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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 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
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 and the y-coordinate by 7. Specifically, My z = SSRP(x, y) dA SSRP(x, y) dA m and M₂ ÿ SSRYP(x, y) DA SSRP(x, y) dA m 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) = xy. 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 Ic= SSRI DA SSR dA 1, m M₂ SSR dA Ye 1. M SSR dA 6 6 Notice that the center of mass 5'5 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) = √y. Find the center of mass. IN || 8-# > X || ONGIN DININ
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 and the y-coordinate by 7. Specifically, My z = SSRP(x, y) dA SSRP(x, y) dA m and M₂ ÿ SSRYP(x, y) DA SSRP(x, y) dA m 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) = xy. 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 Ic= SSRI DA SSR dA 1, m M₂ SSR dA Ye 1. M SSR dA 6 6 Notice that the center of mass 5'5 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) = √y. Find the center of mass. IN || 8-# > X || ONGIN DININ
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