Center of Mass Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by and the y-coordinate by . Specifically, My Szp(x, y) dA SSRP(x, y) DA m and M₂ SSRYP(x, y) DA m SSRP(x, y) dA Example 15.6.3: Center of mass Again consider the same triangular region R with vertices (0, 0), (0, 3), (3, 0) and with density function p(x, y) = zy. Find the center of mass. Show solution If in Example 15.6.3 we choose the density p(x, y) instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid, My SSRZ DA SSR dA 1, m M₂ Ye SSRy da SSR dA 1. m 6 6 Notice that the center of mass 55 is not exactly the same as the centroid (1, 1) of the triangular region. This is due to the variable density of R If the density is constant, then we just use p(x, y) = c (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina. ? Exercise 15.6.3 Again use the same region R as above and use the density function p(x, y) = √zy. Find the center of mass. X 15 || DINGIN DINDIN
Center of Mass Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by and the y-coordinate by . Specifically, My Szp(x, y) dA SSRP(x, y) DA m and M₂ SSRYP(x, y) DA m SSRP(x, y) dA Example 15.6.3: Center of mass Again consider the same triangular region R with vertices (0, 0), (0, 3), (3, 0) and with density function p(x, y) = zy. Find the center of mass. Show solution If in Example 15.6.3 we choose the density p(x, y) instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid, My SSRZ DA SSR dA 1, m M₂ Ye SSRy da SSR dA 1. m 6 6 Notice that the center of mass 55 is not exactly the same as the centroid (1, 1) of the triangular region. This is due to the variable density of R If the density is constant, then we just use p(x, y) = c (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina. ? Exercise 15.6.3 Again use the same region R as above and use the density function p(x, y) = √zy. Find the center of mass. X 15 || DINGIN DINDIN
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Center of Mass
Finally, we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by \( \bar{x} \) and the y-coordinate by \( \bar{y} \). Specifically,
\[ \bar{x} = \frac{M_y}{m} = \frac{\iint_R x \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA} \]
and
\[ \bar{y} = \frac{M_x}{m} = \frac{\iint_R y \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA} \]
#### Example 15.6.3: Center of Mass
Again consider the same triangular region \( R \) with vertices \( (0, 0) \), \( (0, 3) \), \( (3, 0) \) and with density function \( \rho(x, y) = xy \). Find the center of mass.
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If in Example 15.6.3 we choose the density \( \rho(x, y) \) instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,
\[ x_c = \frac{M_y}{m} = \frac{\iint_R x \, dA}{\iint_R \, dA} = \frac{\frac{9}{2}}{\frac{9}{2}} = 1, \]
\[ y_c = \frac{M_x}{m} = \frac{\iint_R y \, dA}{\iint_R \, dA} = \frac{\frac{9}{2}}{\frac{9}{2}} = 1. \]
Notice that the center of mass \( \left( \frac{6}{5}, \frac{6}{5} \right) \) is not exactly the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74b6836e-7075-4604-8b7e-b1441272473c%2Fcb57ebd8-8b8e-4b47-bb52-93d70909f3f9%2Fm40ebt9_processed.png&w=3840&q=75)
Transcribed Image Text:### Center of Mass
Finally, we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by \( \bar{x} \) and the y-coordinate by \( \bar{y} \). Specifically,
\[ \bar{x} = \frac{M_y}{m} = \frac{\iint_R x \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA} \]
and
\[ \bar{y} = \frac{M_x}{m} = \frac{\iint_R y \rho(x, y) \, dA}{\iint_R \rho(x, y) \, dA} \]
#### Example 15.6.3: Center of Mass
Again consider the same triangular region \( R \) with vertices \( (0, 0) \), \( (0, 3) \), \( (3, 0) \) and with density function \( \rho(x, y) = xy \). Find the center of mass.
<div style="border:1px solid #c3c3c3; padding:10px; margin-top:10px; margin-bottom:10px;">
<button style="background-color:#f0f0f0; border:none; padding:5px;">Show solution</button>
</div>
If in Example 15.6.3 we choose the density \( \rho(x, y) \) instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,
\[ x_c = \frac{M_y}{m} = \frac{\iint_R x \, dA}{\iint_R \, dA} = \frac{\frac{9}{2}}{\frac{9}{2}} = 1, \]
\[ y_c = \frac{M_x}{m} = \frac{\iint_R y \, dA}{\iint_R \, dA} = \frac{\frac{9}{2}}{\frac{9}{2}} = 1. \]
Notice that the center of mass \( \left( \frac{6}{5}, \frac{6}{5} \right) \) is not exactly the
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