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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

Given Data:

The radius of the circle is r.

The angle ( $α$ ) subtended by the given segment on the center P can be determined from geometry as,

$α = 2 π - 2 θ$

The value of $θ$ can be determined as,

$\begin{array}{rcl} \cos θ & = & \frac{P Q}{P R} \\ \cos θ & = & \frac{\frac{r}{2}}{r} \\ \cos θ & = & \frac{1}{2} \\ θ & = & 60^{\circ} \times \frac{π rad}{180^{\circ}} \\ = & \frac{π}{3} rad \end{array}$

Step 2

The angle subtended by the segment can be determined as,

$α = 2 π - 2 \times \frac{π}{3} α = \frac{4 π}{3}$

The expression for the polar moment of inertia of the segment of the circle is,

$J = \frac{r^{4}}{4} (α - \sin α + \frac{2}{3} \sin α \sin^{2} \frac{α}{2})$

Substituting the known values in the above equation,

$J = \frac{r^{4}}{4} (\frac{4 π}{3} - \sin \frac{4 π}{3} + \frac{2}{3} \sin \frac{4 π}{3} \sin^{2} \frac{4 π}{3}) J = 1.15 r^{4}$

Thus, the required polar moment of inertia is 1.15r⁴.

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