Find the Center of Mass of a Lamina

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SUMMARY

The discussion focuses on calculating the x-coordinate of the center of mass for a lamina defined in the first quadrant by the lines x = 6 and y = 1, with a density function p(x,y) = x + y + 1. The correct formulas for mass and moments are established as Mass = ∫∫spdσ and My = ∫∫sxpdσ. The user initially miscalculated the moment Myz, leading to an incorrect x-coordinate of 14/3 instead of the correct value of 11/3. The error was identified as a misunderstanding of the moment notation and the boundaries of the two-dimensional object.

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  • Knowledge of the concept of center of mass in two dimensions
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jabar616
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Homework Statement



Find the x-coordinate of the center of mass of the lamina that occupies the region cut from the first quadrant by the lines x = 6 and y = 1 if the density function is p(x,y) = x + y + 1

Homework Equations



Mass = ∫∫spdσ
X Coordinate of center of Mass = Myz/M
Myz = ∫∫sxpdσ

The Attempt at a Solution



I am not sure how to set this one up, as I have not seen one like this without a bounding equation (ie x2 + y2 + z2 = a2 or similar.)

I attempted simply setting up the integral as being from 0≤x≤6 and 0≤y≤1 so M=∫∫x+y+1dydx = 27 and Myz = ∫∫x2 +xy +x dydx = 126

However, dividing 126 by 27 results in 14/3, not 11/3 as the answer theoretically should be.

Any help would be greatly appreciated!
 
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The problem appears to be in your calculation of Myz. If you can't find the problem, please post your work here so it can be reviewed in detail.
 
jabar616 said:

Homework Statement



Find the x-coordinate of the center of mass of the lamina that occupies the region cut from the first quadrant by the lines x = 6 and y = 1 if the density function is p(x,y) = x + y + 1

Homework Equations



Mass = ∫∫spdσ
X Coordinate of center of Mass = Myz/M
Myz = ∫∫sxpdσ
Your two formulas above are a bit confusing. You have a two-dimensional object, so the two moments are with respect to the x-axis or the y axis.

The notation Myz is the moment about the y-z plane, which isn't applicable here because the object is two-dimensional.

The formula to use for the x-coordinate is My/M
jabar616 said:

The Attempt at a Solution



I am not sure how to set this one up, as I have not seen one like this without a bounding equation (ie x2 + y2 + z2 = a2 or similar.)
Here the boundaries are a lot simpler: the region is a rectangle, with x ranging between 0 and 6 and y ranging between 0 and 1.
jabar616 said:
I attempted simply setting up the integral as being from 0≤x≤6 and 0≤y≤1 so M=∫∫x+y+1dydx = 27 and Myz = ∫∫x2 +xy +x dydx = 126

However, dividing 126 by 27 results in 14/3, not 11/3 as the answer theoretically should be.

Any help would be greatly appreciated!
 
SteamKing said:
The problem appears to be in your calculation of Myz.
Yes, I agree. BTW, I get 11/3 for the x-coord. at CM.

Please show us your calculations for My.
 
I did end up finding my error, of course it was just a simple issue that I had missed. Thank you all!
 

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