Find the Center of Mass of a Lamina

In summary, the lamina that occupies the region cut from the first quadrant by the lines x = 6 and y = 1 has a center of mass at (6,1). The x-coordinate of the center of mass is 11/3 of the distance from the origin to the lamina.
  • #1
jabar616
2
0

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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  • #2
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.
 
  • #3
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!
 
  • #4
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.
 
  • #5
I did end up finding my error, of course it was just a simple issue that I had missed. Thank you all!
 

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

The center of mass of a lamina is the point at which its mass is evenly distributed in all directions.

2. How do you find the center of mass of a lamina?

The center of mass of a lamina can be found by using the formula: x̄ = (Σm_i*x_i) / (Σm_i) and ȳ = (Σm_i*y_i) / (Σm_i), where x̄ and ȳ are the coordinates of the center of mass, m_i is the mass of each small element, and x_i and y_i are the respective x and y coordinates of each small element.

3. What are the units of the center of mass of a lamina?

The units of the center of mass of a lamina are the same as the units of the coordinates used in the formula to find it. For example, if the coordinates are in meters, the center of mass will also be in meters.

4. Why is finding the center of mass important?

Finding the center of mass is important because it helps determine the stability and balance of an object. It is also a crucial concept in physics, as it is used to calculate the motion of an object in response to external forces.

5. Can the center of mass of a lamina fall outside of the object?

No, the center of mass of a lamina will always fall within the boundaries of the object. This is because the center of mass is calculated using the mass and coordinates of every small element within the object, so it must lie within the object itself.

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