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UNSW MMAN2300 2013-10-05 Assignment C - c.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.\\ \\ &\sin\phi =\frac{R}{L}\cos\theta,\qquad \cos\phi =\sqrt{1-\left(\frac{R}{L}\right)^2\cos^2\theta} =\frac{\sqrt{L^2-R^2\cos^2\theta}}{L}.\\ \\ &\omega_{BC} =\frac{-\omega_{AB}\sin\theta}{\sqrt{\left(\frac{L}{R}\right)^2-\cos^2\theta}} =\frac{-R\omega_{AB}\sin\theta}{\sqrt{L^2-R^2\cos^2\theta}} =\frac{-R\omega_{AB}\sin\theta}{L\cos\phi} =\frac{-\omega_{AB}\sin\phi\sin\theta}{\cos\theta\cos\phi} =-\omega_{AB}\tan\phi\tan\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{(R^2-L^2)R\cos\theta}{(L^2-R^2\cos^2\theta)^\frac{3}{2}}\cdot\omega_{AB}^2 =\frac{(R^2-L^2)R\cos\theta}{L^3\cos^3\phi}\cdot\omega_{AB}^2\\ &\qquad=\frac{(R^2-L^2)\sin\phi}{L^2\cos^3\phi}\cdot\omega_{AB}^2 =\left[\left(\frac{R}{L}\right)^2-1\right]\frac{\omega_{AB}^2\tan\phi}{\cos^2\phi} .\\ \\ &v_{C_y}=R\cos\theta(\omega_{AB}-\omega_{BC}).\\ \\ &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 =-R\omega_{AB}^2\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}} -R\omega_{AB}^2\frac{\sin^2\theta}{\sqrt{\left(\frac{L}{R}\right)^2-\cos^2\theta}} -R\omega_{AB}^2\sin\theta\\ &\qquad =-R\alpha_{BC}\cos\theta +R\omega_{AB}\omega_{BC}\sin\theta -R\omega_{AB}^2\sin\theta\\ &\qquad=-R\alpha_{BC}\cos\theta +R\omega_{AB}(\omega_{BC}-\omega_{AB})\sin\theta\\ &\qquad=-R\alpha_{BC}\cos\theta -v_{C_y}\omega_{AB}\tan\theta\\ \\ \\ \text{c)}\quad&\text{Determine the horizontal and vertical forces acting on the connecting rod at points B and C as functions of crank angle.}\\ \\ &\text{$F_B$ provides the centripetal force on $B$ towards $A$ and the force pushing the bottom end of the rob around.}\\ &\text{$F_C$ is the result of the force from the left wall in the $\mathbf{i}$ direction, and the weight of the piston in the $-\mathbf{j}$ direction.}\\ &\text{These are the only two forces acting on the connecting rod.}\\ &-F_{C_y}-m_P~g=m_P~a_{C_y}.\\ \\ &\boxed{F_{C_y}=-m_P~a_{C_y}-m_P~g.}\qquad\text{Generally, $g+a_{C_y}>0$. So $F_{C_y}<0$ (downward).}\\ \\ &\text{The centre of gravity $D$ divides the rob into $CD$ and $DB$ in the ratio of $L-H:H$.}\\ &r_D=\frac{H}{L}r_C+\frac{L-H}{L}r_B,\quad v_D=\frac{H}{L}v_C+\frac{L-H}{L}v_B,\quad a_D=\frac{H}{L}a_C+\frac{L-H}{L}a_B.