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UNSW MMAN2300 2013-10-05 Assignment C - f.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large MMAN2300 2013-10-05 Assignment C} \begin{align*} \text{\bf Given:}\quad& \text{$m_P$=440 g, $R$=42 mm, $H$=38 mm, $L$=147 mm, $m_R$=470 g, $\bar{I}_R$=1.75 gm$^2$.}\\ &\text{$\omega_{AB}$=3500 rpm=$3500\times\frac{2\pi}{60}=366.52$ rad/s.}\\ &\frac{L}{R}=3.5,\quad\left(\frac{L}{R}\right)^2=12.25,\quad\frac{R}{L}=0.2857,\quad\left(\frac{R}{L}\right)^2=0.08163,\quad L-H=0.109,\quad\frac{H}{L}=0.2585.\\ \\ &\omega_{BC}=\frac{-\omega_{AB}\sin\theta}{\sqrt{\left(\frac{L}{R}\right)^2-\cos^2\theta}}=\frac{-366.52\sin\theta}{\sqrt{12.25-\cos^2\theta}},\quad \omega_{BC}^2=\frac{1.343\times 10^5\sin\theta}{12.25-\cos^2\theta}.\\ &\sin\phi=\frac{R}{L}\cos\theta=0.2857\cos\theta,\quad \cos\phi=\sqrt{1-\left(\frac{R}{L}\right)^2\cos^2\theta}=\sqrt{1-0.08163\cos^2\theta}.\\ &\alpha_{BC}=\frac{\left[1-\left(\frac{L}{R}\right)^2\right]\cos\theta}{\left[\left(\frac{L}{R}\right)^2-\cos^2\theta\right]^\frac{3}{2}}\cdot\omega_{AB}^2=\frac{-1.511\times 10^6\cos\theta}{\left[12.25-\cos^2\theta\right]^\frac{3}{2}}.\\ &a_{C_y}=-R\omega_{AB}^2\left\{\frac{\left[1-\left(\frac{L}{R}\right)^2\right]\cos^2\theta}{\left[\left(\frac{L}{R}\right)^2-\cos^2\theta\right]^\frac{3}{2}}+\frac{\sin^2\theta}{\sqrt{\left(\frac{L}{R}\right)^2-\cos^2\theta}}+\sin\theta\right\}\\ &\qquad=-5642\left\{\frac{-11.25\cos^2\theta}{\left[12.25-\cos^2\theta\right]^\frac{3}{2}}+\frac{\sin^2\theta}{\sqrt{12.25-\cos^2\theta}}+\sin\theta\right\}\\ \end{align*} % % % \begin{align*} \text{f)}\quad&\text{Find an expression for the kinetic energy of the connecting rod as a function of crank angle.}\\ \\ &T=\frac{1}{2}m_R\bar{v}^2+\frac{1}{2}I_C\omega_{BC}^2.\\ &\qquad\mathbf{\bar{v}}=v_C+\mathbf{\omega_{BC}}\times r_{D/C},\quad \quad I_C=\bar{I}_R+m_R~r_{D/C}^2=1.75\times 10^{-3}+0.470\times 0.109^2=7.334\times 10^{-3}.\\ &v_{C/B}=\mathbf{\omega_{BC}}\times r_{CB} =\mathbf{k}~\omega_{BC}\times L\left(-\mathbf{i}\sin\phi+\mathbf{j}\cos\phi\right) =\omega_{BC}L\left(-\mathbf{j}\sin\phi-\mathbf{i}\cos\phi\right)\\ &=\omega_{BC}L\left(-\mathbf{j}\frac{R}{L}\sin\theta-\mathbf{i}\sqrt{1-\left(\frac{R}{L}\right)^2\cos^2\theta}\right) \quad=-\omega_{BC}\left(\mathbf{j}R\cos\theta+\mathbf{i}\sqrt{L^2-R^2\cos^2\theta}\right).\\ &v_B=v_A+\mathbf{\omega_{AB}}\times r_{B/A} =0+\mathbf{k}~\omega_{AB}\times R(\mathbf{i}\cos\theta+\mathbf{j}\sin\theta) =\omega_{AB}R(\mathbf{j}\cos\theta-\mathbf{i}\sin\theta).\\ &v_C=v_B+v_{C/B}=\omega_{AB}R(\mathbf{j}\cos\theta-\mathbf{i}\sin\theta)-\omega_{BC}\left(\mathbf{j}R\cos\theta+\mathbf{i}\sqrt{L^2-R^2\cos^2\theta}\right).