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Kinetic Energy Body

  1. Oct 9, 2013 #1
    Hi All,
    I have a very simple question on a calculation, I would not doubt it correctness if it were not for the fact I am told by a colleague it is wrong, and I cannot see why (good job college times are gone).

    We have a cylinder of lenght L, density $$\rho$$ constant along the cylinder, one end is clamped, one end is moved according to a function $$\delta(t)$$ describing the imposed displacement over time.
    Each section is supposed to move as described by the function $$u = \frac{\delta}{L} x$$, u being the displacement.
    Then the total kinetic energy equals
    $$E = \frac{1}{2} \rho \int_{0}^{L} \dot{u(x)}^{2} \mathrm{d}x$$
    yielding $$E = \frac{1}{6}\rho L \dot{\delta}^2$$, so 3 times less than the kinetic energy of a translation at speed $$\delta$$.
    It seems all reasonable to me...maybe some rest will help..
    Thanks!
     
  2. jcsd
  3. Oct 9, 2013 #2

    phyzguy

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    You have done it correctly. What you have described is a cylinder rotating about one end by an angular frequency:
    [tex]\Omega = \frac{\dot{\delta}}{L}[/tex]
    It has a kinetic energy given by:
    [tex]KE = \frac{I \Omega^2}{2} = \frac{M L^2 \dot{\delta}^2}{2 \times 3 L^2} = \frac {M \dot{\delta}^2}{6} = \frac {\rho L \dot{\delta}^2}{6}[/tex]
    Where the moment of inertia of the cylinder about one end is given by:
    [tex]I = \frac{M L^2}{3}[/tex]

    This is less than the kinetic energy of the whole cylinder moving at the speed δdot because one end is moving at this speed and the other end is motionless, with the part in between moving at intermediate speeds. Where is the problem?
     
    Last edited: Oct 9, 2013
  4. Oct 9, 2013 #3
    What kind of density is your ρ? Linear density (m/L)?
     
  5. Oct 10, 2013 #4
    Phyzguy, many thanks for your reply. The probelm is that I was not referring to a rotating cylinder, but to one clamped to one end and extended from the other one..A strethcing rod...
    Nasu, yo are correct I should have stated it explicitly, it is a linear density (such that the mass pertaining to a differential element is $$\rho \mathrm{d}x$$.
    thanks
     
  6. Oct 10, 2013 #5

    phyzguy

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    Ah, I see. In that case, the problem with your analysis is that ρ is not constant, but is a function of space and time. You need to write an expression for ρ(x,t) and move it inside the integral.
     
  7. Oct 10, 2013 #6
    Phyzguy,

    I understand (hopefully) what you say, for the strain in the cylinder will alter the density.
    That is arguably true, but I missed to mention the calculation is done in the frame of infinitesimal elasticity, within which strains are <<1 and the variation in density is negligible.

    Thanks
     
  8. Oct 10, 2013 #7

    phyzguy

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    So then what is the question? What does your colleague say is incorrect?
     
  9. Oct 11, 2013 #8
    There is no problem actually, faced with the question it turned out the collegaue was wrong...thanks anyhow!
     
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