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      <text>15.2.1.4 Calculating centre of mass rod with 2 masses</text>
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      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p>Objects of mass \(m\) and \(3m\) are placed at the locations shown in diagram (a). Find the distance \(\bar{x}\) of the centre of mass from the origin \(O\) as illustrated in diagram (b).</p>
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<p>First calculate the sum of the individual moments: [[input:ans1]][[validation:ans1]] \( \textrm{kg}\, \textrm{m}\) [[feedback:prt1]]</p>
<p> The moment of the total mass about \(O\) is \(4 m \times \bar{x}\).</p>
<p> The moment of the total mass is equal to the sum of the moments of the individual masses. What equation do we get by equating the two moments?</p>
<p> [[input:ans2]] [[validation:ans2]] \(\times \; m = \) [[input:ans3]] [[validation:ans3]] \(\times \; m\bar{x}\)</p>
<p> So what is the distance \(\bar{x}\)?
    <br> \(\bar{x}=\) [[input:ans4]] [[validation:ans4]] [[feedback:prt2]] </p>]]></text>
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      <text><![CDATA[<p>The sum of the individual moments is: \(6\times m + 10 \times 3m = 36m.\)</p>
<p>Equating the two moments gives: \(36m = 4m\bar{x}\). And thus \(\bar{x}=\frac{36m}{4m}=9.\)</p>
<p>So the centre of mass is located a distance 9 units along the \(x\)-axis. Note that it is closer to the position of \(3m\) mass than to the position of \(1m\) mass (actually in the ratio 3:1).</p>]]></text>
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      <text>ta1 : 36*m;
ta2 : 36;
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