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Mass of bar

  1. Jul 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A thin uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energy of all these segments.


    2. Relevant equations
    K=Integral of 1/2 times dm v^2
    dm=M/L *dx


    3. The attempt at a solution

    I got stuck when trying to substitute dm for M/L*dx. why does dm/dx=M/L if M is the mass of the entire bar and L is the length of the bar?
     
  2. jcsd
  3. Jul 5, 2010 #2
    dm/dx = M/L
    Because, each of those quantities is the density. The density is an "intensive" property of a material, it doesn't matter if I have an ocean of water or a cup of water, they are the same density. (technically density would vary with depth in an ocean due to the enormous weight, but that is beside the point)

    So your dm/dx is a small portion of the rod but it should give the exact same ratio (same density) as using the entire rod does.

    Try it yourself. 12kg / 2m = 6kg/m
    The density is 6...
    Ok, now use half the rod
    6kg / 1m = 6kg/m
    Or a fourth of the rod
    3kg / 0.5m = 6kg/m
    it will always be 6kg/m even with infinitesimal quantities like dm and dx.

    [EDIT] Also keep in mind that the actual density would have to be in meters cubed, not just meters. But in this case all we are concerned with is "linear density".
     
  4. Jul 5, 2010 #3
    Thank you much, that makes sense :)
     
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