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      <text>12.2.3.4 Crank used to drive a piston example</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4><br>

<h5>Crank used to drive a piston</h5><br>

<h6>Introduction</h6>

<p> A crank is used to drive a piston as in the figure below. </p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/12-2-fig-20-Crank.svg" alt="Diagram of a crank used to drive a piston." width="579" height="171" class="img-responsive atto_image_button_text-bottom"><br></figure>

<h6>Problem</h6>
<p>The angular velocity of the crankshaft is the rate of change of the angle \(\theta,\ \omega= \mathrm{d}\theta/\mathrm{d}t.\) The piston moves horizontally with velocity \(v_p\) and acceleration \(a_p\); \(r\) is the length of the crank and \(l\) is the length of the connecting
    rod. The crankpin performs circular motion with a velocity of \(v_c\) and centripetal acceleration of \(\omega^2r\). The acceleration \(a_p\) of the piston varies with \(\theta\) and is related by \[ a_p =\omega^2 r\left(\cos\theta+\frac{r\cos2\theta}{l}\right)\]</p>
<p>Find the maximum and minimum values of the acceleration \(a_p\) when \(r = 150\) mm and \(l = 375\) mm.</p><br>

<h6>Mathematical statement of the problem</h6>
<p>We need to find the stationary values of \(a_p =\omega^2 r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)\) when \(r = 150\) mm and \(l = 375\) mm. We do this by solving \(\dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}=0\) and then analysing the stationary points to decide whether
    they are a maximum, minimum or point of inflexion. </p><br>

<h6>Mathematical analysis</h6>
<p>\[ a_p =\omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)\;\; \mbox{ so }\;\; \dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}=\omega^2r\left(-\sin\theta-\dfrac{2r\sin2\theta}{l}\right)\]. To find the maximum and minimum acceleration we need to solve \[\dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}=0\Leftrightarrow
    \omega^2 r\left(-\sin\theta-\dfrac{2r\sin2\theta}{l}\right)=0.\]\[\sin\theta+\dfrac{2r}{l}\sin2\theta=0\Leftrightarrow \sin\theta+\dfrac{4r}{l}\sin\theta\cos\theta=0\]\[\Leftrightarrow\quad \sin\theta\left(1+\dfrac{4r}{l}\cos\theta\right)=0\]\[\Leftrightarrow\quad
    \sin\theta=0\;\; \mbox{or}\;\; \cos\theta=-\dfrac{l}{4r}\;\; \mbox{ and as } r=150 \mbox{ mm }\;\; \mbox{ and }\;\; l=375 \mbox{ mm }\]\[\Leftrightarrow\quad \sin\theta=0\;\; \mbox{ or } \;\;\cos\theta=-\dfrac{5}{8}\].</p>

<p><strong>CASE 1: \(\sin\theta = 0\)</strong></p>
<p>If \(\sin\theta=0\) then \(\theta=0\) or \(\theta=\pi\). If \(\theta=0\) then \(\cos\theta=\cos2\theta=1\) so \[ a_p =\omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)=\omega^2r\left(1+\dfrac{r}{l}\right) =\omega^2r\left(1+\dfrac{2}{5}\right)=\dfrac{7}{5}\omega^2r\]If
    \(\theta=\pi\) then \(\cos\theta=-1,\ \cos2\theta=1\) so \[ a_p=\omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)= \omega^2r\left(-1+\dfrac{r}{l}\right)=\omega^2r\left(-1+\dfrac{2}{5}\right)=-\dfrac{3}{5}\omega^2r\]In order to classify the
    stationary points, we differentiate \(\dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}\) with respect to \(\theta\) to find the second derivative: \[ \dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}=\omega^2r\left(-\cos\theta-\dfrac{4r\cos2\theta}{l}\right) =-\omega^2r\left(\cos\theta+\dfrac{4r\cos2\theta}{l}\right)\]</p>
<p>At \(\theta=0\) we get \(\dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}=-\omega^2r\left(1+\dfrac{4r}{l}\right)\) which is negative.</p>
<p>So \(\theta=0\) gives a <strong>maximum</strong> value and \(a_p =\dfrac{7}{5}\omega^2r\) is the value at the maximum.</p>
<p>At \(\theta=\pi\) we get \(\dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}=-\omega^2r\left(-1+\dfrac{4}{l}\right) =-\omega^2r\left(\dfrac{3}{5}\right)\) which is negative.</p>
<p>So \(\theta=\pi\) gives a <strong>maximum</strong> value and \(a_p=-\dfrac{3}{5}\omega^2r\)</p>

<p><strong>CASE 2: \(\cos\theta = -\dfrac{ 5}{ 8}\)</strong></p>
<p>If \(\cos\theta =-\dfrac{5}{8}\) then \(\cos2\theta = 2\cos^2\theta -1 = 2\left(\dfrac{5}{8}\right)^2 -1\) so \(\cos2\theta =-\dfrac{7}{32}\). \[ a_p = \omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)= \omega^2r\left(-\dfrac{5}{8}+-\dfrac{7}{32}\times
    \dfrac{2}{5}\right)=\dfrac{57}{80}\omega^2r.\]At \(\cos\theta=-\dfrac{5}{8}\) we get \(\dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}= \omega^2r\left(-\cos\theta-\dfrac{4r\cos2\theta}{l}\right)=\omega^2r\left(\dfrac{5}{8}+\dfrac{4r}{l}\ \dfrac{7}{32}\right)\) which is
    positive.
</p>
<p>So \(\cos\theta= -\dfrac{5}{8}\) gives a <strong>minimum</strong> value and \(a_p=-\dfrac{57}{80}\omega^2r\)</p>
<p>Thus the values of \(a_p\) at the stationary points are: \[\dfrac{7}{5}\omega^2r\;\mbox{ (maximum), }\;\; -\dfrac{3}{5}\omega^2r\;\mbox{ (maximum) and }\;\; -\dfrac{57}{80}\omega^2r\;\mbox{ (minimum). }\]So the overall maximum value is \(1.4\omega^2r
    = 0.21\omega^2\) and the minimum value is\(-0.7125 \omega^2r = -0.106875\omega^2\) where we have substituted \(r = 150\) mm (\(=\) 0.15 m) and \(l =375\) mm (\(= 0.375\) m).</p><br>

<h6>Interpretation</h6>
<p>The maximum acceleration occurs when \(\theta=0\) and \(a_p=0.21\omega^2\).</p>
<p>The minimum acceleration occurs when \(\cos\theta =-\dfrac{5}{8}\) and \(a_p = -0.106875\omega^2\).</p>]]></text>
<file name="12-2-diagram-crank.svg" path="/" encoding="base64">[encoded data, 33412 characters]</file>
<file name="12-2-fig-20-Crank.svg" path="/" encoding="base64">[encoded data, 32488 characters]</file>
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