10-16 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \Delta t_{1\rarr2}&=6.00\ut{s}\\ \theta_1&=10.0\ut{rad}\\ \theta_2&=70.0\ut{rad}\\ \omega_2&=15.0\ut{rad/s}\\ \end{cases} $$ $$\ab{a}$$ $$\Delta \theta=\frac{1}{2}\(\omega+\omega_0\)t,$$ $$\Delta \theta_{1\rarr2}=\frac{1}{2}\(\omega_1+\omega_2\)t,$$ $$ \begin{aligned} \omega_1&=\frac{2\Delta \theta}{t}-\omega_2\\ &=5\ut{rad/s}\\ &=5.00\ut{rad/s}\\ \end{aligned} $$ $$\ab{b}$$ $$\Delta \theta=\omega t-\frac{1}{2}\alpha t^2,$$ $$\Delta \theta_{1\rarr2}=\omega_2 t-\frac{1}{2}\alpha t^2$$ $$ \begin{aligned} \alpha&=\frac{2(\omega_2t-\Delta \theta)}{t^2}\\ &=\frac{5}{3}\ut{rad/s^2}\\ &\approx 1.6666666666666667\ut{rad/s^2}\\ &\approx 1.67\ut{rad/s^2}\\ \end{aligned} $$ $$\ab{c}$$ $$2\alpha \Delta \theta_{2\rarr 0}={\omega_2}^2-{\omega_0}^2,$$ $$\theta_2-\theta_0=\frac{{\omega_2}^2}{2\alpha}$$ $$ \begin{aligned} \theta_0&=\theta_2-\frac{{\omega_2}^2}{2\alpha}\\ &=\frac{5}{2}\ut{rad}\\ &=2.50\ut{rad} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \theta&=\omega_0t+\frac{1}{2}\alpha t^2,\\ &=\frac{5}{6} t^2\\ \end{aligned} $$ $$ \begin{aligned} \omega&=\omega_0+\alpha t,\\ &=\frac{5}{3} t\\ \end{aligned} $$