10-14 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} r&=0.20\ut{m}\\ m&=2.0\ut{kg}\\ \theta&=20\degree\\ a&=2.0\ut{m/s^2}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} \Sigma F&=ma\\ \Sigma F&=mg\sin\theta-T\\ a&=r\alpha\\ \tau&=I\alpha\\ \tau&=rT \end{cases} $$ $$ \begin{aligned} I&=\frac{g\sin\theta-a}{a}mr^2\\ &=\frac{g\sin20\degree -2}{25}\\ &\approx 5.416287354178676\times10^{-2}\ut{kg\cdot m^2}\\ &\approx 5.4\times10^{-2}\ut{kg\cdot m^2}\\ &\approx 54\ut{g\cdot m^2}\\ \end{aligned} $$