10-4 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \omega &=2000\ut{rev/min}=\frac{200\pi}{3}\ut{rad/s}\\ r_1&=1.50\ut{m}\\ r_2&=0.150\ut{m}\\ \end{cases} $$ $$ \begin{aligned} a_R(r)&=r\omega^2\\ &=\frac{4\pi^2r}{9}\times10^4 \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} a(r_1)&=\frac{4\pi^2r_1}{9}\times10^4\\ &=\frac{2\pi^2}{3}\times10^4\ut{m/s^2}\\ &\approx 6.579736267392906\times10^4\ut{m/s^2}\\ &\approx 6.58\times10^4\ut{m/s^2}\\ &\approx 65.8\ut{km/s^2}\\ \end{aligned} $$ $$ \begin{aligned} a(r_2)&=\frac{4\pi^2r_2}{9}\times10^4\\ &=\frac{2\pi^2}{3}\times10^3\ut{m/s^2}\\ &\approx 6.579736267392906\times10^3\ut{m/s^2}\\ &\approx 6.58\times10^3\ut{m/s^2}\\ &\approx 6.58\ut{km/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} a_R(r)&=\frac{4\pi^2r}{9}\times10^4,\\ \frac{\dd a_R}{\dd r}&=\frac{\dd }{\dd r}\(\frac{4\pi^2r}{9}\times10^4\)\\ &=\frac{4\pi^2}{9}\times10^4\ut{/s^2}\\ &\approx 4.3864908449286035\times10^4\ut{/s^2}\\ &\approx 4.39\times10^4\ut{/s^2}\\ \end{aligned} $$