Find the Y component of the Center of Mass

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SUMMARY

The discussion focuses on calculating the Y component of the Center of Mass for a triangular shape using calculus. The key equation utilized is ∫(r*dm)/M, where the user attempts to derive the area of horizontal strips for integration. The user initially calculates the Y component as (2/3)B but realizes that due to the mass distribution, the correct value should be (1/3)B. The confusion arises from the definition of area and the integration limits, particularly in determining the relationship between x and y along the hypotenuse of the triangle.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of Center of Mass in physics.
  • Knowledge of geometric properties of triangles and linear equations.
  • Ability to manipulate equations involving mass density and area.
NEXT STEPS
  • Learn how to derive the equation of a line from geometric shapes.
  • Study the application of integration in finding the Center of Mass for various shapes.
  • Explore the concept of mass density and its implications in physics problems.
  • Review techniques for setting up and solving integrals involving variable limits.
USEFUL FOR

Students studying physics or engineering, particularly those focusing on mechanics and calculus applications in determining the Center of Mass for irregular shapes.

johnnyboy53
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Homework Statement


Find the Y component of the Center of Mass

http://img607.imageshack.us/img607/7122/83370796.png


Homework Equations


∫(r*dm)/M


The Attempt at a Solution


I keep coming up with (2/3)B but i know since that there is more mass near the origin axis, it should be 1/3B. I think it is my definition of area that is wrong. I am using horizontal strips with A * dy but i am unsure how to substitute A since it is a varying as y increases

∫(r*dm)/M
σ=dm/da
σ*da=dm

∫(r*σ*da)/M
da= area of small strip = dy*Mslope=
Y=Mslope*x=
Mslope=B/A

∫(y*σ*dy*y(A/B))/M
y(A/B)=X

(σA/BM )∫y^2dy=

σ= M/Area
Area = 1/2AB
σAB=2M

(AσB2/3M)=
(σAB/3M)*B =
(2MB/3M)=
(2/3)B

Which is wrong. I think the error is somewhere around the da part. I am a little bit unsure of what the area of each horizontal strip would be since the left side of the shape is a function of x, or is it a function of y?
 
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If you write out what you mean by A, B, Mslope etc. it might become apparent to you what's wrong. But if it doesn't, it will give the rest of us a chance to spot it.
 
oh wow i forgot the picture. Sorry!

The lines are the dy's
 
It looks like σ is meant to represent the mass density of the material comprising the triangle, and that it is assumed to be constant (triangle material is of uniform density). If that's the case, then you can ignore it (or assume σ = 1 in whatever units you prefer) and just work with the areas.

If you can find an equation of the line for the hypotenuse of the triangle then you can determine x in terms of y (that is, x(y) ) for the line. With x(y) you can find the length of your area element for a given y (its thickness is dy and its length lies between x(y) and x = A).

So, begin by finding the equation of the line on which the hypotenuse "lives". It's a linear equation of the form y = m*x. Solve for x in terms of y and carry on...
 
Hypotenuse of the triangle would be:
y=Mslope*x
Mslope = B/A
y=(B/A)x
x=(A/B)y

is that right? That's what i did on my attempt
 
johnnyboy53 said:
Hypotenuse of the triangle would be:
y=Mslope*x
Mslope = B/A
y=(B/A)x
x=(A/B)y
OK. What do you have for the mass (or area) of the strip width dy at height y?
 

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