Q 37. Let 聽be triangular region with ... [FREE SOLUTION]

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Chapter 13: Q 37. (page 1039)

Let be triangular region with vertices $(1, 0), (2, 1), and (2, - 1)$
If the density at each point in $T_{2}$ is proportional to the square of the point鈥檚 distance from the $y$ -axis, find the
moments of inertia about the $x$ - and $y$ -axes. Use these
answers to find the radii of gyration of $T_{2}$ about the
$x$ - and $y$ -axes.

Short Answer

Expert verified

The moment of inertia is $I_{y} = \frac{129}{15} k, I_{x} = \frac{49}{30} k$

The mass is $m = \frac{5}{3} k$

Radius of gyration is $R_{y} = \sqrt{\frac{147}{50}} and R_{x} = \sqrt{\frac{147}{150}}$

Step by step solution

Given Information

Vertices of triangle are $(1, 0), (2, 1), and (2, - 1)$

$蚁 (x, y) = k x^{2}$

Calculating IY

It can be calculated as $I_{y} = {鈭�}_{惟} x^{2} 蚁 (x, y) d A$

Putting limits gives

$I_{y} = {鈭�}_{1}^{2} {鈭�}_{- x + 1}^{x - 1} x^{2} 蚁 (x, y) d y d x$

$I_{y} = {鈭�}_{1}^{2} {鈭�}_{- x + 1}^{x - 1} x^{2} k x^{2} d y d x [蚁 (x, y) = k x^{2}]$

$I_{y} = k {鈭�}_{1}^{2} {鈭�}_{- x + 1}^{x - 1} x^{4} d y d x$

Solving inner integral

$I_{y} = k {鈭�}_{1}^{2} [y]_{- x + 1}^{x - 1} x^{4} d x = k {鈭�}_{1}^{2} [x - 1 - (- x + 1)] x^{4} d x k {鈭�}_{1}^{2} 2 (x^{5} - x^{4}) d x$

Solving further

$I_{y} = 2 k {[\frac{x^{6}}{6} - \frac{x^{5}}{5}]}_{1}^{2} = 2 k [(\frac{2^{6}}{6} - \frac{2^{5}}{5}) - (\frac{1}{6} - \frac{1}{5})] = \frac{129}{15} k$

$\frac{49}{30} k$

Calculating Ix

Similarly, $I_{x} = {鈭�}_{惟} y^{2} 蚁 (x, y) d A$

Putting limits

$I_{x} = {鈭�}_{1}^{2} {鈭�}_{- x + 1}^{x - 1} y^{2} 蚁 (x, y) d y d x$

$I_{x} = {鈭�}_{1}^{2} {鈭�}_{- x + 1}^{x - 1} y^{2} k x^{2} d y d x [蚁 (x, y) = k x^{2}]$

$I_{x} = k {鈭�}_{1}^{2} x^{2} {[\frac{y^{3}}{3}]}_{- x + 1}^{x - 1} d x = k {鈭�}_{1}^{2} x^{2} [\frac{(x - 1)^{3} - (- x + 1)^{3}}{3}] d x = k {鈭�}_{1}^{2} x^{2} [\frac{2 (x - 1)^{3}}{3} d x$

$I_{x} = \frac{2}{3} k {鈭�}_{1}^{2} [x^{5} - 3 x^{4} + 3 x^{3} - x^{2}] d x$

Solving further

$I_{x} = \frac{2}{3} k {[\frac{x^{6}}{6} - \frac{3 x^{5}}{5} + \frac{3 x^{4}}{4} - \frac{x^{3}}{3}]}_{1}^{2}$

$I_{x} = \frac{2}{3} k [(\frac{2^{5}}{5} - \frac{3 脳 2^{4}}{4} + \frac{3 脳 2^{3}}{3} - \frac{2^{2}}{2}) - (\frac{1}{5} - \frac{3}{4} + \frac{3}{3} - \frac{1}{2})]$

$I_{x} = \frac{2}{3} k [\frac{49}{20}] = \frac{49}{30} k$

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Short Answer

Step by step solution

Given Information

Calculating IY

Calculating Ix

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91影视

Let be triangular region with vertices (1,0),(2,1), and(2,-1)If the density at each point in T2is proportional to the square of the point鈥檚 distance from the y-axis, find themoments of inertia about the x- and y-axes. Use theseanswers to find the radii of gyration of T2about thex- and y-axes.

Short Answer

Step by step solution

Given Information

Calculating IY

Calculating Ix

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Statistics

Decision Maths

Geometry

Discrete Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.