4-32 할리데이 11판 솔루션 일반물리학

$$\begin{cases} R&=4.00\ut{m}\\ T&=20.0\ut{s} \end{cases}$$ $$\begin{aligned} \vec O &= R\j\\ &= 4\j \end{aligned}$$ $$\begin{aligned} \omega &= \frac{2\pi}{T}\\ &=\frac{\pi}{10}\ut{rad/s} \end{aligned}$$ $$\begin{aligned} \vec R &= R\cos\theta\i+R\sin\theta\j\\ &= R\cos(\theta_0+\Delta\theta)\i+R\sin(\theta_0+\Delta\theta)\j\\ &= R\cos(\theta_0+\omega t)\i+R\sin(\theta_0+\omega t)\j\\ &= 4\cos\(-\frac{\pi}{2}+\frac{\pi}{10} t\)\i+4\sin\(-\frac{\pi}{2}+\frac{\pi}{10} t\)\j\\ &= 4\cos\frac{\pi(t-5)}{10}\i+4\sin\frac{\pi(t-5)}{10}\j\\ &= 4\sin\frac{\pi t}{10}\i-4\cos\frac{\pi t}{10}\j\\ \end{aligned}$$ $$\begin{aligned} \vec r&=\vec O + \vec R,\\ &=4\sin\frac{\pi t}{10}\i+\(4-4\cos\frac{\pi t}{10}\)\j\\ &=4\sin\frac{\pi t}{10}\i+8\sin^2\frac{\pi t}{20}\j\\ \end{aligned}$$ $$\ab{a,b,c}$$ $$\begin{aligned} r&=\sqrt{\(4\sin\frac{\pi t}{10}\)^2+\(8\sin^2\frac{\pi t}{20}\)^2}\\ &=\abs{8\sin\frac{\pi t}{20}} \end{aligned}$$ $$\begin{aligned} \theta_r&=\tan^{-1}\frac{8\sin^2\frac{\pi t}{20}}{4\sin\frac{\pi t}{10}}\\ &=\tan^{-1}\(\tan\frac{\pi t}{20}\)\\ &=\frac{\pi t}{20} \end{aligned}$$ $$\begin{aligned} \therefore \vec r(t)&=\la r,\theta \ra\\ &=\la\abs{8\sin\frac{\pi t}{20}},\frac{\pi t}{20}\ra \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} \vec r(5.00\ut{s})&=\la\abs{8\sin\frac{5\pi}{20}},\frac{5\pi}{20}\ra\\ &=\la4\sqrt2\ut{m},\frac{\pi}{4}\ut{rad}\ra\\ &\approx \la5.656854249492381\ut{m},0.7853981633974483\ut{rad}\ra\\ &\approx \la5.66\ut{m},0.785\ut{rad}\ra\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \vec r(7.50\ut{s})&=\la\abs{8\sin\frac{7.5\pi}{20}},\frac{7.5 \pi }{20}\ra\\ &=\la8 \cos\frac{\pi }{8}\ut{m},\frac{3\pi}{8}\ut{rad}\ra\\ &\approx \la7.391036260090294\ut{m},1.178097245096172\ut{rad}\ra\\ &\approx \la7.39\ut{m},1.18\ut{rad}\ra\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \vec r(10.0\ut{s})&=\la\abs{8\sin\frac{10\pi}{20}},\frac{10 \pi }{20}\ra\\ &=\la8\ut{m},\frac{\pi}{2}\ut{rad}\ra\\ &\approx \la8\ut{m},1.570796326794897\ut{rad}\ra\\ &\approx \la8.00\ut{m},1.57\ut{rad}\ra\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \vec r(t)&= 4\sin\frac{\pi t}{10}\i+8\sin^2\frac{\pi t}{20}\j\\ \Delta \vec r &= \vec r(10)-\vec r(5)\\ &=(8\j)-(4\i+4\j)\\ &=-4\i+4\j\ut{m}\\ &=\la4\sqrt2\ut{m},\frac{3\pi}{4}\ut{rad}\ra\\ &\approx \la5.66\ut{m},2.36\ut{rad}\ra \end{aligned}$$ $$\ab{e}$$ $$\begin{aligned} \vec v&=\frac{\Delta \vec r }{t}\\ &=\frac{-4\i+4\j}{5}\ut{m/s}\\ &=\la\frac{4\sqrt2}{5}\ut{m/s},\frac{3\pi}{4}\ut{rad}\ra\\ &\approx\la1.13\ut{m/s},2.36\ut{rad}\ra\\ \end{aligned}$$ $$\ab{f,g}$$ $$\begin{aligned} \vec r&= 4\sin\frac{\pi t}{10}\i+8\sin^2\frac{\pi t}{20}\j\\ \vec v(t) &= \dxt{\vec r(t)}\\ &=\frac{2}{5}\pi\cos\frac{\pi t}{10}\i+\frac{2}{5}\pi\sin\frac{\pi t}{10}\j\\ \end{aligned}$$ $$\ab{f}$$ $$\begin{aligned} \vec v(5.00\ut{s})&=\frac{2}{5}\pi\cos\frac{\pi (5)}{10}\i+\frac{2}{5}\pi\sin\frac{\pi (5)}{10}\j\\ &=\frac{2\pi}{5}\j\ut{m/s}\\ &=\la\frac{2\pi}{5}\ut{m},\frac{\pi}{2}\ut{rad}\ra\\ &\approx\la1.26\ut{m},1.57\ut{rad}\ra \end{aligned}$$ $$\ab{g}$$ $$\begin{aligned} \vec v(10.0\ut{s})&=\frac{2}{5}\pi\cos\frac{\pi (10)}{10}\i+\frac{2}{5}\pi\sin\frac{\pi (10)}{10}\j\\ &=-\frac{2 \pi }{5}\i\ut{m/s}\\ &=\la\frac{2\pi}{5}\ut{m},\pi\ut{rad}\ra\\ &\approx\la1.26\ut{m},3.14\ut{rad}\ra \end{aligned}$$ $$\ab{h,i}$$ $$\begin{aligned} \vec v&= \frac{2}{5}\pi\cos\frac{\pi t}{10}\i+\frac{2}{5}\pi\sin\frac{\pi t}{10}\j\\ \vec a(t) &= \dxt{\vec v(t)}\\ &=-\frac{1}{25}\pi^2\sin\frac{\pi t}{10}\i+\frac{1}{25}\pi^2\cos\frac{\pi t}{10}\j\\ \end{aligned}$$ $$\ab{h}$$ $$\begin{aligned} \vec a(5.00\ut{s})&=-\frac{1}{25}\pi^2\sin\frac{\pi (5)}{10}\i+\frac{1}{25}\pi^2\cos\frac{\pi (5)}{10}\j\\ &=-\frac{\pi^2}{25}\i\\ &=\la\frac{\pi^2}{25}\ut{m},\pi\ut{rad}\ra\\ &\approx \la0.395\ut{m},3.14\ut{rad}\ra\\ \end{aligned}$$ $$\ab{i}$$ $$\begin{aligned} \vec a(10.0\ut{s})&=-\frac{1}{25}\pi^2\sin\frac{\pi (10)}{10}\i+\frac{1}{25}\pi^2\cos\frac{\pi (10)}{10}\j\\ &=-\frac{\pi^2}{25}\j\\ &=\la\frac{\pi^2}{25}\ut{m},-\frac{\pi}{2}\ut{rad}\ra\\ &\approx \la0.395\ut{m},-1.57\ut{rad}\ra\\ \end{aligned}$$