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Static friction of ration of wheel

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    A constant horizontal force F of magnitude 10N is applied to a wheel of mass 10 kg and radius 0.3 m. The wheel rolls smoothly and the acceleration of its center of mass is 0.6 m/s^2. What is that frictional force on the wheel?


    3. The attempt at a solution
    ok so the frictional force is static friction and assuming it rolls to the right and taking the right as positive, we have (S = static friction force):

    ma=F-S
    S=F-ma = 10- (10)(0.6) = 4 N

    and I know that that's the right answer from my book. But can't you also find this using torques? (in the following, I=rotational inertia and R=radius, and taking clockwise as negative torque):

    [itex]\tau[/itex]=I[itex]\alpha[/itex]=-SR

    S= (-I/R)[itex]\alpha[/itex]

    [itex]\alpha[/itex]=-a/R

    I=(1/2)MR^2

    S=(1/2)Ma = (1/2)(10)(0.6) = 3N ....which is not the right answer...why?



    (sry meant "rotation" in title of thread not ration lol)
     
  2. jcsd
  3. Oct 23, 2011 #2

    Delphi51

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    It is an odd question. I don't think the friction it is asking for is anything to do with static friction of wheel on floor. The wheel is turning nicely on the floor, not trying to slip. The static friction is the potential grip the wheel has on the floor and would apply if you put on the brakes and tried to drag the wheel.

    It seems to me the question is asking for the overall friction on the motion as you initially worked it out to 4 N. It is likely due mostly to kinetic friction in the wheel bearing.
     
  4. Oct 23, 2011 #3
    Shouldn't the force depend on the rotational inertia of the wheel which was not given?
     
  5. Oct 23, 2011 #4

    Delphi51

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    Thanks Spinnor! You are absolutely right; rotational inertia should be worked into it, and the 4 N answer is an oversimplification. Curious that it is the "correct" answer.

    Oh, it doesn't give the moment of inertia (or info that could be used to calculate it). I guess we are suppose to ignore it.
     
    Last edited: Oct 23, 2011
  6. Oct 24, 2011 #5
    I did work rotational inertia into my second calculations where i got 3 N. But I think my mistake came in assuming that the rotational inertia is 1/2MR^2, which should be right since this is a wheel, but since they dont give it i guess you have to work with stuff they do give you.....
     
  7. Oct 24, 2011 #6
    according to my book, for a wheel to spin smoothly, there must be static friction between it and the surface of contact...and i am positive that they dont mean anything to do with the wheel bearing..
     
  8. Oct 24, 2011 #7

    Delphi51

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    Yeah, if there wasn't any friction it would slide instead of turning. If there was any moment of inertia. But it certainly isn't correct to use the whole static coefficient in the way you did. That would make it harder to pull the thing with turning wheels than dragging with locked wheels (kinetic coefficient).

    The moment of inertia depends on the construction of the wheel. If most of the mass is at the hub, the moment is very small.
     
  9. Oct 26, 2011 #8

    if it certainly isnt correct then how come it gives the correct answer? I saw the solution manual in class and they do it same way. and every similar problem is done using the whole static coefficient..
     
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