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Homework Help: Find M, Mx, and My for a thin plate if density=ky

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Set up integrals to find M, Mx, and My for the thin plate if d(x)=ky

    bounded by y=2-x and y = 4-x^2 or x = sqrt(4-y)

    2. Relevant equations

    M = [itex]\int[/itex] dm = [itex]\int[/itex]d(x)dA
    Mx = [itex]\int[/itex]ydm
    My = [itex]\int[/itex]xdm

    3. The attempt at a solution
    I broke this down into three equations. x1 = 2-y; x2 = -sqrt(4-y), x3= sqrt(4-y)
    I used horizontal elements so that d(x) is constant across the entire element.

    When calculating M, I got [itex]\int[/itex](ky)(x2-x1)dy = [itex]\int[/itex](0→3)(ky)(sqrt(4-y))-(2-y)dy + [itex]\int[/itex](3→4)(ky)(sqrt(4-y))-(-sqrt(4-y))dy

    I got this down to:

    k[itex]\int[/itex](0→3)(y2-2y+ysqrt(4-y))dy + 2k[itex]\int[/itex](3→4)(ysqrt(4-y))dy

    The problem is basically that I can't figure out how to integrate these to find the total mass. Please help thanks!
     
  2. jcsd
  3. Nov 19, 2012 #2

    haruspex

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    I feel it would be much simpler to handle this as a double integral, integrating wrt y first. That will avoid surds in the bounds.
     
  4. Nov 19, 2012 #3
    It very well may, I just haven't learned anything about that yet.

    EDIT -- I am looking up double integrals, how would you apply it here?
     
    Last edited: Nov 19, 2012
  5. Nov 19, 2012 #4

    haruspex

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    In x-y co-ordinates, an element of area is dxdy (or dydx - either will do).
    So the mass of lamina covering a region A is [itex]\int_{(x,y) \in A}ρ(x,y) dx dy[/itex]. If the area is bounded by f(x)<y<g(x) and a<x<b then you can write M = [itex]\int_{x=a}^b\int_{y=f(x)}^{g(x)}\rho(x,y) dy dx[/itex]
     
  6. Nov 19, 2012 #5
    Ok I'm in my first calculus class and some of unfortunately I don't even recognize some of your notation. But I do understand somewhat and will try to use this. Thanks very much.

    ok i am trying to figure this out. but my density function is ky, where k is some constant. it doesnt have an x value
     
    Last edited: Nov 19, 2012
  7. Nov 20, 2012 #6

    HallsofIvy

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    The plate is bounded by [itex]y= 4- x^2[/itex] and by [itex]y= 2- x[/itex] which intersect when [itex]y= 4- x^2= 2- x[/itex] or [itex]x^2- x- 2= (x- 2)(x+ 1)= 0[/itex]

    With density ky, the mass is given by
    [tex]M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx[/tex]
     
  8. Nov 20, 2012 #7

    ok I watched a couple videos on youtube and this is what I came up with:

    [tex]M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx[/tex]
    ⇔ k[itex]\int_{y=2-x}^{4- x^2} ydy[/itex]
    = k (y2/2) = k [(4-x^2)/2 - (2-x)/2]
    = k/2 [4-x^2 - 2+x]
    = k/2 [-x^2+x-2]

    then

    [tex]M= \int_{x=-1}^2 (k/2 [-x^2+x-2]) dx[/tex]
    = (k/2)[(-x^3)/3 + (x^2)/2 +2x]from -1->2
    = ... = 9k/4
    how bad is it?
     
    Last edited: Nov 20, 2012
  9. Nov 20, 2012 #8

    haruspex

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    That's My, right?
    No, you've dropped a power of 2. Setting y = 4- x2 in y2/2 gives (4- x2)2/2
     
  10. Nov 21, 2012 #9
    Ok I did it again making sure to get the powers correct and I got M = 44k/5.
    I calculated My to be 9k/2.

    When I divided My/M to get xbar I got 88/45 which can't be right!
     
  11. Nov 21, 2012 #10

    haruspex

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    I don't get quite the same. I have M = 31/6, My = 549/310, giving ybar (not xbar!) about 1.8. That's quite close to what you have, and seems perfectly reasonable to me.
     
  12. Nov 21, 2012 #11
    That's strange. I thought xbar = My/M ? Maybe Im getting my limits messed up, I evaluated My from 0->3, because from 3->4 is symmetric.
     
  13. Nov 21, 2012 #12

    haruspex

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    No. ∫∫ydydx/∫∫dydx must be computing an average value of y.
    What is from 0 to 3? x? The x range is determined by where the two curves meet.
     
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