6-10 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} T_{\max}&=35\ut{N}\\ m&=0.40\ut{kg}\\ R&=0.91\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$a_R=\frac{v^2}{R},$$ $$ \begin{aligned} \Sigma F_R&=\frac{mv^2}{R}\\ T+\frac{mg}{\sin\theta}&=\frac{mv^2}{R}\\ \end{aligned} $$ $$ \begin{aligned} T(\theta)&=\frac{mv^2}{R}-\frac{mg}{\sin\theta}\\ \frac{\dd T}{\dd\theta}&=\frac{\dd}{\dd\theta}\(\frac{mv^2}{R}-\frac{mg}{\sin\theta}\)\\ &= mg \cot\theta \csc\theta=0\\ \end{aligned} $$ $$\theta = \pm\frac{\pi}{2}\ut{rad}$$ $$ \begin{cases} T\(\frac{\pi}{2}\)&=\frac{mv^2}{R}-mg\\ T\(-\frac{\pi}{2}\)&=\frac{mv^2}{R}+mg\\ \end{cases} $$ $$T_{\max}=T\(-\frac{\pi}{2}\)$$ $$\therefore \text{Bottom}$$ $$\ab{b}$$ $$T_B =\frac{mv^2}{R}+mg$$ $$ \begin{aligned} v&=\sqrt{\frac{R (T-mg)}{m}}\\ &=\frac{1}{10}\sqrt{\frac{91(175-2 g)}{2}}\\ &\approx 8.408385606048286\ut{m/s}\\ &\approx 8.4\ut{m/s}\\ \end{aligned} $$