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Angular Momentum of a rotating mass

  1. May 28, 2016 #1
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    1. The problem statement, all variables and given/known data

    When the 3.2-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force F on the cord is increased, the bob rises and then rotates around the horizontal circular path B. (picture included)

    2. Relevant equations
    L = I ω v = rω

    3. The attempt at a solution
    The solution from the book goes into a long procedure with summation of forces to find the angle etc. I tried this

    Since there is no moment about the z axis angular momentum is conserved about the z axis. I1ω1 = I2ω2
    For a particle I= mr2 so initially, I = (3.2)(0.6sinθ)2 and after I = (3.2)(0.3sinθ)2. For ω, initially it is (0.6)(sinθ)(1.5m/s) at the end is is (0.3sinθ)(v2)

    Setting these equal to each other and solving gives v=12m/s But the answer is 2m/s Not sure where im going wrong. Thank you!
     

    Attached Files:

  2. jcsd
  3. May 28, 2016 #2

    Merlin3189

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    Gold Member

    You don't seem to mark theta on your diagram. I assume it is the angle the chord makes with the z axis. If so, it will not be the same in both positions.
    As the tension increases, the "orbit" becomes more horizontal and theta changes.
     
  4. May 28, 2016 #3
    Ah that would explain it. I think i need to find theta using newton's laws. Then I can apply it to the conservation of angular momentum.

    Thank You.
     
  5. May 29, 2016 #4

    Merlin3189

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    Gold Member

    Remember that the reason the ball is not just spinning with the chord horizontal, is gravity. The tension in the chord is providing the opposition to gravity as well as centripetal force.
     
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