11-1 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=10.0\ut{kg}\\ R&=0.400\ut{m}\\ r&=0.300\ut{m}\\ I&=0.600\ut{kg\cdot m^2}\\ \theta&=30.0\degree\\ d&=2.00\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} -\Delta y&=d\sin\theta,\\ -\Delta \GE &= mg(-\Delta y)\\ &= mgd\sin\theta\\ &= 10g\taag1\\ \end{aligned} $$ $$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\ \KE : \text{Translational Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \begin{aligned} {v_\com}&=r\omega\\ {v_\com}^2&=r^2\omega^2\\ \frac{1}{2}m{v_\com}^2&=\frac{1}{2}mr^2\omega^2\\ &=\frac{mr^2}{I}\cdot\frac{1}{2}I\omega^2\\ \KE&=\frac{mr^2}{I}\RE\\ &=\frac{3}{2}\cdot \RE\\ \end{aligned} $$ $$\therefore \KE : \RE =3:2$$ $$ \begin{cases} \KE&=\cfrac{3}{5}(\KE+\RE)\\ \RE&=\cfrac{2}{5}(\KE+\RE)\\ \end{cases} $$ $$\Sigma \Delta E=0,$$ $$\Delta \KE+\Delta \RE=-\Delta \GE,$$ $$ \begin{cases} \KE_f&=\cfrac{3}{5}(-\Delta \GE)\\ \RE_f&=\cfrac{2}{5}(-\Delta \GE)\\ \end{cases} $$ $$ \begin{cases} \KE_f&=\cfrac{3}{5}\cdot 10g\\ \RE_f&=\cfrac{2}{5}\cdot 10g\\ \end{cases} $$ $$ \begin{cases} \KE_f&=6g\\ \RE_f&=4g\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \RE_f&=4g\\ &=39.2266\ut{J}\\ &\approx 39.2\ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \KE_f&=6g\\ &=58.8399\ut{J}\\ &\approx 58.8\ut{J}\\ \end{aligned} $$