10-48 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \omega_i&=33\frac{1}{3}\ut{rev/min}\\ t&=15.0\ut{s}=\frac{1}{4}\ut{min}\\ \omega_f&=0\\ \end{cases} $$ $$\ab{a}$$ $$\omega=\omega_0+\alpha t,$$ $$ \begin{aligned} \alpha&=\frac{\omega_f-\omega_i}{t}\\ &=\frac{0-\(33+\frac{1}{3}\ut{rev/min}\)}{\frac{1}{4}\ut{min}}\\ &=-\frac{400}{3}\ut{rev/min^2}\\ &\approx -133.33333333333334\ut{rev/min^2}\\ &\approx -133\ut{rev/min^2}\\ \end{aligned} $$ $$\ab{b}$$ $$\Delta \theta=\frac{1}{2}\(\omega+\omega_0\)t,$$ $$ \begin{aligned} \Delta \theta&=\frac{0+\(33+\frac{1}{3}\ut{rev/min}\)}{2}\cdot \frac{1}{4}\ut{min}\\ &=\frac{25}{6}\ut{rev}\\ &\approx 4.166666666666667\ut{rev}\\ &\approx 4.17\ut{rev}\\ \end{aligned} $$