Answered: A circular agger of diameter A is cut…

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

A circular agger of diameter A is cut into a uniform square metal plate with equal sides 2a. Where is the center of mass of the remaining portion? Consider mass per area

Expert Solution

Step 1

Let's assume that there are two sections in the structure on is the whole square and the other one is the circular section:

Let's assume the density of the square plate is = $ρ$

Then the mass of the whole plate is,

$\begin{array}{rcl} M & = & ρ \cdot {(2 a)}^{2} \\ = & 4 ρ \cdot a^{2} \end{array}$

Step 2

Now the mass of the circular section is,

$\begin{array}{rcl} m & = & ρ \cdot \frac{π}{4} {(a)}^{2} \\ = & ρ \cdot \frac{{πa}^{2}}{4} \end{array}$

The center of mass of the circle is at the point, $O_{1}$ , and the center of mass of the square plate is at the point, $O$ .

Now calculate the distance of the point $O_{1}$ form the point $O$ as:

$\begin{array}{rcl} x_{1} & = & \sqrt{a^{2} + a^{2}} - \sqrt{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} \\ = & \sqrt{2} a - \frac{a}{\sqrt{2}} \\ = & \frac{a}{\sqrt{2}} \end{array}$

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