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Circular motion and friction?

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A penny of mass 3.1g rests on a small 20.0g block supported by a spinning disk of radius(r) 12cm.If the coefficients of friction between block and disk are 0.75(static) and 0.64(kinetic) while those for the penny and block are 0.45(kinetic) and 0.52(static).What is the maximum rate of rotation (in revolutions per minute) that the disk can have before either the block or the penny starts to slip?

    2. Relevant equations

    Fr=[tex]\mu[/tex]R
    F=m(omega)2r
    where
    omega=angular velocity

    3. The attempt at a solution

    Here's the force diagram I drew,
    http://img24.imageshack.us/img24/8028/cicularmotion.png" [Broken]

    but I'm not very sure what I should do next.
    I guess I should use Fr or Fr'=[tex]\mu[/tex]R(or S as labelled in the diagram)
    but I don't know which [tex]\mu[/tex] to use,the coefficient of static friction or kinetic friction ? and what is the difference between the 2 types of coefficients?
    I know that static friction is the friction acting on the block when it is just about to move and is slightly more than the dynamic friction,but I'm not sure what to do with this info?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 30, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Apply Newton's 2nd law to the penny. What's the only centripetal force acting on it? What's the maximum value of that force before slipping occurs? Use that to find the maximum ω.

    Since you are trying to avoid slipping, static friction is all you need. (Assuming I understand the problem correctly.)
     
  4. Jul 30, 2009 #3
    Does this mean, the penny slips first?If so how do we decide it?

    The frictional force between the penny and the block,I guess?

    If my answer to the 1st question was correct,then it would be,
    Fr' = [tex]\mu[/tex]s * S which would be equal to mω2r
    =[tex]\mu[/tex]s*m'g = mω2r
    ω2 = (0.52*10)/(12*10-2)
    ω= 6.58rads-1

    =(6.58*60)/2pi = 62.83 rpm ?
     
  5. Jul 30, 2009 #4

    Doc Al

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    Staff: Mentor

    Assume the penny does not slip first, then do a similar analysis for the block. Compare the two speeds.

    Sure.

    Looks good. (Are you supposed to use g = 10 m/s^2, instead of the usual 9.8?)
     
  6. Jul 30, 2009 #5
    We're asked to take g=10 if the question doesn't state g as equal to 9.8m/s^2

    OK.So
    Fr= [tex]\mu[/tex]s" * R = Mω2r
    But now I'm a bit confused.Is the reaction force(R) on the block equal to the weight of the block alone or equal to the weights of both block and disk?
    and is M= m+m' ?
    where m' is the mass of the penny and m the mass of the block
     
  7. Jul 30, 2009 #6

    Doc Al

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    Staff: Mentor

    I assume you mean "penny", not "disk". The normal force exerted by the disk on the block will equal the combined weight of block + penny.
    Yes, treat the "block+penny" as a single system. (Assume the penny doesn't slip.)
     
  8. Jul 30, 2009 #7
    Yes,I meant disk.

    Assuming the penny doesn't slip,I get ω= 25 rads-1 which is more than the maximum angular velocity for the penny which would mean the penny slips first.

    So in all such cases is it safe to assume the object on top slips first?
     
  9. Jul 30, 2009 #8

    Doc Al

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    Staff: Mentor

    Yes, you meant penny. :wink:

    Right.
    Not at all. It depends on the coefficient of friction. (Swap the μs and see what happens.)
     
  10. Jul 30, 2009 #9
    Oops!I did it again.

    Tried it.now I understand it.

    THANK YOU VERY MUCH!
     
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