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Minimum coefficient of static friction

  1. Nov 15, 2013 #1
    1. The problem statement, all variables and given/known data
    I know the diameter is 30 m and the radius is 15
    The period for one revolution I suppose is some multiple of 7.03


    2. Relevant equations



    3. The attempt at a solution

    I couldn't think of an equation where friction was involved where rotational motion was concerned. I used 2pir/7.03 to answer b but of course it was wrong.
     

    Attached Files:

  2. jcsd
  3. Nov 15, 2013 #2
    There is a downward force due to gravity which is balanced by a frictional force. This frictional force is proportional to the central force caused by the rotation (the constant of proportionality is the friction coefficient. Can you find an equation for the central force?
     
  4. Nov 15, 2013 #3
    Fkn ???
     
  5. Nov 15, 2013 #4
    This equation will be used. But first you must calculate the central force. It is given by
    F=m v2 / R. This gives the force normal to the rotating surface.
     
  6. Nov 15, 2013 #5
    So the central force is the same as the centripetal force???
     
  7. Nov 15, 2013 #6
    Yes. Different names for the same thing
     
  8. Nov 15, 2013 #7
    okay because I've never heard central force. So how can I use that equation when I don't know the mass or velocity?
     
  9. Nov 15, 2013 #8
    You know that he returns to the same point after 7.03 seconds. How far has he travelled? So what was his speed?

    Leave mass in as an unknown for now. Maybe you will have to deal with it later, maybe not...
     
  10. Nov 18, 2013 #9
    how do I calculate how many rotations he made in 7.03 seconds or do I assume it was one?
     
  11. Nov 19, 2013 #10
    I would assume it's 1.
     
  12. Nov 19, 2013 #11
    okay so V= (2piR)/t ??
     
  13. Nov 19, 2013 #12
  14. Nov 19, 2013 #13
    okay so the Normal force is equal to mass * acceleration, a= (v^2)/r so N= mv^2 * 1/r .....right?
     
  15. Nov 19, 2013 #14
    Yes that's correct
     
  16. Nov 19, 2013 #15
    okay friction= μN

    μ((mv^2)/r)=mg

    m cancels out on both sides and μ=gr/v^2

    μ=gr/v^2

    μ= ((9.8)(15))/((2∏*15)/(7.03))^2

    μ= .818

    what about part B??
    I really appreciate your help also
     
    Last edited: Nov 19, 2013
  17. Nov 19, 2013 #16
    Is the guy accelerating downward? So is there a vertical force acing on the guy? You have worked out the equation for the centripetal force (or acceleration), how fast would he have to be going for this force to equal 4 G's?
     
  18. Nov 19, 2013 #17
    I don't understand what 4 G's is...how would I know if he is accelerating downward?
     
  19. Nov 19, 2013 #18
    G is the gravitational acceleration at the Earth's surface. i.e 9.8 m/s^2.

    What would be the consequence of accelerating downward in a room with no floor? Does this happen to the man?
     
  20. Nov 19, 2013 #19
    oh okay I didn't deduce that from the question, don't laugh at me but I thought he was referring to the price of the special suit at 4g's meaning $4000. This question totally doesn't make any sense to me. The consequence of accelerating downward in a room with no floor would be a fall right? Where would I find the clue to tell if it had happened to the man.
     
  21. Nov 19, 2013 #20
    so actually it's 4*9.8???
     
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