\\ &a_C=\mathbf{j}a_{C_y},\quad r_B=R(\mathbf{i}\cos\theta+\mathbf{j}\sin\theta),\quad v_B=R\omega_{AB}(-\mathbf{i}\sin\theta+\mathbf{j}\cos\theta),\quad a_B=-R\omega_{AB}^2(\mathbf{i}\cos\theta+\mathbf{j}\sin\theta).\\ &a_D=\frac{H}{L}\mathbf{j}a_{C_y}-\frac{L-H}{L}R\omega_{AB}^2(\mathbf{i}\cos\theta+\mathbf{j}\sin\theta) =-\mathbf{i}\frac{L-H}{L}R\omega_{AB}^2\cos\theta+\mathbf{j}\left(\frac{H}{L}a_{C_y}-\frac{L-H}{L}R\omega_{AB}^2\sin\theta\right).\end{align*} % % \begin{align*} &\text{Alternatively, }a_D=a_C+\mathbf{\alpha_{BC}}\times r_{D/C}-\omega_{BC}^2\cdot r_{D/C}\\ &\quad=\mathbf{j}a_{C_y}+\mathbf{k}~\alpha_{BC}\times[\mathbf{i}(L-H)\sin\phi-\mathbf{j}(L-H)\cos\phi]-\omega_{BC}^2\cdot[\mathbf{i}(L-H)\sin\phi-\mathbf{j}(L-H)\cos\phi]\\ &\quad=\mathbf{j}a_{C_y}+\alpha_{BC}\cdot[\mathbf{j}(L-H)\sin\phi+\mathbf{i}(L-H)\cos\phi]-\omega_{BC}^2\cdot[\mathbf{i}(L-H)\sin\phi-\mathbf{j}(L-H)\cos\phi].\\ &\quad=\mathbf{j}a_{C_y}+\alpha_{BC}\mathbf{j}(L-H)\sin\phi+\alpha_{BC}\mathbf{i}(L-H)\cos\phi-\omega_{BC}^2\mathbf{i}(L-H)\sin\phi+\omega_{BC}^2\mathbf{j}(L-H)\cos\phi.\\ &\quad=\mathbf{i}\left[\alpha_{BC}(L-H)\cos\phi-\omega_{BC}^2(L-H)\sin\phi\right]+\mathbf{j}\left[a_{C_y}+\alpha_{BC}(L-H)\sin\phi+\omega_{BC}^2(L-H)\cos\phi\right].\\ &\text{This can be proven to be equivalent to $a_D$ above.} \end{align*} % % \begin{align*} &F_{B_y}+F_{C_y}-m_R~g=m_R~a_{D_y}.\\ &F_{B_y} =m_R~a_{D_y}+m_R~g-F_{C_y} =m_R~a_{D_y}+m_R~g+m_P~g+m_P~a_{C_y}\\ &\qquad=m_R\left[a_{C_y}+\alpha_{BC}(L-H)\sin\phi+\omega_{BC}^2(L-H)\cos\phi\right]+m_R~g+m_P~g+m_P~a_{C_y}\\ &\qquad=(m_R+m_P)(a_{C_y}+g)+m_R\left[\alpha_{BC}(L-H)\sin\phi+\omega_{BC}^2(L-H)\cos\phi\right]\\ &\qquad=(m_R+m_P)(a_{C_y}+g)+m_R\left[\frac{(R^2-L^2)\sin\phi}{L^2\cos^3\phi}\cdot\omega_{AB}^2\cdot(L-H)\sin\phi+\frac{R^2\omega_{AB}^2\sin^2\theta}{L^2\cos^2\phi}(L-H)\cos\phi\right]\\ &\qquad=(m_R+m_P)(a_{C_y}+g)+\frac{m_R\omega_{AB}^2(L-H)}{L^2\cos^2\phi}\left[(R^2-L^2)\frac{\sin^2\phi}{\cos\phi}+R^2\sin^2\theta\cos\phi\right]\\ &\qquad=(m_R+m_P)(a_{C_y}+g)+\frac{m_R\omega_{AB}^2(L-H)}{L^2\cos^2\phi}\left[(R^2-L^2)\frac{\sin^2\phi}{\cos\phi}+(R^2-L^2\sin^2\phi)\cos\phi\right]\\ &\qquad=(m_R+m_P)(a_{C_y}+g)+\frac{m_R\omega_{AB}^2(L-H)}{L^2\cos^3\phi}\left[R^2\sin^2\phi-L^2\sin^2\phi+R^2\cos^2\phi-L^2\sin^2\phi\cos\phi^2\right]\\ &\qquad=(m_R+m_P)(a_{C_y}+g)+\frac{m_R\omega_{AB}^2(L-H)}{L^2\cos^3\phi}\left[R^2-L^2\sin^2\phi(1+\cos\phi^2)\right]\\ \\ \\ \\ &\text{FOLLOWINGS ARE UNFINISHED}\\ % %m_R~\left(\frac{H}{L}a_{C_y}-\frac{L-H}{L}R\omega_{AB}^2\sin\theta\right)+(m_R+m_P)g\\ &=m_R~\left(-\frac{H}{L} 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\} -\frac{L-H}{L}R\omega_{AB}^2\sin\theta\right)+(m_R+m_P)g\\ &=-m_R\frac{H}{L}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+\frac{L-H}{H}\sin\theta \right\}+(m_R+m_P)g\\ &\boxed{F_{B_y}=-685.5~\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}}+3.868\sin\theta \right\}+8.927.}\\ \end{align*} % % % \begin{align*} &\text{Moment about $B$:}\quad -L\sin\phi\cdot F_{C_y}-L\cos\phi\cdot F_{C_x}=\left(\bar{I}_R+m_R~H^2\right)\alpha_{BC}.\\ &F_{C_x}=\frac{\left(\bar{I}_R+m_R~H^2\right)\alpha_{BC}+L\sin\phi\cdot F_{C_y}}{-L\cos\phi} =\frac{1}{-L\cos\phi}\left[\left(\bar{I}_R+m_R~H^2\right)\alpha_{BC}-L\frac{R}{L}\cos\theta\cdot m_P~g\right],\\ &=\frac{-1}{\sqrt{L^2-R^2\cos\theta}}\left[\left(\bar{I}_R+m_R~H^2\right)\cdot \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 -R\cos\theta\cdot m_P~g\right]\\ &=\frac{-R\cos\theta}{\sqrt{L^2-R^2\cos\theta}}\left[\left(\bar{I}_R+m_R~H^2\right)\cdot \frac{\left(R^2-L^2\right)}{\left(L^2-R^2\cos^2\theta\right)^\frac{3}{2}}\cdot\omega_{AB}^2 -m_P~g\right]\\ &=\frac{-R\left(\bar{I}_R+m_R~H^2\right)\cdot\left(R^2-L^2\right)\omega_{AB}^2}{\left(L^2-R^2\cos^2\theta\right)^2}\cdot \cos\theta +\frac{R~m_P~g}{\sqrt{L^2-R^2\cos\theta}}\cos\theta\\ \\ &\boxed{F_{C_x}=\frac{582.3}{(1-0.08163\cos^2\theta)^2}\cdot\cos\theta+\frac{1.233}{\sqrt{1-0.08163\cos^2\theta}}\cdot\cos\theta.}\\ \\ &\text{Horizontal force:}\quad F_{C_x}+F_{B_x}=m_R~a_{D_x}.\\ &F_{B_x}=m_R~a_{D_x}-F_{C_x} =-m_R\frac{L-H}{L}R\omega_{AB}^2\cos\theta+\frac{R\left(\bar{I}_R+m_R~H^2\right)\cdot\left(R^2-L^2\right)\omega_{AB}^2}{\left(L^2-R^2\cos^2\theta\right)^2}\cdot \cos\theta -\frac{R~m_P~g}{\sqrt{L^2-R^2\cos\theta}}\cos\theta\\ &=R\omega_{AB}^2\left[-m_R\frac{L-H}{L}+\frac{\left(\bar{I}_R+m_R~H^2\right)\cdot\left(R^2-L^2\right)}{\left(L^2-R^2\cos^2\theta\right)^2}\right]\cos\theta -\frac{R~m_P~g}{\sqrt{L^2-R^2\cos\theta}}\cos\theta.\\ &\boxed{F_{B_x}=-1966\cos\theta-\frac{582.3}{(1-0.08163\cos^2\theta)^2}\cos\theta-\frac{1.233}{\sqrt{1-0.08163\cos^2\theta}}\cos\theta.}\\ \end{align*} \end{document}