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      <text>15.2.2.2 COM of symmetric triangular lamina</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise <br></p>

<p>Consider the plane sheet, or lamina, shown below. In this exercise we will find the location of its centre of mass \((\bar{x}, \bar{y})\).</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/15-2-2-2-fig.svg" alt="Section 15.2.2 exercise 1" width="300" height="327" class="img-responsive atto_image_button_text-bottom"><br></figure>



<ol class="HELM_parts">

    <li>
        <p>First, by noting the symmetry of the lamina we can already determine the \(y\) coordinate, \(\bar{y}\), of the centre of mass.</p>
        <p>\(\bar{y} = \) [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p>
    </li>



    <li>
        <p>Let \(\rho\) stand for the mass per unit area of the lamina hence the total area is \(3\) units and the total mass is \(3\rho\). Its moment about the \(y\)-axis is \(3\rho\bar{x}\).
            <br><br>To find \(\bar{x}\) first divide the lamina into a large number of thin vertical slices. In the figure below a typical slice has been highlighted. Note that the slice has been drawn from the point \(P\) on the line \(y = 3x\). The
            point \(P\) has coordinates \((x, 3x)\). The thickness of the slice is \(\delta x\).&nbsp;A typical slice of this sheet has been shaded in the left image below.</p>
        <p></p>
        <figure class="HELM_figure"><img src="@@PLUGINFILE@@/15-2-2-2-fig2.svg" alt="Slice of lamina and rectangular approximation with P on left" width="300" height="327" role="presentation" class="img-responsive atto_image_button_middle"></figure>
        <p>
        </p>
        <p> Approximate the area of the slice by a rectangle as indicated in the right of the image above.</p>
        <p>Approximate area of slice: [[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p>
    </li>



    <li>
        <p>Writing \(\rho\) as the mass per unit area, approximate the mass of the slice using the area from the previous part.</p>
        <p>Note: write <code>rho</code> for \(\rho\).</p>
        <p>Approximate mass of slice: [[input:ans3]] [[validation:ans3]] [[feedback:prt3]]</p>
    </li>



    <li>
        <p>The centre of mass of this slice lies on the \(x\)-axis. We will take the limit as the thickness approaches zero so the slice can be assumed to be a point mass located a distance \(x\) from \(O\). Write down the approximate moment of the mass
            of the slice about the \(y\)-axis using the mass above.</p>
        <p>[[input:ans4]] [[validation:ans4]] [[feedback:prt4]]</p>
    </li>



    <li>
        <p>Now to consider the moment of the lamina we add up contributions from all such slices in the lamina. Following which we take a limit by&nbsp;<span style="font-size: 0.9375rem;">letting \(\delta x \to 0\) to obtain the integral:</span></p>
        <p></p>\(3\rho\bar{x}=\)
        <table>
            <tbody>
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                    <td>\(\displaystyle \int\)</td>
                    <td>
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                                <tr>
                                    <td>[[input:ans9]][[validation:ans9]]</td>
                                </tr>
                                <tr>
                                    <td>[[input:ans8]][[validation:ans8]]</td>
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                    </td>
                    <td><br></td>
                    <td><br></td>
                    <td>[[input:ans5]][[validation:ans5]]</td>
                    <td><br></td>
                    <td>\(~\mathrm{d}x\)</td>
                    <td><br></td>
                    <td></td>
                    <td><br></td>
                </tr>
            </tbody>
        </table>[[feedback:prt8]] [[feedback:prt9]]&nbsp;[[feedback:prt5]]&nbsp;</li>
    <li>
        <p>By evaluating the integral we can find</p>
        <p>\(\bar{x}=\)[[input:ans7]] [[validation:ans7]] [[feedback:prt7]]</p>
    </li>


</ol>]]></text>
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<file name="15.2.2.2 picture 2b.png" path="/" encoding="base64">[encoded data, 71884 characters]</file>
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      <text><![CDATA[<ol class="HELM_parts">
    <li>\(\bar{y}=0\) since the lamina is symmetric through reflection in the \(x\)-axis.</li>
    <li>The width is \(\delta\) and height \(3x+3x=6x\) so we have \(6x\delta\).</li>
    <li>The mass is uniform so we simply have \(\rho \times 6x\delta\).</li>
    <li>We are told to take the distance from the \(y\)-axis to the centre of mass of this rectangle to be \(x\) so the moment of inertia is \(x \times 6x\rho\delta = 6x^2\rho\delta\).</li>
    <li>The lamina sits above the interval \([0,1]\), taking the limit as \(\delta \rightarrow 0\) give the integral \(\int_{0}^{1}6\rho x^2 dx\)</li>
    <li>Evaluating the integral \(\int_{0}^{1}6\rho x^2 dx = \left[2\rho x^3 \right]_{0}^{1} = 2\rho\).<br><br>This must equal the moment of the total mass acting at the centre of mass so \(3\rho\bar{x} = 2\rho\) giving \(\bar{x} =2/3\). <br><br>Now the coordinates
        of the centre of mass are \((2/3, 0)\).</li>
</ol>]]></text>
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ta1: 0;
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ta5: 6*rho*x*x;
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