Finding the center of mass of laminae

In summary: R^2). So the x-coordinate of the centroid is ( pi*R^2)/2 and the y coordinate is ( pi*R^2)/R.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 isM_
  • #1
dlivingston
16
0
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
[itex]M_k=∫_R\left(x-k\right) σ\left(P\right) dA[/itex]

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,
[itex]y=4-\frac{4}{3}x[/itex]
For the semicircle,
[itex]x^2+y^2=1[/itex]
And the horseshoe,
[itex]x^2+y^2=25[/itex]
[itex]x^2+y^2=9[/itex]

The Attempt at a Solution


Welp, I'm lost. As I said, I can figure it out easily using the calculus 2 method of [itex]x_{cm}=\frac{M_y}{M}[/itex], 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?
 
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  • #2
*cricket*
Anyone?
 
  • #3
The [itex]M_k[/itex] gives, roughly speaking, the potential torque on the lamina if one were to try to balance it on the line [itex]x=k[/itex]. In order for the lamina to be balanced on that line, this "tourque" would need to be zero.

If you set [itex]M_k=0[/itex] and solve for [itex]k[/itex], I think you'll see what the "[itex]x-k[/itex] nonsense" is for.
 
  • #4
dlivingston said:
*cricket*
Anyone?
Wow, you waited a whole hour before bumping? Sorry, I was eating supper!
 
  • #5
dlivingston said:
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
[itex]M_k=∫_R\left(x-k\right) σ\left(P\right) dA[/itex]

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.

[itex]For the triangle,
[itex]y=4-\frac{4}{3}x[/itex]
For the semicircle,
[itex]x^2+y^2=1[/itex]
And the horseshoe,
[itex]x^2+y^2=25[/itex]
[itex]x^2+y^2=9[/itex]

The Attempt at a Solution


Welp, I'm lost. As I said, I can figure it out easily using the calculus 2 method of [itex]x_{cm}=\frac{M_y}{M}[/itex], but I'm not sure the way he wants it done.
Well, you mean [itex]x_{cm}=\frac{M_x}{M}[/itex], and [itex]y_{cm}=\frac{M_y}{M}[/itex]

What's the density formula? Is it just the equations I listed above? What's this (x-k) nonsense?
x- k is, just your quote above says, the horizontal distance from point (x, y) to the line x= k.
1) A triangle with sides 3", 4", and 5" respectively.
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 [itex](1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx[/itex] and the y coordinate is [itex](1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx[/itex].

2) A semicircle
The area of a semicircle of radius R (you don't give a radius) is [itex]\pi R^2/2[/itex]. It's probably best to use polar coordinates to do the integrals:
[tex](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[/tex]
(It might occur to you that, by symmetry, the x coordinate of the centroid is 0)
and
[tex](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[/tex]

3) A "horseshoe", i.e. a half-ring, whose inner and outer edges are composed of circular arcs of radii 3" and 5" respectively.
So this is the area between two semi-circles. Its area is [itex](pi(5)^2/2- \pi(3)^2/2= 8\pi[/itex].The coordinates of the centroid are
[tex](1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r cos(\theta)d\theta dr[/tex]
(Again, by symmetry, this should be 0.)
and
[tex](1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r sin(\theta)d\theta dr[/tex]
 
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  • #6
Ah, you both are absolutely fantastic! Thank you.
 

1. What is the center of mass of a lamina?

The center of mass of a lamina, also known as the centroid, is a point that represents the average position of the mass of the lamina. It is the point where the lamina can be balanced and has equal weight distribution on all sides.

2. How is the center of mass of a lamina calculated?

The center of mass of a lamina can be calculated by dividing the sum of the product of the mass of each part of the lamina and its distance from a chosen reference point by the total mass of the lamina. This can be expressed as x̅ = (∑mixi) / (∑mi) for the x-coordinate and y̅ = (∑miyi) / (∑mi) for the y-coordinate.

3. What is the importance of finding the center of mass of a lamina?

Finding the center of mass of a lamina is important because it helps in understanding the stability and balance of the lamina. It is also useful in various engineering and physics applications, such as in designing structures and predicting their behavior under different conditions.

4. Can the center of mass of a lamina be outside of the lamina?

No, the center of mass of a lamina will always be located within the boundaries of the lamina. This is because the center of mass is a representation of the lamina's mass distribution and cannot exist outside of it.

5. How does the shape of a lamina affect its center of mass?

The shape of a lamina can greatly affect its center of mass. For example, a symmetrical lamina will have its center of mass at its geometric center, while an irregularly shaped lamina will have its center of mass shifted towards the heavier side. The distribution of mass in a lamina also plays a role in determining its center of mass.

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