6-63 할리데이 11판 솔루션 일반물리학

(풀이자주:문제의 의도가 3주기를 원주율로 두라는 경우로 보입니다. 해당 풀이를 별첨합니다.) $$ \begin{cases} m&=2.0\ut{g}\\ R&=5.0\ut{cm}\\ 3T_1&=3.14\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v_1&=\frac{2\pi R}{T_1}\\ &=\frac{1500\pi}{157}\ut{cm/s}\\ &\approx 30.015216435571272\ut{cm/s}\\ &\approx 30\ut{cm/s}\\ \end{aligned} $$ $$\ab{b}$$ $$a_R = \frac{{v_1}^2}{R},$$ $$ \begin{aligned} a_R&=\frac{450000\pi^2}{24649}\ut{cm/s^2}\\ &\approx 180.18264353483758\ut{cm/s^2}\\ &\approx 0.18\ut{m/s^2} \end{aligned} $$ $$\ab{c}$$ $$\text{in}$$ $$\ab{d}$$ $$ \begin{aligned} f&=\Sigma F_R=ma_R\\ &=\frac{9 \pi ^2}{246490}\ut{N}\\ &\approx 0.00036036528706967513\ut{N}\\ &\approx 0.36\ut{mN}\\ \end{aligned} $$ $$\ab{e}$$ $$\text{in}$$ $$\ab{f}$$ $$R=10\ut{cm},$$ $$ \begin{aligned} \mu m g &=\frac{mv^2}{R}\\ \mu &=\frac{v^2}{gR}\\ &\approx 0.09186758145484827\\ &\approx 9.2\times10^{-2}\\ \end{aligned} $$ $$\title{Case Pi}$$ $$ \begin{cases} m&=2.0\ut{g}\\ R&=5.0\ut{cm}\\ 3T_2&=\pi\ut{s}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v&=\frac{2\pi R}{T_2}\\ &=30\ut{cm/s}\\ \end{aligned} $$ $$\ab{b}$$ $$a_R = \frac{{v_2}^2}{R},$$ $$ \begin{aligned} a_R&=180\ut{cm/s^2}\\ &= 0.18\ut{m/s^2} \end{aligned} $$ $$\ab{c}$$ $$\text{in}$$ $$\ab{d}$$ $$ \begin{aligned} f&=\Sigma F_R=ma_R\\ &=\frac{9}{25000}\ut{N}\\ &= 0.00036\ut{N}\\ &= 0.36\ut{mN}\\ \end{aligned} $$ $$\ab{e}$$ $$\text{in}$$ $$\ab{f}$$ $$R=10\ut{cm},$$ $$ \begin{aligned} \mu m g &=\frac{mv^2}{R}\\ \mu &=\frac{v^2}{gR}\\ &=\frac{9}{10g}\\ &\approx 0.09177445916801354\\ &\approx 9.2\times10^{-2}\\ \end{aligned} $$