Q. 59 Let 惟聽be a lamina in the xy-pl... [FREE SOLUTION]

Chapter 13: Q. 59 (page 1040)

Let $惟$ be a lamina in the xy-plane. Suppose $惟$ is composed of two non-overlapping lamin $惟_{1}$ and $惟_{2}$ , as follows:
Show that if the masses and centers of masses of $惟_{1}$ and $惟_{2}$ are $m_{1}$ and $m_{2},$ and $({\overset{炉}{x}}_{1}, {\overset{炉}{y}}_{1}) and ({\overset{炉}{x}}_{2}, {\overset{炉}{y}}_{2})$ respectively, then the center of mass of $惟$ is $(\overset{炉}{x}, \overset{炉}{y}),$ where
$\overset{炉}{x} = \frac{m_{1} {\overset{炉}{x}}_{1} + m_{2} {\overset{炉}{x}}_{2}}{m_{1} + m_{2}} and \overset{炉}{y} = \frac{m_{1} {\overset{炉}{y}}_{1} + m_{2} {\overset{炉}{y}}_{2}}{m_{1} + m_{2}}$

Short Answer

Expert verified

center of mass $x & y$ is the ratio of the sum of linear moment of the mass about the y-axis and x-axis respectively of both regions to the sum of both masses.

So the center of mass of $惟$ is $(x, y) = (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}) .$

Step by step solution

Step 1. Given information.

The Center of mass of $惟_{1}$ is $m_{1}$ at $(x_{1}, y_{1}) .$

The Center of mass of $惟_{2}$ is $m_{2}$ at $(x_{2}, y_{2}) .$

The Center of mass of

惟

is at

(x, y) .

Step 2. moment of the mass.

The x-coordinate of the center of mass $x$ of $惟_{1}$ is role="math" localid="1650320187800" $x_{1} = \frac{M_{y}}{m_{1}} .$

So linear moment of the mass about the y-axis in the region $惟_{1}$ is role="math" localid="1650320718738" $M_{y 1} = m_{1} x_{1} .$

The y-coordinate of the center of mass $y$ of $惟_{1}$ is $y_{1} = \frac{M_{x}}{m_{1}} .$

So linear moment of the mass about the x-axis in the region $惟_{1}$ is role="math" localid="1650320725214" $M_{x 1} = m_{1} y_{1} .$

The x-coordinate of the center of mass of $惟_{2}$ is role="math" localid="1650320259892" $x_{2} = \frac{M_{y}}{m_{2}} .$

So linear moment of the mass about the y-axis in the region $惟_{2}$ is role="math" localid="1650320732056" $M_{y 2} = m_{2} x_{2} .$

The y-coordinate of the center of mass of $惟_{2}$ is $y_{2} = \frac{M_{x}}{m_{2}} .$

So linear moment of the mass about the x-axis in the region $惟_{2}$ is role="math" localid="1650320738823" $M_{x 2} = m_{2} y_{2} .$

Step 3. Center of mass.

center of mass $x$ is the ratio of the sum of linear moment of the mass about the y-axis of both regions to the sum of both masses.

$x = \frac{M_{y 1} + M_{y 2}}{m_{1} + m_{2}} x = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

center of mass $y$ is the ratio of the sum of linear moment of the mass about the y-axis of both regions to the sum of both masses.

$y = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}$

So center of mass is $(x, y) = (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}) .$

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. moment of the mass.

Step 3. Center of mass.

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