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UNSW MMAN1300 2012-10-15 AQ1.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large MMAN1300 2012-10-15 AQ1} \begin{align*} \text{\bf Given:}\quad& \text{$m$=12.6 Kg, $v$=5.8 m/s, $M$=4.2 Kg, $r$=1.3m}\\ \\ \\ &\text{For the block,}\quad T-mg=ma,\quad T=ma+mg\ldots(1)\\ &\text{For the bar, there is $T$ on the left, $A$ on the right and the weight $Mg$ in the middle.}\\ &\text{The acceleration of the bar is $\frac{-a}{2}$ at the centre of gravity (the mid-point).}\\ &T+A_y-Mg=M\frac{-a}{2},\quad A_y=M\left(g-\frac{a}{2}\right)-T.\\ &\text{Sub (1) into the above:}\quad A_y=Mg-\frac{Ma}{2}-(ma+mg),\\ &A_y=(M-m)g-\left(\frac{M}{2}+m\right)a\ldots(2)\\ &\text{There is also a rotation on the bar about A with an anti-clockwise $\alpha=\frac{a}{r}$.}\\ &\text{The moment of the bar about $A$ is}\quad I=\frac{Mr^2}{3}.\\ &\tau=Mg\cdot\frac{r}{2}-Tr=I\alpha=\frac{Mr^2}{3}\cdot\frac{a}{r}.\\ &\times\frac{6}{r}:\quad 3Mg-6T=2Ma\\ &\text{Sub (1) into the above:}\quad 3Mg-6(ma+mg)=2Ma.\\ &(2M+6m)a=(3M-6m)g,\quad a=\frac{3M-6m}{2M+6m}g\approx -7.350\text{ ms$^{-2}$}\ldots(3).\\ &\text{Sub (3) into (2):}\quad A_y=(M-m)g-\left(\frac{M}{2}+m\right)\cdot\left(\frac{3M-6m}{2M+6m}g\right)=25.725 N\\ \\ \\ &\text{For $A_x$, it can be determined by $a_x$ of the bar at its centre of gravity, the mid-point.}\\ &\text{At the tip, }v_t=-v,\quad v_t=-\omega r.\quad\text{At the mid-point, }v_m=-\omega\cdot\frac{r}{2}=\frac{v_t}{2}=-\frac{v}{2},\\ &v_x=v_m\sin(-\theta)=\frac{v}{2}\sin\theta.\quad\text{($\theta$ positive for anti-clockwise; $x$ points to the right.)}\\ &a_x=\frac{d\:v_x}{dt}=\frac{v}{2}\cos\theta\cdot\frac{d\theta}{dt}=\frac{v}{2}\cos\theta\cdot\omega=\frac{v}{2}\cos\theta\cdot\frac{v}{r}=\frac{v^2}{2r}.\qquad(\theta=0.)\\ &A_x=Ma_x=\frac{Mv^2}{2r}\approx 54.342 N.\quad\text{(Towards the right into the wall)}\\ \\ \\ &A=\sqrt{A_x^2+A_y^2}\approx 60.123 N.\\ \end{align*} % % % \begin{align*} \text{Alternative Method:}&\\ \\ \text{\bf Given:}\quad& \text{$m$=12.6 Kg, $v$=5.8 m/s, $M$=4.2 Kg, $r$=1.3m}\\ \\ \\ &\text{Total energy = PE(Block) + KE(Block) + PE(Bar) + KE(Bar) + RE(Bar).}\\ &E=mgh+\frac{1}{2}mv^2+MgH+\frac{1}{2}MV^2+\frac{1}{2}I\omega^2.\\ &\text{The centre of gravity of the bar is in its mid-point. $\theta$ is positive for anti-clockwise.}\\ &H=\frac{r}{2}\sin(-\theta),\quad V=\frac{r(-\theta)}{2},\quad I=\frac{Mr^2}{3},\quad\omega=\frac{v}{r}.\\ &E=mgh+\frac{1}{2}mv^2-Mg\frac{r\sin(\theta)}{2}+\frac{1}{2}M\left(\frac{r\theta}{2}\right)^2+\frac{1}{2}\cdot\frac{Mr^2\omega^2}{3}\\ &\frac{d}{dt}E=mg\frac{dh}{dt}+\frac{m}{2}\cdot\frac{dv^2}{dt}-\frac{Mgr}{2}\cdot\frac{d\sin(\theta)}{dt}+\frac{Mr^2}{8}\cdot\frac{d\theta^2}{dt}+\frac{M}{6}\cdot\frac{dv^2}{dt}\\ &\quad=mgv+\frac{m}{2}\cdot 2v\frac{dv}{dt}-\frac{Mgr\cos(\theta)}{2}\cdot\frac{d\theta}{dt}+\frac{Mr^2}{8}\cdot 2\theta\frac{d\theta}{dt}+\frac{M}{6}\cdot 2v\frac{dv}{dt}\\ &\quad=mgv+mva-\frac{1}{2}Mgr\cos(\theta)\cdot\omega+\frac{1}{4}Mr^2\theta\cdot\omega+\frac{1}{3}Mv\cdot a\\ &\quad=mgv+mva-\frac{1}{2}Mgv\cos(\theta)+\frac{1}{4}Mr\theta\cdot v+\frac{1}{3}Mv\cdot a\\ &\quad=0\\ &mg+ma-\frac{1}{2}Mg\cos(\theta)+\frac{1}{4}Mr\theta+\frac{1}{3}Ma=0.\\ &\text{When }\theta=0,\quad mg+ma-\frac{1}{2}Mg+\frac{1}{3}Ma=0,\quad 6mg+6ma-3Mg+2Ma=0,\\ &a=\frac{3M-6m}{2M+6m}g\approx -7.350\text{ ms$^{-2}$}.\\ \\ \\ &\text{$-a$ is the tangential acceleration of the bar at the loose end.}\\ &\text{Since we chose an anti-clockwise direction,}\quad -a=r\:(-\alpha),\quad a=r\:\alpha.\\ &\text{At the centre of gravity of the bar (mid-point),}\\ &\text{tangential acceleration}\quad a_T=\alpha\cdot\frac{r}{2}=\frac{a}{2}\approx -3.675\text{ ms$^{-2}$ (downward decelerating), and}\\ &\text{normal acceleration}\quad a_N=\frac{r}{2}\omega^2=\frac{r}{2}\left(\frac{v}{r}\right)^2=\frac{v^2}{2r}\approx 12.938\text{ ms$^{-2}$ (leftward acceleration)}.\\ &T-mg=ma,\quad T=ma+mg.\\ &\text{Downward force: }Mg-T-A_y=Ma_T,\quad A_y=Mg-(ma+mg)-\frac{Ma}{2}\approx 25.725\text{ ms$^{-2}$}.\\ &A_x=Ma_N=\frac{Mv^2}{2r}\approx 54.342\text{ ms$^{-2}$}.\\ &A=\sqrt{A_x^2+A_y^2}\approx 60.123\text{ N}.\\ \end{align*} \end{document}