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Why rolling is less than kinetic

  1. Dec 8, 2004 #1
    The order is
    rolling < kinetic < Static.

    I understand why rolling is less than kinetic(Sliding).
    When a body rolls, the velocity of the point of contact is directed at an angle from the surface. Hence, the friction acts at an angle. Hence, rolling < kinetic. Am I right??

    Why is kinetic < Static???
     
  2. jcsd
  3. Dec 8, 2004 #2

    kreil

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    Friction is calculated by:

    [tex]F_f={\mu}F_N[/tex]
    [tex]{\mu}=\frac{F_f}{F_N}[/tex]

    Since you need to apply more force to an object to make it start moving than you do to keep it moving, Ff would be larger for a static mu than for a kinetic mu, making the fraction value larger for a calculated static coefficient.
     
  4. Dec 8, 2004 #3
    Why should you need more force to start an object than to keep it moving?
     
  5. Dec 8, 2004 #4

    arildno

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    1) Well, it really is unphysical if it were strictly otherwise:
    Think if you pushed a non-moving object with EXACTLY the force maximum static friction can produce (is).
    Static friction then yields, and kinetic friction takes over the stage:
    But if that force EXCEEDS maximum static friction (and, hence, your own pushing force), the object would start accelerating TOWARDS you rather than AWAY from you.
    That doesn't seem likely, or what? :wink:
    2) But, you could ask:
    Why couldn't maximum static friction be EQUAL in magnitude to kinetic friction?
    The answer to that is:
    It might well be, and certain material DO have that property almost realized
    (I think teflon is one of those material, but beware: I am not sure on that point!!)
    3) Lastly, let's look at the significance of kinetic friction being almost always strictly less than maximal static friction:
    Assume that your pushing force just balances maximal static friction (is greater than it by an ARBITRARILY SMALL quantity if you like!) :
    Friction then switches into kinetic, and Newton's 2.law states, in the horizontal, flat contact surface case:
    [tex]\mu_{s}mg-\mu_{k}mg=ma_{0}[/tex]
    (where [tex]\mu_{s},\mu_{k}[/tex] are friction coefficients, m mass of object, [tex]a_{0}[/tex] (initial) acceleration, and g acceleration of gravity.

    That is, given any pushing force strictly GREATER than maximal static friction, the acceleration "a" it produces necessarily obeys the inequality:
    [tex]a\geq{a}_{0}=(\mu_{s}-\mu_{k})g>0[/tex]
    That is, the object starts moving with a JOLT, not smoothly.
    (That can only be realized in the case where the coefficients of friction are EQUAL, where we can get ARBITRARILY SMALL initial accelerations)

    If you think about it, that's a fairly good modelling of our experience.
     
    Last edited: Dec 8, 2004
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