\\ &\because v_{C_x}=0,\quad\therefore v_C=\mathbf{j}~v_{C_y}=\mathbf{j}(\omega_{AB}-\omega_{BC})R\cos\theta.\\ &r_{D/C}=(L-H)(\mathbf{i}\sin\phi-\mathbf{j}\cos\phi).\\ %=(L-H)\left(\mathbf{i}\frac{R}{L}\sin\theta-\mathbf{j}\sqrt{1-\left(\frac{R}{L}\right)^2\cos^2\theta}\right).\\ &\mathbf{\bar{v}}=v_C+\mathbf{\omega_{BC}}\times r_{D/C} =\mathbf{j}(\omega_{AB}-\omega_{BC})R\cos\theta+\mathbf{k}~\omega_{BC}\times(L-H)(\mathbf{i}\sin\phi-\mathbf{j}\cos\phi)\\ &=\mathbf{j}(\omega_{AB}-\omega_{BC})R\cos\theta+\omega_{BC}(L-H)(\mathbf{j}\sin\phi+\mathbf{i}\cos\phi).\\ &\bar{v}^2=\left[\omega_{BC}(L-H)\cos\phi\right]^2+\left[(\omega_{AB}-\omega_{BC})R\cos\theta+\omega_{BC}(L-H)\sin\phi\right]^2\\ &\quad=\left[\frac{-366.52\sin\theta}{\sqrt{12.25-\cos^2\theta}}\cdot 0.109\cdot \sqrt{1-0.08163\cos^2\theta}\right]^2\\ &\qquad+\left[\left(366.52+\frac{366.52\sin\theta}{\sqrt{12.25-\cos^2\theta}}\right)\cdot 0.042\cos\theta-\frac{366.52\sin\theta}{\sqrt{12.25-\cos^2\theta}}\cdot 0.109\cdot 0.2857\sin\theta\right]^2\\ &\quad=\left[\frac{-39.95\sin\theta}{\sqrt{12.25-\cos^2\theta}}\sqrt{1-0.08163\cos^2\theta}\right]^2 +\left[15.39\left(1+\frac{\sin\theta}{\sqrt{12.25-\cos^2\theta}}\right)\cos\theta-\frac{11.47\sin^2\theta}{\sqrt{12.25-\cos^2\theta}}\right]^2\\ &\quad=\frac{1596\sin^2\theta}{12.25-\cos^2\theta}(1-0.08163\cos^2\theta) +\left[15.39\cos\theta+\frac{15.39\sin\theta\cos\theta}{\sqrt{12.25-\cos^2\theta}}-\frac{11.47\sin^2\theta}{\sqrt{12.25-\cos^2\theta}}\right]^2\\ &\quad=\frac{1596\sin^2\theta}{12.25-\cos^2\theta}(1-0.08163\cos^2\theta) +\left[15.39\cos\theta+\frac{\sin\theta}{\sqrt{12.25-\cos^2\theta}}(15.39\cos\theta-11.47\sin\theta)\right]^2\\ \\ &\text{Putting all together}\ldots\\ &T=\frac{1}{2}m_R\bar{v}^2+\frac{1}{2}I_C\omega_{BC}^2=0.53~\bar{v}^2+7.334\times 10^{-3}~\omega_{BC}^2.\\ &\quad=0.53\left\{\frac{1596\sin^2\theta}{12.25-\cos^2\theta}(1-0.08163\cos^2\theta) +\left[15.39\cos\theta+\frac{\sin\theta}{\sqrt{12.25-\cos^2\theta}}(15.39\cos\theta-11.47\sin\theta)\right]^2\right\}\\ &\qquad+7.334\times 10^{-3}\left(\frac{-366.52\sin\theta}{\sqrt{12.25-\cos^2\theta}}\right)^2.\\ &\quad=\frac{845.9\sin^2\theta}{12.25-\cos^2\theta}(1-0.08163\cos^2\theta) +\left[8.157\cos\theta+\frac{\sin\theta}{\sqrt{12.25-\cos^2\theta}}(8.157\cos\theta-6.079\sin\theta)\right]^2\\ &\qquad+\frac{985.2\sin^2\theta}{12.25-\cos^2\theta}.\\ &\quad=\frac{1831\sin^2\theta-69.05\sin^2\theta\cos^2\theta}{12.25-\cos^2\theta} +\left[8.157\cos\theta+\frac{\sin\theta}{\sqrt{12.25-\cos^2\theta}}(8.157\cos\theta-6.079\sin\theta)\right]^2.\\ \end{align*} \end{document}