6-19 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=0.060\ut{kg}\\ L&=0.90\ut{m}\\ 2\pi R&=0.50\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$R=\frac{0.5}{2\pi}=\frac{1}{4\pi}$$ $$ \begin{aligned} h&=\sqrt{L^2-R^2}\\ &=\sqrt{\frac{81}{100}-\frac{1}{16 \pi ^2}} \end{aligned} $$ $$$$ $$ \begin{cases} \sin\theta=\cfrac{R}{L}\\ \cos\theta=\cfrac{h}{L} \end{cases} $$ $$ \begin{cases} \Sigma F_y&=F_T\sin\theta-mg=0\\ \Sigma F_R&=F_T\cos\theta=\cfrac{mv^2}{R} \\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} F_T&=\frac{mg}{\sin\theta}\\ &=\frac{mgL}{R}\\ &=\frac{27\pi}{125}g\\ &\approx 6.65463591280649\ut{N}\\ &\approx 6.7\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v&=\sqrt{\frac{F_TR\cos \theta}{m}}\\ &=\sqrt{\frac{(\frac{mgL}{R})\(R\)\(\frac{h}{L}\)}{m}}\\ &=\sqrt{g h}\\ \end{aligned} $$ $$ \begin{aligned} T&=\frac{2\pi R}{v}\\ &=\frac{2\pi R}{\sqrt{g h}}\\ &=\frac{1}{2 \sqrt[4]{\frac{81}{100}-\frac{1}{16 \pi ^2}} \sqrt{g}}\\ &\approx 0.1686322268415886\ut{s}\\ &\approx 0.17\ut{s}\\ \end{aligned} $$