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Laws of motion and circular motion

  1. Aug 30, 2011 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img687.imageshack.us/img687/5055/imagelrl.jpg [Broken]


    2. Relevant equations

    Static friction = (9.81sin60 x 0.2)N
    Centripetal force = mr(w)²



    3. The attempt at a solution

    I am only abl to get part (i), need hints for part (ii) and (iii).

    Thanks in advanced
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 30, 2011 #2

    Hootenanny

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    What can you say about the forcing acting on the slider if it doesn't move, relative to the rod?
     
  4. Aug 30, 2011 #3
    If it doesn't move, I assume the centripetal force= the static friction
    Which is (cos30)mrw^2= (0.2)(sin60)9.81

    But this doesn't give me the answer given which is 4.07rad/s. Can highlight where did I go wrong?
     
  5. Aug 31, 2011 #4

    Hootenanny

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    You have only considered one component of the centripetal force: the component that acts parallel to the inclined rod. What about the other component that acts perpendicular to the inclined rod?
     
  6. Sep 4, 2011 #5
    Hi, how do I go abt finding the force by rod on the slider in part(iii)?
     
  7. Dec 21, 2013 #6
    Did I have anything wrong?
    mrω2=μkmgcos30+mgtan60
     
  8. Dec 21, 2013 #7

    adjacent

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    @haha1234, this thread is about 2 years old and @Hootenanny is no longer active on PF.
     
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