6-8 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=0.250\ut{kg}\\ m_B&=0.010\ut{kg}\\ m&=m_A+m_B=0.260\ut{kg}\\ R&=0.650\ut{m}\\ T_{\max}&=38.0\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$a=\frac{v^2}{R},$$ $$ \begin{cases} T:\text{Top}\\ B:\text{Bottom}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_T &=\cfrac{mv^2}{R}\\ \Sigma F_B &=\cfrac{mv^2}{R}\\ \end{cases} $$ $$ \begin{cases} T_T+mg &=\cfrac{mv^2}{R}\\ T_B-mg &=\cfrac{mv^2}{R}\\ \end{cases} $$ $$ \begin{cases} T_T &=\cfrac{mv^2}{R}-mg\\ T_B &=\cfrac{mv^2}{R}+mg\\ \end{cases} $$ $$\therefore T_B \gt T_T$$ $$\text{Bottom}$$ $$\ab{b}$$ $$T_B =\frac{mv^2}{R}+mg=38\ut{N}$$ $$ \begin{aligned} vv&=\sqrt{\frac{R (T-mg)}{m}}\\&=\frac{1}{2} \sqrt{380-\frac{13 g}{5}}\\ &\approx 9.41412117512835\ut{m/s}\\ &\approx 9.41\ut{m/s}\\ \end{aligned} $$