11-54 할리데이 11판 솔루션 일반물리학

(풀이자주:풀이에 도형의 회전관성이 필요합니다. 관련한 내용은 별도의 링크로 분리했습니다. 해당 도형의 회전관성을 구하는 법에 관한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.) https://solutionpia.tistory.com/800 $$I_{\text{Solid Sphere}}=\frac{2}{5}MR^2,$$ $$ \put \begin{cases} i : \text{Start}\\ Q : \text{Q Point}\\ H : \text{Highest Point}\\ \end{cases} $$ $$ \begin{cases} m&=0.340\ut{g}\\ v_i&=0\\ R&=21.0\ut{cm}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$N_H\ge0,$$ $$ \begin{aligned} \Sigma F_H&={m{v_H}^2\over R}\\ mg+N_{H}&={m{v_H}^2\over R}\\ N_{H}&={m{v_H}^2\over R}-mg\ge0\\ \end{aligned} $$ $$ \begin{aligned} {{v_H}^2\over R}-g&\ge0\\ {{v_H}^2}&\ge gR\taag1\\ \end{aligned} $$ $$\Delta \Sigma E=0,$$ $$v=\omega r,$$ $$ \begin{aligned} 0&=mg\Delta y_{i\rarr H}+{1\over2}I\Delta \br{{\omega}^2}+{1\over2}m\Delta \br{{v}^2}\\ &=mg\br{2R-h}+{1\over2}I {{\omega_H}^2}+{1\over2}m{{v_H}^2}\\ &=mg\br{2R-h}+{1\over2}\br{\frac{2}{5}mr^2} {\br{v_H\over r}^2}+{1\over2}m{{v_H}^2}\\ &=mg\br{2R-h}+{7\over10}m{{v_H}^2}\\ &=g\br{2R-h}+{7\over10}{{v_H}^2}\\ \end{aligned} $$ $${{v_H}^2}={10\over7}g\br{h-2R}\taag2$$ $$ \begin{aligned} {{v_H}^2}={10\over7}g\br{h-2R}&\ge gR\\ {10\over7}\br{h-2R}&\ge R\\ \end{aligned} $$ $$ \begin{aligned} h&\ge \br{2+{7\over10}}R={27\over10}R\\ &\ge {576\over10}\ut{cm}= 57.6\ut{cm}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} 0&=mg\Delta y_{i\rarr Q}+{1\over2}I\Delta \br{{\omega}^2}+{1\over2}m\Delta \br{{v}^2}\\ &=mg\br{R-h}+{1\over2}I {{\omega_Q}^2}+{1\over2}m{{v_Q}^2}\\ &=mg\br{R-6R}+{1\over2}\br{\frac{2}{5}mr^2} {\br{v_Q\over r}^2}+{1\over2}m{{v_Q}^2}\\ &=mg\br{-5R}+{7\over10}m{{v_Q}^2}\\ &=-5gR+{7\over10}{{v_Q}^2}\\ \end{aligned} $$ $$ {v_Q}^2={50\over7}gR, $$ $$\Sigma F_Q={m{v_Q}^2\over R}$$ $$ \begin{aligned} N_{Q} &={m\over R}\cdot{50\over7}gR\\ &={50\over7}mg\\ &=23.81615\times10^{-3}\ut{N}\\ &\approx 23.8\times10^{-3}\ut{N}\\ &\approx 23.8\ut{mN}\\ \end{aligned} $$ $$\ab{c}$$ $$\text{Left}$$