A lamina with constant density $\rho(x, y) = \rho$ occupies the given region. Find the moments of inertia $I_x$ and $I_y$ and the radii of gyration $\bar{x}$ and $\bar{y}$. The region is the rectangle $0 \leq x \leq 2b$, $0 \leq y \leq 4h$.

Neetesh Kumar | November 27, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 15.4 Question Number 11
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Step-by-step solution:

Step 1: Definitions of moments of inertia and radii of gyration

1. Moment of inertia about the $x$-axis: $I_x = \iint_D y^2 \rho \, dA$

2. Moment of inertia about the $y$-axis: $I_y = \iint_D x^2 \rho \, dA$

3. Radii of gyration: $\bar{x} = \sqrt{\frac{I_y}{m}}, \quad \bar{y} = \sqrt{\frac{I_x}{m}}$

The mass $m$ is given by: $m = \iint_D \rho \, dA$

Step 2: Compute the mass $m$

The region $D$ is the rectangle $0 \leq x \leq 2b$ and $0 \leq y \leq 4h$. The area element is $dA = dx \, dy$.

$m = \int_0^{2b} \int_0^{4h} \rho \, dy \, dx$

Evaluate the $y$-integral: $\int_0^{4h} \rho \, dy = \rho \int_0^{4h} 1 \, dy = \rho \cdot (4h - 0) = 4h\rho$

Now, integrate with respect to $x$: $m = \int_0^{2b} 4h\rho \, dx = 4h\rho \cdot (2b - 0) = 8bh\rho$

Thus, the mass is: $m = 8bh\rho$

Step 3: Compute $I_x$

The moment of inertia about the $x$-axis is: $I_x = \int_0^{2b} \int_0^{4h} y^2 \rho \, dy \, dx$

First, evaluate the $y$-integral: $\int_0^{4h} y^2 \rho \, dy = \rho \cdot \left[\frac{y^3}{3}\right]_0^{4h} = \rho \cdot \frac{(4h)^3}{3} = \rho \cdot \frac{64h^3}{3}$

Substitute into $I_x$: $I_x = \int_0^{2b} \frac{64h^3\rho}{3} \, dx$

Factor out constants: $I_x = \frac{64h^3\rho}{3} \cdot (2b - 0)$

Simplify: $I_x = \frac{128bh^3\rho}{3}$

Step 4: Compute $I_y$

The moment of inertia about the $y$-axis is: $I_y = \int_0^{2b} \int_0^{4h} x^2 \rho \, dy \, dx$

First, evaluate the $y$-integral: $\int_0^{4h} \rho \, dy = \rho \cdot (4h - 0) = 4h\rho$

Substitute into $I_y$: $I_y = \int_0^{2b} x^2 (4h\rho) \, dx$

Factor out constants: $I_y = 4h\rho \int_0^{2b} x^2 \, dx$

Evaluate the $x$-integral: $\int_0^{2b} x^2 \, dx = \left[\frac{x^3}{3}\right]_0^{2b} = \frac{(2b)^3}{3} = \frac{8b^3}{3}$

Substitute back: $I_y = 4h\rho \cdot \frac{8b^3}{3} = \frac{32b^3h\rho}{3}$

Step 5: Compute $\bar{x}$ and $\bar{y}$

The radii of gyration are: $\bar{x} = \sqrt{\frac{I_y}{m}}, \quad \bar{y} = \sqrt{\frac{I_x}{m}}$

Substitute $I_y = \frac{32b^3h\rho}{3}$ and $m = 8bh\rho$ into $\bar{x}$: $\bar{x} = \sqrt{\frac{\frac{32b^3h\rho}{3}}{8bh\rho}} = \sqrt{\frac{32b^2}{3 \cdot 8}} = \sqrt{\frac{4b^2}{3}} = \frac{2b}{\sqrt{3}}$

Substitute $I_x = \frac{128bh^3\rho}{3}$ and $m = 8bh\rho$ into $\bar{y}$: $\bar{y} = \sqrt{\frac{\frac{128bh^3\rho}{3}}{8bh\rho}} = \sqrt{\frac{128h^2}{3 \cdot 8}} = \sqrt{\frac{16h^2}{3}} = \frac{4h}{\sqrt{3}}$

Final Answers:

$I_x = \boxed{\frac{128bh^3\rho}{3}}$

$I_y = \boxed{\frac{32b^3h\rho}{3}}$

$\bar{x} = \boxed{\frac{2b}{\sqrt{3}}}$

$\bar{y} = \boxed{\frac{4h}{\sqrt{3}}}$

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