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Velocity in a circular path

  1. Mar 30, 2012 #1
    1. All in the picture



    2.v2=u2+2as



    3. I have no idea where to start, I went through all the equations but i dont have time or velocity to sub in :(

    It also has circular motion which may or may not introduce pi. I didn't think this would be relevant but anyway.
     

    Attached Files:

  2. jcsd
  3. Mar 30, 2012 #2

    gneill

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    Staff: Mentor

    The question concerns energy and energy conservation. What forms of energy can you identify in the scenario?
     
  4. Mar 30, 2012 #3
    There is kinetic and potential gravitational energy
    so Ek=1/2mv2 but that needs velocity so:
    Ep=mgh but that only gives horizontal velocity whereas, I need all velocity.
     
  5. Mar 30, 2012 #4

    gneill

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    Staff: Mentor

    Why do you need velocity? What will velocity tell you?

    Why do you suppose that the height of the object only reaches 0.14 m after it breaks the pencil rather than returning to its starting height of 0.30 m? What has changed?
     
  6. Mar 30, 2012 #5
    for q 25 velocity is the answer
    and in 26 I need the work done
     
  7. Mar 30, 2012 #6

    gneill

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    Staff: Mentor

    Both questions involve changes in energy. Consider the energy changes taking place.
     
  8. Mar 30, 2012 #7
    Ep=mgh=Ek=1/2mv2
    1.176=0.4*9.8*0.3=1.176=0.5(0.4)(v)2
    (1.176/0.5)/0.4=v2=5.88
    √(5.88)=2.42=ANS!!!!
    YAY thanks gneill :)
     
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