drum and the ground)??
    ∫F dr=T2-T1
    Now the only force doing work is weight, because friction acts over zero distance, and normal force is perpendicular to the displacement. But, with this reference point, what is the kinetic energy?? I mean, what is the speed I shoud consider? The linear speed of the center of mass? The linear speed of that point? Both the rotational and the linear speed(in this case, what linear speed and why?)? Maybe the linear speed is zero (because this point is instantaneously at rest)?...

    Regards.
     
  15. Dec 14, 2015 #14

    Doc Al

    User Avatar

    Staff: Mentor

    One must be careful about interpreting the W-E theorem. Despite the name, it's not really a statement about "real" work, but of pseudo-work (sometimes called "center of mass" work). It is an application of Newton's 2nd law:
    [tex]F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)[/tex]
    Note that you are multiplying the net force times the displacement of the center of mass, which allows you to calculate the resulting change in the kinetic energy of the center of mass. In Morin's example, the friction provides the net force and you can use the theorem to calculate the change in KE of the car. The friction causes the car to slow, but it does no work on the car. (The real work would involve friction times the displacement of the point of application--which is zero, since there is no slipping.)

    The same analysis, of course, applies to an accelerating car. Friction causes the car to accelerate, but does no (real) work on the car (assuming no slipping). Nonetheless, the W-E theorem applies as before. The energy comes from within the car, not from the road. (But that external force from the road is certainly needed to create the motion of the car.)
     
  16. Dec 20, 2015 #15
    Here there's nothing like choosing a reference , its all about choosing a system - centre of mass or
    - Whole body
     
  17. Dec 20, 2015 #16

    All this is explaining that change in internal energy of the system is manifesting in form of energy , cause there's friction inspiring that !!
     
  18. Dec 20, 2015 #17
    Thanks fr your reply, Shreyas Samudra.

    I will try to work it out again keeping this in mind.

    What I still do not understand is the fact that, if friction is nor doing work, why it is sais that friction from the ground on the tires causes the car to slow down...
    In my textbooks, nobody mention pseudo-work, they just tlak about work. So by W-E theorem, if frction does no work, it should not cause the car to slow down...

    Thanks.
     
  19. Dec 21, 2015 #18
    Look,
    For a complicated system(where motion of all the parts of the system is not same), we can apply work energy theorem in 2 ways
    1] WE theorem for COM, i.e you treate the whole body as a point particle, and find work done by all the forces on COM i.e integral of force times dx(COM)
    So you will equate all this to change in KE of COM (here you will not include rotational KE of the body as, now your body is a point particle)
    2]Apply WE theorem for whole body , here rotational KE will be considered and work done by those forces which are not causing displacement at the point of contact will be zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook