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Uniform Wire Question

  1. Feb 20, 2005 #1
    Problem here:
    http://www.quantumninja.com/hw/random/problem4.jpg

    I was trying to figure out how to go about this problem. So far I have come up with

    center of mass=
    [tex] \frac{L}{4}cos\frac{\beta}{2} [/tex]

    Torque=t
    [tex] \sum t=I\alpha [/tex]
     
  2. jcsd
  3. Feb 20, 2005 #2
    Are you sure this is right
    [tex]cm = \frac{L}{4}cos\frac{\beta}{2} [/tex]
    If [itex]\beta[/itex] goes to zero then the cos will go to one and you will be left with [itex] \frac{L}{4}[/itex] which does not seem right to me for a stright rod.
     
  4. Feb 20, 2005 #3
    I thought that was right
     
  5. Feb 20, 2005 #4
    Actually maybe it would be
    L/4*cos(b/2)+L/2
    center of mass
    [tex] \frac{L}{4}cos\frac{b}{2}+\frac{L}{2} [/tex]
     
    Last edited: Feb 20, 2005
  6. Feb 20, 2005 #5
    I am sorry you are right. If beta goes to zero the rod becomes lengh of L/2 and the center of mass would the be a L/4.
     
  7. Feb 20, 2005 #6
    Which one is right the first or 2nd one
     
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