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

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In summary: That is, by solving 4- x2= 2- x.)Yes from 0 to 3 is x. x = y2+ y- 2= (y+ 2)(y- 1). From y= 1 to y= 2, x= 0 to 3. From 3 to 4, x= 2 to 3. Symmetry would say that xbar would be the x value of the point of intersection of the two curves. I don't know what that is (I'm not going to try to find it using the "quadratic equation" - I'm too lazy for that!) but it certainly is not 88/
  • #1
cloudboy
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Homework Statement


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)

Homework Equations



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

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!
 
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  • #2
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.
 
  • #3
haruspex said:
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.

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?
 
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  • #4
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]
 
  • #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 doesn't have an x value
 
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  • #6
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]
 
  • #7
HallsofIvy said:
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]
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?
 
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  • #8
cloudboy said:
[tex]M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx[/tex]
That's My, right?
⇔ k[itex]\int_{y=2-x}^{4- x^2} ydy[/itex]
= k (y2/2) = k [(4-x^2)/2 - (2-x)/2]
No, you've dropped a power of 2. Setting y = 4- x2 in y2/2 gives (4- x2)2/2
 
  • #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!
 
  • #10
cloudboy said:
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!
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.
 
  • #11
haruspex said:
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.

That's strange. I thought xbar = My/M ? Maybe I am getting my limits messed up, I evaluated My from 0->3, because from 3->4 is symmetric.
 
  • #12
cloudboy said:
That's strange. I thought xbar = My/M ?
No. ∫∫ydydx/∫∫dydx must be computing an average value of y.
Maybe I am getting my limits messed up, I evaluated My from 0->3, because from 3->4 is symmetric.
What is from 0 to 3? x? The x range is determined by where the two curves meet.
 

1. What is the meaning of "M" in this equation?

In this equation, "M" refers to the mass of the thin plate. It is a measure of the amount of matter present in the plate.

2. What does "Mx" represent in this equation?

"Mx" represents the mass of the thin plate in the x-direction. This means that it is the amount of mass present along the horizontal axis of the plate.

3. How is "My" different from "Mx" in this equation?

"My" represents the mass of the thin plate in the y-direction. This means that it is the amount of mass present along the vertical axis of the plate. "Mx" and "My" are different because they represent different directions in which the mass is distributed.

4. What is the value of "k" in this equation?

In this equation, "k" represents the density of the thin plate. It is a constant value that is specific to the material of the plate and is used to calculate the mass of the plate.

5. How can I find the values of M, Mx, and My for a specific thin plate?

To find the values of M, Mx, and My for a specific thin plate, you will need to know the dimensions of the plate and the density of the material it is made of. You can then plug these values into the equation M = kxy, where "x" and "y" are the dimensions of the plate, to calculate the mass and its distribution in the x and y directions.

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