4-13 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&=4.5\ut{m}\\ a_R&=6.5g\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$a=\frac{v^2}{R},$$ $$ \begin{aligned} v&=\sqrt{aR}\\ &=\sqrt{(6.5g)(4.5)}\\ &=\frac{3 }{2}\sqrt{13g}\\ &=\frac{3 }{200}\sqrt{\frac{2549729}{2}}\ut{m/s^2}\\ &\approx 16.93648465591369\ut{m/s^2}\\ &\approx 17\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} f &=\frac{\omega}{2\pi}\\ &= \frac{v}{2\pi R}\\ &= \frac{\sqrt{aR}}{2\pi R}\\ &=\frac{\sqrt{13g}}{6 \pi }\ut{Hz}\\ &=\frac{\sqrt{13g}}{6 \pi }\ut{/s}\cdot\frac{60\ut{s}}{1\ut{min}}\\ &=\frac{10\sqrt{13g}}{\pi}\ut{/min}\\ &\approx 35.94033668784937\ut{/min}\\ &\approx 36\ut{/min}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} T&=\frac{1}{f}\\ &=600 \sqrt{\frac{2}{2549729}} \pi\ut{s}\\ &\approx 1.669433442460896\ut{s}\\ &\approx 1.67\ut{s}\\ \end{aligned} $$