8-32 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L&=0.80\ut{m}\\ m&=0.12\ut{kg}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \put \begin{cases} 0 : \text{Vertical Peak Point}\\ 1 : \text{Horizontal Point}\\ 2 : \text{Lowest Point}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{cases} v_0&=0 \end{cases} $$ $$\Sigma E_0=\Sigma E_2,$$ $$ \begin{aligned} \KE_0+\GE_0&=\KE_2+\GE_2\\ \KE_2&=-\Delta \GE_{0\rarr2}\\ \frac{1}{2}m{v_2}^2&=-\Delta (mgh)_{0\rarr2}\\ \end{aligned} $$ $$ \begin{aligned} v_2&=\sqrt{2g(-\Delta h_{0\rarr2})}\\ &=\sqrt{2g(2L)}\\ &=\sqrt{4gL}\\ &=4\sqrt{\frac{g}{5}}\\ &\approx 5.601899677787884\ut{m/s}\\ &\approx 5.6\ut{m/s} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma F_{2y}&=\Sigma F_{2R}\\ T-mg&=\frac{m{v_2}^2}{R}\\ \end{aligned} $$ $$ \begin{aligned} T&=\frac{m(4gL)}{L}+mg\\ &=5mg\\ &=\frac{3g}{5}\\ &=5.88399\ut{N}\\ &\approx 5.9\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{cases} v_1&=0 \end{cases} $$ $$ \begin{aligned} \cos\theta&=\frac{-\Delta h_{1\rarr f}}{L}\\ -\Delta h_{1\rarr f}&=L\cos\theta \end{aligned} $$ $$\Sigma E_1=\Sigma E_f,$$ $$ \begin{aligned} \GE_1+\KE_1&=\GE_f+\KE_f\\ \GE_1&=\GE_f+\KE_f\\ \KE_f&=-\Delta \GE_{1\rarr f}\\ \frac{1}{2}m{v_f}^2&=-\Delta (mgh)_{1\rarr f}\\ {v_f}^2&=2g(-\Delta h_{1\rarr f})\\ {v_f}^2&=2gL\cos\theta\\ \end{aligned} $$ $$ \Sigma F_R=T-mg\cos\theta =\frac{mv^2}{R} $$ $$ \begin{aligned} T&=mg\cos\theta+\frac{m(2gL\cos\theta)}{L}\\ &=3mg\cos\theta \end{aligned} $$ $$T=3mg\cos\theta=mg$$ $$3\cos\theta=1$$ $$ \begin{aligned} \theta&=\cos^{-1}\frac{1}{3}\\ &\approx 1.2309594173407747\ut{rad}\\ &\approx 1.2\ut{rad}\\ \end{aligned} $$ $$\ab{d}$$ $$\theta=\cos^{-1}\frac{1}{3},$$ $$\text{Constant about }m$$