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Friction on a casing on a metal ring

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A casing of mass m can glide along a thin metal ring lying horizontally. It has an initial speed of v0. How long a distance s will it glide before stopping if its friction coefficient is μ?

    The answer is given as

    s = r/2μ * ln( v02 + √(v02 + g2r2)/ g * r )

    The task is to provide the deductive path to reach that answer.

    2. Relevant equations

    Centripetal force: mv2/r

    Gravitational force: mg

    Kinetic energy: mv2/2

    3. The attempt at a solution

    The casing will stop once all the kinetic energy has turned into friction work. The friction is a sum of that caused by the gravitational force and the centripetal force. The friction work Wf is thus

    Wf = s * F = s * μ * ( mv2/r + mg )

    Since Wf is equal to the kinetic energy we can write the following equation:

    s = Kinetic energy / F = (mv2/2) / (μ * ( mv2/r + mg ))

    This can be quite cleanly simplified into something that looks similar to the answer I am supposed to reach:

    s = r/2μ * ( v2/ (gr + v2) )

    But this is where I run into a brick wall. I can't figure out how to integrate it properly, all the similar equations I have seen integrated seem to include arctan somewhere in the answer.

    I apologise any mistakes I made in the translation of the problem and in laying them out clearly, and thank you for your attention!
    Last edited: Sep 24, 2013
  2. jcsd
  3. Sep 24, 2013 #2


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    I haven't solved this, just starting now, but already I am thinking the two forces must be combined as vectors not simply added. We have mg vertically and mv²/R horizontally so the combined force magnitude will be the common m times the square root of g² + (v²/R)². Also, the velocity is a variable, and it has a complicated non-linear acceleration μF so we are in for some trouble - can we even find the velocity function?

    This problem looks tough for me; I have long since forgotten my differential equations class 40 years ago so I hope others will pitch in!
  4. Sep 24, 2013 #3
    I am pretty sure that friction of this kind means that we do not have to look on the forces as vectors. What the friction coefficient is, as far as I am aware, is the portion of a perpendicular force that is turned into friction, and friction is by definition a force that is directly opposed to the movement of the object.

    If that made any sense.

    The velocity function sounds tricky to find. The way I figured we would get the value for s is by integrating the function with regards to v, with v going from v0 to 0. Since we know the start and end points of the interval I was hoping we would not need any closer information on how quickly the casing actually decelerates, but that depends on actually getting an analytical answer to the integral.

    I really appreciate your input by the way!
  5. Sep 24, 2013 #4


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    Friction depends on the normal force, pressing the slider against the ring. That is the vector sum of the horizontal and vertical forces so I'm pretty sure the normal force magnitude is square root of {g² + (v²/R)²} all right.

    I'm playing with this, writing F = ma, μFn = ma,
    a = dv/dt = μF/m where F = the square root above.
    This looks like it might be possible to integrate with a substitution. If you could get it to say
    dw/sqrt(1 + w²) = constants * dt
    could you integrate it? If so, try to come up with a w substitution that incorporates the R and a g so you can factor out the g from the square root and leave just (1 + w²) under the square root. If you need a hint, try letting w = v*sqrt(R) and see how it simplifies.
    Last edited: Sep 24, 2013
  6. Sep 24, 2013 #5


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    I did the integration and was able to solve it for v and integrate again to get x.
    But I didn't get anything like the answer given for the question!
    It is probably worth you working it through, though - I'm having a lot of trouble with calculations these days.
    Your turn to help me! Hopefully somebody else will pitch in.
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