10-18 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \omega_A&=9.5\ut{rad/s}\\ \omega_{B0}&=0\\ \alpha_B&=2.2\ut{rad/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \theta_A&=\theta_B\\ \omega_A t&=\omega_0t+\frac{1}{2}\alpha t^2\\ \omega_A&=\frac{1}{2}\alpha t\\ \end{aligned} $$ $$ \begin{aligned} t&=\frac{2\omega_A}{\alpha}\\ &=\frac{95}{11}\ut{s}\\ &\approx 8.636363636363637\ut{s}\\ &\approx 8.6\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_A-\theta_B&=\omega_A t-\(\omega_0t+\frac{1}{2}\alpha t^2\)\\ \theta_{A\larr B}&=\omega_A t-\frac{1}{2}\alpha t^2\\ &=\frac{19}{2}t-\frac{11}{10}t^2\\ \end{aligned} $$ $$ \begin{aligned} \frac{\max(\theta_{A\larr B})}{2\pi}&=\frac{1805}{176\pi}\\ &\approx 3.2644849122826263\\ &\approx 3.3\\ \end{aligned} $$ $$\theta_{A\larr B}=2n\pi,$$ $$n=1,2,3,3,2,1,0$$ $$\therefore \text{not 1st time}$$