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    <name>
      <text>27.1.6.2  Applications of Surface Integration Over Rectangular Areas Task</text>
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    <questiontext format="moodle_auto_format">
      <text><![CDATA[<p>Given a rectangular lamina of length \(\ell \), width \(b\), thickness \(t\) (small) and density \(ρ\) (see the below figure), find the the moment of inertia of this lamina about the \(y\)-axis. Give your answer in terms of \(M\) and \(\ell \), (which can be entered as M and L respectively) noting that the mass of the lamina is \(M=\ell b t \rho\).</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/27.1-fig-9-xyplane.svg" alt="A diagram showing the xy plane representing a rectangular lamina" width="249" height="115" class="img-fluid atto_image_button_text-bottom"></figure></p>
<p> Moment of inertia = [[input:ans1]] [[validation:ans1]]</p>]]></text>
<file name="27.1-fig-9-xyplane.svg" path="/" encoding="base64">[encoded data, 37812 characters]</file>
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    <generalfeedback format="moodle_auto_format">
      <text><![CDATA[From a property previously, we know that the moment of inertia (or second moment of area) is given by the integral
\begin{align}
\int_{0}^l \int_0^b x^2 \rho t\;dydx &= \rho t (b-0) \int_0^l x^2\;dx \\
&= b\rho t\left[\frac{x^3}{3} \right]_0^l \\
& = b\rho t \frac{l^3}{3}
\end{align}
As the mass of the lamina is \(M = lbt\rho \), the moment of inertia simplifies to \(\frac{1}{3}Ml^2\). Again, the \(t\) and \(\rho\) are included in the integral to make it a moment of inertia rather than a simple second moment. So we have </p>
<p> Moment of inertia \(= \frac{1}{3}Ml^2\). </p>]]></text>
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    <questionvariables>
      <text>ta1: (1/3)*M*L^2;</text>
    </questionvariables>
    <specificfeedback format="html">
      <text>[[feedback:prt1]]</text>
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    <questionnote format="html">
      <text><![CDATA[<p> Moment of inertia = {@ta1@} </p>]]></text>
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