11-45 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} M : \text{motor}\\ P : \text{ship}\\ \end{cases} $$ $$ \begin{cases} I_M&=3.0\times10^{-3}\ut{kg\cdot m^2}\\ I_P&=12\ut{kg\cdot m^2}\\ \Delta \theta_P&=30\degree={1\over12}\ut{rev}\\ \end{cases} $$ $$\vec L = I\vec \omega,$$ $$ \begin{aligned} \vec L_\net&=\Sigma \vec L\\ \Sigma I\cdot\vec\omega_\net&=\Sigma \br{I\vec\omega}\\ \end{aligned} $$ $$ \therefore \vec \omega_\net={\Sigma \br{I\vec\omega}\over \Sigma I} $$ $$ \begin{aligned} \Delta \theta_\net &= \int_0^t\omega_\net \dd t,\\ &= \int_0^t{\Sigma \br{I\vec\omega}\over \Sigma I} \dd t\\ &= {1 \over \Sigma I}\int_0^t\Sigma \br{I\vec\omega} \dd t\\ &= {1 \over \Sigma I}\sum\br{\int_0^t I\vec\omega \dd t}\\ &= {1 \over \Sigma I}\sum\br{I\int_0^t \vec\omega \dd t}\\ &= {\Sigma\br{I\Delta \theta} \over \Sigma I}\\ \end{aligned} $$ $$\Delta \theta_\net=0,$$ $$ \begin{aligned} 0 &={\Sigma\br{I\Delta \theta} \over \Sigma I}\\ &={I_M\Delta \theta_M+I_P\Delta \theta_P \over I_M+I_P}\\ &=I_M\Delta \theta_M+I_P\Delta \theta_P\\ \end{aligned} $$ $$\therefore \theta_M=-{I_P\over I_M}\Delta \theta_P$$ $$ \begin{aligned} \abs{ \theta_M} &={I_P\over I_M}\Delta \theta_P\\ &=4000\Delta \theta_P\\ &={1000\over3}\ut{rev}\\ &\approx 3.333333333333333\times10^2\ut{rev}\\ &\approx 3.3\times10^2\ut{rev}\\ \end{aligned} $$