Finding the center of mass of laminae

[dlivingston]

Calculus 3 project – any and all help is appreciated.
We've just gone over center of mass with double integrals, so it's a bit peculiar to see this project feature only one integral. I went over that in calc 2 – and as a result, know how to calculate it that way.

However (as you will see in the formula below), I don't think he wants me to calculate it that way. I have no idea how to proceed and the book is of absolutely no help.

Anything you could do to help me out would be fantastic.

And here's the problem:

Homework Statement

There are three type of laminae we are trying to find:
1) A triangle with sides 3", 4", and 5" respectively.
2) A semicircle
3) A "horseshoe", i.e. a half-ring, whose inner and outer edges are composed of circular arcs of radii 3" and 5" respectively.

Definition 1:
The first moment of a planar mass distribution about a line l is the integral over the region of the (area) density σ(P) times the distance of P from the line. For example, the first moment about the line x=k is
M_k=∫_R\left(x-k\right) σ\left(P\right) dA

Definition 2:
In order for a line to be a balancing line, the first moment about this line must be zero.

Homework Equations

See the M_k above
I found, I believe, the equations for the laminae:
For the triangle,
y=4-\frac{4}{3}x
For the semicircle,
x^2+y^2=1
And the horseshoe,
x^2+y^2=25
x^2+y^2=9

The Attempt at a Solution

Welp, I'm lost. As I said, I can figure it out easily using the calculus 2 method of x_{cm}=\frac{M_y}{M}, but I'm not sure the way he wants it done.

What's the density formula? Is it just the equations I listed above? What's this (x-k) nonsense?

 

[dlivingston]

*cricket*
Anyone?

 

[gopher_p]

The M_k gives, roughly speaking, the potential torque on the lamina if one were to try to balance it on the line x=k. In order for the lamina to be balanced on that line, this "tourque" would need to be zero.

If you set M_k=0 and solve for k, I think you'll see what the "x-k nonsense" is for.

 

[HallsofIvy]

*cricket*
Anyone?

Wow, you waited a whole hour before bumping? Sorry, I was eating supper!

 

[HallsofIvy]

Calculus 3 project – any and all help is appreciated.
We've just gone over center of mass with double integrals, so it's a bit peculiar to see this project feature only one integral. I went over that in calc 2 – and as a result, know how to calculate it that way.

However (as you will see in the formula below), I don't think he wants me to calculate it that way. I have no idea how to proceed and the book is of absolutely no help.

Anything you could do to help me out would be fantastic.

You are not given any "density" function so this is a purely mathematical problem, not physics. You are really looking for the "centroid", not the "center of mass". That is the same as assuming the "density" is a constant.

And here's the problem:

Homework Statement

There are three type of laminae we are trying to find:
1) A triangle with sides 3", 4", and 5" respective
2) A semicircle
3) A "horseshoe", i.e. a half-ring, whose inner and outer edges are composed of circular arcs of radii 3" and 5" respectively.

Definition 1:
The first moment of a planar mass distribution about a line l is the integral over the region of the (area) density σ(P) times the distance of P from the line. For example, the first moment about the line x=k is
M_k=∫_R\left(x-k\right) σ\left(P\right) dA

Definition 2:
In order for a line to be a balancing line, the first moment about this line must be zero.

Homework Equations

See the M_k above
I found, I believe, the equations for the laminae:

This is a single lamina. "laminae" is the plural.

For the triangle, <br /> y=4-\frac{4}{3}x<br /> For the semicircle,<br /> x^2+y^2=1<br /> And the horseshoe,<br /> x^2+y^2=25<br /> x^2+y^2=9<br /> <br /> <h2>The Attempt at a Solution</h2><br /> Welp, I&#039;m lost. As I said, I can figure it out easily using the calculus 2 method of x_{cm}=\frac{M_y}{M}, but I&#039;m not sure the way he wants it done.

<br /> Well, you mean x_{cm}=\frac{M_x}{M}, and y_{cm}=\frac{M_y}{M}<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> What&#039;s the density formula? Is it just the equations I listed above? What&#039;s this (x-k) nonsense? </div> </div> </blockquote> x- k is, just your quote above says, the horizontal distance from point (x, y) to the line x= k.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 1) A triangle with sides 3&quot;, 4&quot;, and 5&quot; respectively. </div> </div> </blockquote> So this is a right triangle. Set up a coordinate system so the right angle is at the origin, one vertex is to (3, 0), and another at (0, 4). The equation of the hypotenuse is y= -(4/3)x+ 4. The area of the triangle is, of course (1/2)(3)(4)= 6 so the x-coordinate of the centroid is (1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx and the y coordinate is (1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 2) A semicircle </div> </div> </blockquote> The area of a semicircle of radius R (you don&#039;t give a radius) is \pi R^2/2. It&#039;s probably best to use polar coordinates to do the integrals:<br /> (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} x d\thetadr= (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} rcos(\theta) d\thetadr<br /> (It might occur to you that, by symmetry, the x coordinate of the centroid is 0)<br /> and<br /> (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} y d\thetadr= (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} rsin(\theta) d\thetadr<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 3) A &quot;horseshoe&quot;, i.e. a half-ring, whose inner and outer edges are composed of circular arcs of radii 3&quot; and 5&quot; respectively. </div> </div> </blockquote> So this is the area between two semi-circles. Its area is (pi(5)^2/2- \pi(3)^2/2= 8\pi.The coordinates of the centroid are<br /> (1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r cos(\theta)d\theta dr <br /> (Again, by symmetry, this should be 0.)<br /> and<br /> (1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r sin(\theta)d\theta dr

 

[dlivingston]

Ah, you both are absolutely fantastic! Thank you.