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Can't use formula for kinetic energy 1/2*mv^2?

  1. Nov 19, 2015 #1
    upload_2015-11-19_15-50-29.png

    Question number 1; find the angular speed.

    I imagine a point mass at the very top, it has potential energy mgL, this is converted to kinetic energy 1/2*mv^2.
    I put them equal to each other and say that v=ωL. This gives me ω=sqrt(2g/L), the answer is supposed to be ω=sqrt(3g/L).


    upload_2015-11-19_15-55-36.png

    Why cant I do it like I did it? Is it wrong to use the formula for kinetic energy (1/2)mv^2?
     
  2. jcsd
  3. Nov 19, 2015 #2

    phyzguy

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    Your solution assumes all of the mass is at the top of the bar. Actually the mass is distributed along the bar. You could do it your way by assuming a string of point masses, each of length dL and mass m/L dL, then integrating along the length of the bar.
     
  4. Nov 19, 2015 #3
    You can't use the formular for translational kinetic energy, because every point of the rod has a different velocity depending on its position. You would have to integrate the kinetic energy of all the infinte small slices of mass along the rod.
     
  5. Nov 19, 2015 #4
    But if mgL=1/2*mv^2, then I can just cross out m on both sides, and it shouldnt have anything to say?
     
  6. Nov 19, 2015 #5
    No, that's not possible, because in the formula for translational kinetic energy the entire mass (every mass point) is supposed to have the same velocity.
     
  7. Nov 19, 2015 #6

    phyzguy

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    You are right that the m cancels out, but each point along the rod has a different L and v. Think of a point 1/2 way out. It is at a distance L/2 from the pivot.
     
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