11-55 할리데이 11판 솔루션 일반물리학

$$\begin{cases} R&=0.48\ut{m}\\ h&=0.34\ut{m}\\ N&=2.00mg\\ I&=\beta mr^2 \end{cases}$$ $$\begin{aligned} {m{v_f}^2\over R}&=\Sigma F_R\\ &=N-mg\\ &=2mg-mg\\ &=mg\\ \end{aligned}$$ $${{v_f}^2}=gR$$ $$v=\omega r,$$ $$\Delta \Sigma E=0$$ $$\begin{aligned} 0 &=mg\Delta y+{1\over2}I\Delta \br{\omega^2}+{1\over2}m\Delta \br{v^2}\\ &=mg\br{-h}+{1\over2}\br{\beta mr^2}\br{v_f\over r}^2+{1\over2}m{v_f}^2\\ &=-2gh+{\beta }{v_f}^2+{v_f}^2\\ \end{aligned}$$ $$\begin{aligned} 2gh &=\br{1+\beta }{v_f}^2\\ &=\br{1+\beta }gR\\ \end{aligned}$$ $$\begin{aligned} \beta &={2h\over R}-1\\ &={5\over 12}\\ &\approx 0.4166666666666667\\ &\approx 0.42\\ \end{aligned}$$