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Centre of mass question.

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    determine the position of the centre of mass of a flat sheet of metal of homogeneous density rho. The sheet lies within the area bounded by the equations y=x^4 and y=5

    2. Relevant equations
    integral centre mass

    3. The attempt at a solution
    a slightly trivial question, i know how to set up the correct integral when finding the centre of mass below the function curve bound by an x=? but with the area above the curve in this manner im finding problems. Should i treat the graph on its side or change the graph form to fit in the other direction. or find the integrals below both functions and do a subtraction to find the positions?
  2. jcsd
  3. Sep 16, 2010 #2


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    First, draw a picture!

    Then either divide the area into horizontal slices of thickness [itex]dy[/itex], express the length of a slice at a give height [itex]y[/itex] as a function of [itex]y[/itex] and integrate, or divide it into veritical slices of width [itex]dx[/itex], express the height of a slice at a given [itex]x[/itex] as a function of [itex]x[/itex] and integrate.
  4. Sep 16, 2010 #3
    thats the method im used to, on which ive found simple enough with other examples but i cant seem to build the correct integral with this set up.
  5. Sep 16, 2010 #4


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    Show your attempt.
  6. Sep 17, 2010 #5
    ok so im not sure how to integrate for the area above the line, so here's my attempt.
    i dont have a scanner so i dont have a diagram to show you. I turned the graph on its side and reversed the axis (tried to cheat a bit here) so my x co ords would be my y.

    y=x^4, y=5

    switching it, x=4th root (y) and x=5 but where i have "y" its my new x axis.

    so using this i did the integration for the horizontal position as normal,
    [integral xydx]/[integralydx]
    doing this i get 5 as my horizontal co-ord which is actually my vertical since i switched it, here i can already see ive gone wrong. but i followed through with the vertical anyway and obtained 0.66 as my other position.

    i made a second attempt with the limits +/- 4th root 5, integrating first y=5 then y=x^4 in both x and y directions then subtracting one from the other to find the positions in the area bounded. but i can see already from the first integration that this is also wrong.

    how do i obtain the position of the centre of mass in this bounded area above the function line??

    thanks in advance.
  7. Sep 17, 2010 #6


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    What were your limits of integration?
  8. Sep 17, 2010 #7
    for this attempt i used 5, 0 as my limits treating the y=5 as x=5 bound by the 4th root curve and the origin.
  9. Sep 17, 2010 #8


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    Your problem is that the length of each ribbon with thickness [itex]dx[/itex] isn't just [itex]y[/itex], you need to use a double integral. For the area of the region, you should have

    [tex]A=\int_0^5 \int_{-x^{1/4}}^{x^{1/4}}dydx[/tex]

    Do you see why?

    For the horizontal and vertical positions of the CoM, you just insert either x or y (and rho) into that double integral and then divide by the area.
  10. Sep 17, 2010 #9
    sorry i think im missing something can you explain why i need to use the double integral, ive not been working for a long time now my maths is rusty. so i need to use the double integral but in the same way with the surface area density process, for the area should i sub in for y, y=x^4 as normal you should i do an integral subtraction to get the area of the enclosed space.
  11. Sep 18, 2010 #10
    rho=m/A, so then dm=rho*dA, which means that
    [tex]m=\int \int \rho dA[/tex]


    [tex]x_{cm}=\frac{\int \int \rho x dA}{m}[/tex]


    [tex]y_{cm}=\frac{\int \int \rho y dA}{m}[/tex]

    Oh yeah, and hint: before you even start, what should x_cm be?
  12. Sep 19, 2010 #11
    should there be a integral on the bottom as well? [tex]\int[/tex]dm? by xcm do you mean the limits of integration for the x position or should i be able to intuitively understand where the x position should be? at a guess i would say the origin on the axis, and therefore do i need to only calculate the y position?
  13. Sep 19, 2010 #12
    Yes, exactly, intuitively you should be able to guess that the x_cm will be on the axis, and mathematically you see it very soon because you are integrating an odd function over symmetric bounds.

    There is an integral on the bottom, yes, it's the integral for m I have at the beginning of my post.

    Oh yeah, and also you don't have to do the exact integral that gabba posted, you could integrate with respect to x first if you really wanted. You'd have to do slightly more work to get the right x bounds.
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