Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In the following lamina, all angles are right angles and the density is constant:

The Center of mass of lamina is $(0, \frac{15}{2}) .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

Given lamina is a composition of rectangles.

density is constant.

Step 2. x coordinate Center of mass of the horizontal lamina

Determine x coordinate Center of mass of the horizontal lamina where x varies from $- 3 to 3$ and y varies from $4 to 6 .$

role="math" localid="1650346135185" $\bar{x} = \frac{\iint_{Ω} x ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A} \begin{aligned} \bar{x_{1}} & = \frac{k \int_{- 3}^{3} \int_{4}^{6} x d y d x}{k \int_{- 3}^{3} \int_{4}^{6} d y d x} \\ \bar{x_{1}} & = \frac{\int_{- 3}^{3} [y]_{4}^{3} x d x}{\int_{- 3}^{3} [y]_{4}^{6} d x} \\ \bar{x_{1}} & = \frac{2 \int_{- 3}^{3} x d x}{2 \int_{- 3}^{3} d x} \\ \bar{x_{1}} & = \frac{2 ({[\frac{x^{2}}{2}]}_{- 3}^{3})}{2 ([x]_{- 3}^{3})} \\ \bar{x_{1}} & = 0 \end{aligned}$

Step 3. y coordinate the Center of mass of the horizontal lamina

Determine y coordinate Center of mass of the horizontal lamina where x varies from $- 3 to 3$ and y varies from $4 to 6 .$

role="math" localid="1650346174670" $\begin{aligned} \bar{y} & = \frac{\iint_{Ω} y ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A} \\ \bar{y_{1}} & = \frac{\int_{- 3}^{3} \int_{4}^{6} k y d y d x}{\int_{- 3}^{3} \int_{4}^{6} k d y d x} \\ \bar{y_{1}} & = \frac{\int_{- 3}^{3} {[\frac{y^{2}}{2}]}_{- 4}^{6} d x}{\int_{- 3}^{3} [y]_{4}^{6} d x} \\ \bar{y_{1}} & = \frac{10 \int_{- 3}^{3} d x}{2 \int_{- 3}^{3} d x} \\ \bar{y_{1}} = & \frac{10 ([x]_{- 3}^{3})}{2 ([x]_{- 3}^{3})} \\ \bar{y_{1}} = & 5 \end{aligned}$

So the center of mass of horizontal lamina is $(x_{1}, y_{1}) = (0, 5) .$

Step 4. x coordinate Center of mass of the vertical lamina

Determine x coordinate Center of mass of the horizontal lamina where x varies from $- 1 to 1$ and y varies from $1 to 4 .$

role="math" localid="1650346297780" $\bar{x} = \frac{\iint_{Ω} x ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A} \begin{aligned} \bar{x_{2}} & = \frac{\int_{- 1}^{1} \int_{1}^{4} k x d y d x}{\int_{- 1}^{1} \int_{1}^{4} k d y d x} \\ \bar{x_{2}} & = \frac{\int_{- 1}^{1} [y]_{1}^{4} x d x}{\int_{- 1}^{1} [y]_{1}^{4} d x} \\ \bar{x_{2}} & = \frac{3 \int_{- 1}^{1} x d x}{3 \int_{- 1}^{1} d x} \\ \bar{x_{2}} & = \frac{3 ({[\frac{x^{2}}{2}]}_{- 1}^{1})}{3 ([x]_{- 1}^{1})} \\ \bar{x_{2}} & = 0 \end{aligned}$

Step 5. y coordinate the Center of mass of the vertical lamina

Determine y coordinate Center of mass of the horizontal lamina where x varies from $- 1 to 1$ and y varies from $1 to 4$

$\begin{aligned} \bar{y} & = \frac{\iint_{Ω} y ρ (x, y) d A}{\iint_{Ω} ρ (x, y) d A} \\ {\bar{y}}_{2} & = \frac{\int_{- 1}^{1} \int_{1}^{4} y k d y d x}{\int_{- 1}^{1} \int_{1}^{4} k d y d x} \\ {\bar{y}}_{2} & = \frac{\int_{- 1}^{1} {[\frac{y^{2}}{2}]}_{1}^{4}}{\int_{- 1}^{1} [y]_{1}^{4} d x} \\ {\bar{y}}_{2} & = \frac{\frac{15}{2} \int_{- 1}^{1} d x}{3 \int_{- 1}^{1} d x} \\ {\bar{y}}_{2} & = \frac{\frac{15}{2} [x]_{- 1}^{1}}{3 [x]_{- 1}^{1}} \\ {\bar{y}}_{2} & = \frac{5}{2} \end{aligned}$

So the center of mass of vertical lamina is $(x_{2}, y_{2}) = (0, \frac{5}{2}) .$

Step 6. Center of mass of composition of the lamina.

add coordinates of all center of mass of all laminas to determine the Center of mass of composition of the lamina.

$\begin{aligned} (\bar{X}, \bar{Y}) & = ({\bar{x}}_{1}, {\bar{y}}_{1}) + ({\bar{x}}_{2}, {\bar{y}}_{2}) \\ (\bar{X}, \bar{Y}) & = (0, 5) + (0, \frac{5}{2}) \\ (\bar{X}, \bar{Y}) & = ((0 + 0), (5 + \frac{5}{2})) \\ (\bar{X}, \bar{Y}) & = (0, \frac{15}{2}) \end{aligned}$

So the center of mass of lamina is $(0, \frac{15}{2}) .$

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Calculus

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Short Answer

In the following lamina, all angles are right angles and the density is constant:

Step by Step Solution

Step 1. Given information.

Step 2. x coordinate Center of mass of the horizontal lamina

Step 3. y coordinate the Center of mass of the horizontal lamina

Step 4. x coordinate Center of mass of the vertical lamina

Step 5. y coordinate the Center of mass of the vertical lamina

Step 6. Center of mass of composition of the lamina.

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Calculus

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Short Answer

In the following lamina, all angles are right angles and the density is constant:

Step by Step Solution

Step 1. Given information.

Step 2. x coordinate Center of mass of the horizontal lamina

Step 3. y coordinate the Center of mass of the horizontal lamina

Step 4. x coordinate Center of mass of the vertical lamina

Step 5. y coordinate the Center of mass of the vertical lamina

Step 6. Center of mass of composition of the lamina.

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