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Kinetic energy

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A thin square (4 ft side) metal sheet of homogeneous density ([tex]\sigma = M/A[/tex]is rotating around one of its diagonals at 10 rev/s. Develop a definite integral to express its kinetic energy.

    2. Relevant equations

    [tex]dK = \frac{1}{2}(r\omega)^{2}\sigma dA[/tex]

    3. The attempt at a solution

    I am using one side of the sheet, and plotting it as the area enclosed between:

    [tex]y_{1}=x[/tex]
    [tex]y_{2}=-x + 4\sqrt{2}[/tex]

    [tex]0\leq x \leq 2\sqrt{2}[/tex]

    Then:

    [tex]v^{2}=(20\pi x)^{2}[/tex]

    and my integral will be:

    [tex]200 \pi^{2}\sigma\int_{0}^{2\sqrt{2}} x^{2}(-x + 4\sqrt{2}-x) \text{d}x[/tex]

    This is half the total kinetic energy, by symmetry, so double the above should be the total.

    Is this correct?

    Thanks!
     
  2. jcsd
  3. Oct 30, 2009 #2

    Redbelly98

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    It looks pretty good, but I have 2 minor objections:

    1. The term (-x + 4√2 -x) can be simplified.

    2. The units in your expression,
    [/URL]
    would be lb-ft^2, which is not a unit of energy.
     
    Last edited by a moderator: May 4, 2017
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