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Circular Motion

  1. Apr 19, 2009 #1
    Circular Motion!!

    1. The problem statement, all variables and given/known data
    A bead of mass m moves on a smooth circular ring of radius a which is fixed in a vertical plane. Its speed at A, the highest point of its path, is v and its speed at B, the lowest part of its path, is 7v.

    a) show that v= [square root](ag/12)

    b) find the reaction of the ring on the bead, in terms of m and g, when the bead is at A


    2. Relevant equations
    F=mv 2 /r
    EK= 1/2mv2
    EP= mgh

    3. The attempt at a solution

    a)
    At point A
    EK= 1/2mv2
    EP= 2mga

    At point B
    GPE= 0
    EK= 1/2m(7v)2

    (Conservation of energy)
    1/2m(7v)2 = 1/2mv2 + 2mga
    (m's cancel)(multiply through by 2)
    49v2= v2 + 4ag
    48v2= 4ag
    v2= 4ag/48 = ag/12
    v=[square root](ag/12)

    That im fine with, its part b which confuses me!

    b)
    mv2/r + mg = R
    (r=a)(v2= ag/12)
    mag/12a + mg= R
    mg/12 + mg= R
    R=13mg/12

    But that isnt correct!! any help would be greatly appreciated =]
     
    Last edited: Apr 19, 2009
  2. jcsd
  3. Apr 19, 2009 #2
    Re: Circular Motion!!

    Is there a sketch to the problem?
    I think R= mgcos(P) +....
    while we need to find angle P, or I'm imaging the case all wrong.
    but I'm not sure, If You could show me a sketch I could be greater help.
     
  4. Apr 19, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Circular Motion!!

    Careful with signs. I'll use positive = up and negative = down:

    ΣF = ma
    -mg + R = -mv²/a

    (and so on..)
     
  5. Apr 19, 2009 #4
    Re: Circular Motion!!

    is that becuase v^2/r = acceleration
    ahhh that would give you an answer of 11mg/12 which is the correct answer!! Thanks for your help =]
     
    Last edited: Apr 19, 2009
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