11-25 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \omega_{Ai}&=900\ut{rev/min}=30\pi\ut{rad/s}\\ I_B&=2I_A\\ \omega_{Bi}&=0\\ \end{cases}$$ $$\put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\ \end{cases}$$ $$\ab{a}$$ $$\Delta \Sigma \vec L=0,$$ $$\begin{aligned} 0&=\Delta L_A+\Delta L_B\\ &=I_A\Delta \omega_A+I_B\Delta \omega_B\\ &=I_A (\omega_f-\omega_{Ai})+2I_A \omega_f\\ &=3\omega_f -\omega_{Ai}\\ \end{aligned}$$ $$\begin{aligned} \omega_f&=\frac{\omega_{Ai}}{3}\\ &=10\pi\ut{rad/s}\\ &\approx 31.41592653589793\ut{rad/s}\\ &\approx 31.4\ut{rad/s}\\ &\approx 31\ut{rad/s}\\ &\approx 3\times10\ut{rad/s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \Ans &=-\frac{\Sigma \Delta \RE}{\Sigma \RE_i}\\ &=\frac{ \frac{1}{2}I_A\Delta ({\omega_{A}}^2)+\frac{1}{2}I_B\Delta ({\omega_{B}}^2)}{- \frac{1}{2}I_A{\omega_{Ai}}^2}\\ &=\frac{ I_A ({\omega_{f}}^2-{\omega_{Ai}}^2)+2I_A{\omega_{f}}^2}{- I_A{\omega_{Ai}}^2}\\ &=\frac{ 3{\omega_{f}}^2-{\omega_{Ai}}^2}{- {\omega_{Ai}}^2}\\ &=\frac{ 3{\(\frac{\omega_{Ai}}{3}\)}^2-{\omega_{Ai}}^2}{ -{\omega_{Ai}}^2}\\ &=\frac{2}{ 3}\\ &\approx 66.66666666666666\ut{\%}\\ &\approx 66.7\ut{\%}\\ &\approx 67\ut{\%}\\ \end{aligned}$$