4-16 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&=20.0\ut{cm}=0.2\ut{m}\\ T&=5.00\ut{ms}=5\times10^{-3}\ut{s}\\ h&=1.20\ut{m}\\ d&=3.50\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$[\text{get }\theta_0]$$ $$ \begin{aligned} \theta_{5\text{o'clock}\larr x+}&= -\frac{2}{12}\cdot2\pi\ut{rad}\\ &=-\frac{\pi}{3}\ut{rad}\\ \end{aligned} $$ $$\theta_{5\text{o'clock}\larr x+}\perp\theta_0,$$ $$ \begin{aligned} \theta_0&=-\frac{\pi}{3}+\frac{\pi}{2}\\ &=\frac{\pi}{6}\ut{rad}\\ \end{aligned} $$ $$[\text{get }v_0 ]$$ $$ \begin{aligned} v_0&=R\omega\\ &=R\cdot\frac{2\pi}{T}\\ &=(0.2)\cdot \frac{2\pi}{5\times10^{-3}}\\ &=80\pi\ut{m/s} \end{aligned} $$ $$[\text{get }\vec v_0 ]$$ $$ \begin{aligned} \vec v_0 &= 80\pi\cos\frac{\pi}{6}\i+80\pi\sin\frac{\pi}{6}\j\\ &=\(40\pi\sqrt3\i+40\pi\j\)\ut{m/s} \end{aligned} $$ $$[\text{get }t ]$$ $$ \begin{aligned} t&=\frac{\Delta x}{v_x}\\ &=\frac{3.5}{40\pi\sqrt3}\ut{s}\\ &=\frac{7}{80 \sqrt{3} \pi }\ut{s}\\ \end{aligned} $$ $$[\text{get }\Delta y ]$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} \Delta y&=v_{y0}t+\frac{1}{2}(-g)t^2\\ &=(40\pi)\(\frac{7}{80\sqrt{3}\pi}\)-\frac{1}{2}g\(\frac{7}{80\sqrt{3}\pi}\)^2\\ &=\(\frac{7}{2 \sqrt{3}}-\frac{49 g}{38400 \pi ^2}\)\ut{m}\\ \end{aligned} $$ $$[\text{get }\Ans ]$$ $$ \begin{aligned} \Ans &= h+\Delta y\\ &=1.20+\(\frac{7}{2 \sqrt{3}}-\frac{49 g}{38400 \pi ^2}\)\\ &\approx 3.219458039874817\ut{m}\\ &\approx 3.22\ut{m}\\ \end{aligned} $$