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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P.

Expert Solution

Step 1

Draw the diagram of the shaded area with a horizontal strip,

From the above diagram, the value of x is written as

$\begin{array}{rcl} x & = & \frac{a}{2} + \frac{a}{2} (\frac{y}{a}) \\ = & \frac{1}{2} (a + y) \end{array}$

Write the expression of an area of the strip,

$\begin{array}{rcl} d A & = & x d y \\ = & \frac{1}{2} (a + y) d y \end{array}$

Calculate the shaded area,

$\begin{array}{rcl} d A & = & \frac{1}{2} (a + y) d y \\ For shaded area, \\ \frac{A}{2} & = & \int_{}^{} d A \\ = & \int_{0}^{a} \frac{1}{2} (a + y) d y \\ = & \frac{1}{2} {[a y + \frac{y^{2}}{2}]}_{0}^{a} \\ = & \frac{1}{2} [a (a) + \frac{a^{2}}{2}] \\ = & \frac{3}{4} a^{2} \\ A & = & \frac{3}{2} a^{2} \end{array}$

Step 2

Write the expression of a moment of inertia about the x-axis,

$\begin{array}{rcl} \frac{1}{2} I_{x} & = & \int_{}^{} y^{2} d A \\ = & \int_{}^{} y^{2} [\frac{1}{2} (a + y)] d y \\ = & \frac{1}{2} \int_{}^{} (a y^{2} + y^{3}) d y \\ I_{x} & = & \int_{0}^{a} (a y^{2} + y^{3}) d y \\ = & {[\frac{a y^{3}}{3} + \frac{y^{4}}{4}]}_{0}^{a} \\ = & [\frac{a {(a)}^{3}}{3} + \frac{a^{4}}{4}] \\ I_{x} & = & \frac{7}{12} a^{4} \end{array}$

Write the expression of a moment of inertia about the y-axis,

$\begin{array}{rcl} \frac{1}{2} I_{y} & = & \frac{1}{3} \int_{}^{} x^{2} d A \\ = & \int_{}^{} x^{2} (x d y) \\ = & \frac{1}{3} \int_{}^{} x^{3} d y \\ = & \frac{1}{3} \int_{0}^{a} {[\frac{1}{2} (a + y)]}^{3} d y \\ = & \frac{1}{24} \int_{0}^{a} (a^{3} + y^{3} + 3 a^{2} y + 3 a y^{2}) d y \\ = & \frac{1}{24} {[a^{3} y + \frac{y^{4}}{4} + \frac{3}{2} a^{2} y^{2} + a y^{3}]}_{0}^{a} \\ \frac{1}{2} I_{y} & = & \frac{1}{24} [a^{3} (a) + \frac{a^{4}}{4} + \frac{3}{2} a^{2} {(a)}^{2} + a {(a)}^{3}] \\ = & \frac{5}{32} a^{4} \\ I_{y} & = & \frac{5}{16} a^{4} \end{array}$

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