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Homework Help: Mass of chain in shape of parabola

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known dataA chain in the shape of [tex]y = x^{2}[/tex] between x = -1 and x = 1, has density of |x|. Find M, and CM.

    2. Relevant equations

    3. The attempt at a solution

    [tex]\int^{1}_{-1}|x|dx = \int^{0}_{-1}-xdx + \int^{1}_{0}xdx = 1[/tex]

    I got this far and realized that I did nothing with the function itself. :redface:

    The formulas I've used before deal with masses bounded by curves, not the mass of the curves themselves.
  2. jcsd
  3. Mar 23, 2010 #2
    Does this help? are you ok with finding the length, THEN using the weight of chain per foot, finding total weight?


    Good Luck,
    LarryR : )

  4. Mar 23, 2010 #3
    No, no. You're on the right tack. Notice that the density of the chain is independent of y, so you don't need to know the function to get the mass.
  5. Mar 24, 2010 #4
    So was I right the first time? How do I find the center of mass?

    I redid the problem using

    [tex]\int^{1}_{-1} \rho\sqrt{1+(y')^{2}} dx[/tex]

    I then threw an x in there to find the moment and the integral became really nasty.

    Intuition tells me that it would be at 0,0 because the function and the density function are both symmetric across the y axis so both sides should weigh the same. Right?
  6. Mar 24, 2010 #5
    Yes, you got it right the first time; the mass of the chain is 1.

    Use the symmetry of the problem to help solve for the center of mass. By symmetry, the center of mass in the x direction is at y-axis. The center of mass for each leg, in the y direction will be the same for each leg so you only need concern yourself with one leg of the parabola.

    You are given rho as a function of x, what is rho as a function of y for the first quadrant? Integrate the mass times displacement from y=0 to y=1.
  7. Mar 24, 2010 #6


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    No, the infinitesimal length element of the parabola is not just [itex]dx[/itex]. To get the mass, you need to integrate the density over the length of the parabola.
    Last edited: Mar 24, 2010
  8. Mar 24, 2010 #7


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    This is the correct method. Break the moment integral into two intervals and use the substitution [itex]x\to-x[/itex] on the negative interval.

    You are half correct, the x-coordinate of the CoM will be zero, because the density at the chain are symmetric about the y-axis. The y-coordinatee need not be zero, and your intuition should tell you that it will be positive since most of the chain is at y>0 and its density is always positive. You can verify this by evaluating both the x and y moment intervals.
  9. Mar 24, 2010 #8
    reread the problem statement
  10. Mar 24, 2010 #9


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    I did, and my comment still stands. What makes you think otherwise?
  11. Mar 25, 2010 #10
    I took my own advise and reread the problem statement. I been more than just unhelpful, spreading confusion. I'm going to go attend to my depleted ego now.
  12. Mar 25, 2010 #11


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    Lol, we all make mistakes. I know I've made my fair share.
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