솔루션피아

$$\title{Rotational Inertia of Hollow Cylinder}$$ $$r=R,$$ $$ \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{LH}=\frac{M}{2\pi R H} \taag1$$ $$ \begin{aligned} l&=r\theta,\\ \dd l&=r\dd\theta\\ &=R\dd\theta\taag2 \end{aligned} $$ $$ \begin{aligned} \dd a&=\dd l\cdot\dd h \\ &=(R\dd\theta)\cdot\dd h \\ &=R\cdot\dd\theta\dd h \\ \end{aligned} $$ $$ \begin{aligned} \dd m&=\sigma\cdot\dd a\\ &=\sig..

$$\title{Rotational Inertia of Hoop}$$$$r=R,$$$$ \lambda=\frac{\dd m}{\dd l}=\frac{M}{L}=\frac{M}{2\pi r}=\frac{M}{2\pi R} \taag1$$$$ \begin{aligned}l&=r\theta,\\\dd l&=r\dd\theta\\&=R\dd\theta\taag2\end{aligned} $$$$$$$$ \begin{aligned} \dd m&=\lambda\cdot\dd l\\ &=\lambda\cdot (R\dd\theta)\\ &=\lambda R\cdot\dd\theta\taag3\\ \end{aligned} $$$$ \begin{aligned} \dd I&=r^2\cdot \dd m\\ &=R^2\cdot..

(풀이자주:풀이에 도형의 회전관성이 필요합니다. 관련한 내용은 별도의 링크로 분리했습니다. 해당 도형의 회전관성을 구하는 법에 관한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.)https://solutionpia.tistory.com/796 가는 막대의 회전 관성$$\title{Rotational Inertia of Staight Rod}$$ $$ \lambda=\frac{\dd m}{\dd L}=\frac{M}{L}\taag1$$ $$ \dd m=\lambda\cdot\dd L $$ $$ \begin{aligned} \dd I&={r}^2\cdot \dd m\\ &={L}^2\cdot (\lambda\cdot\dd L)\\ &=\lambda{L}^2\cdot \dd L\taa..

$$ \begin{cases} r&=2.3\times10^4\ut{ly}\\v&=250\ut{km/s}=2.5\times10^5\ut{m/s}\\t&=4.5\times10^9\ut{y}\\\end{cases} $$$$ \begin{cases} c&=299~792~458\ut{m/s}\\1\ut{d}&=86400\ut{s}\\1\ut{y}&=365.25\ut{d}=31~557~600\ut{s}\\1\ut{ly}&=9~460~730~472~580~800\ut{m}\end{cases} $$$$\ab{a}$$$$ \begin{aligned}T&=\frac{2\pi r}{v}\\&=\frac{8703872034774336 \pi }{5}\ut{s}\\&=\frac{6895226534 \pi }{125}\ut{y}..

$$ \begin{cases} v_x&=480\ut{km/h}=\frac{400}{3}\ut{m/s}\\r&=1.5\ut{m}\\\omega &=2000\ut{rev/min}=\frac{200\pi}{3}\ut{rad/s}\end{cases} $$$$\ab{a}$$$$ \begin{aligned}v_t&=r\omega\\&=100\pi\ut{m/s}\\&\approx 3.141592653589793\times10^2\ut{m/s}\\&\approx 3.1\times10^2\ut{m/s}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}v&=\sqrt{{v_x}^2+{v_t}^2}\\&\approx 3.4128261278399657\times10^2\ut{m/s}\\&\ap..

$$ \begin{cases} \omega_A&=9.5\ut{rad/s}\\\omega_{B0}&=0\\\alpha_B&=2.2\ut{rad/s^2}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\theta_A&=\theta_B\\\omega_A t&=\omega_0t+\frac{1}{2}\alpha t^2\\\omega_A&=\frac{1}{2}\alpha t\\\end{aligned} $$$$ \begin{aligned}t&=\frac{2\omega_A}{\alpha}\\&=\frac{95}{11}\ut{s}\\&\approx 8.636363636363637\ut{s}\\&\approx 8.6\ut{s}\\\end{aligned} $$$$\ab{b}$$$$ \begin..

$$ \begin{cases} r&=2.40\ut{cm}=2.4\times10^{-2}\ut{m}\\I&=7.40\times10^{-4}\ut{kg\cdot m^2}\\m&=6.20\ut{kg}\\v_i&=0\\t&=91.0\ut{ms}=9.1\times10^{-2}\ut{s}\\\Delta \theta&=0.130\ut{rad}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$\ab{a}$$$$\Delta \theta=\omega_0t+\frac{1}{2}\alpha t^2,$$$$ \begin{aligned}\alpha &=\frac{2\Delta \theta}{t^2}\\&=\frac{20000}{637}\ut{rad/s^2}\\&\approx 31.39717425431711\ut..

$$ \begin{cases} \Delta t_{1\rarr2}&=6.00\ut{s}\\\theta_1&=10.0\ut{rad}\\\theta_2&=70.0\ut{rad}\\\omega_2&=15.0\ut{rad/s}\\\end{cases} $$$$\ab{a}$$$$\Delta \theta=\frac{1}{2}\(\omega+\omega_0\)t,$$$$\Delta \theta_{1\rarr2}=\frac{1}{2}\(\omega_1+\omega_2\)t,$$$$ \begin{aligned}\omega_1&=\frac{2\Delta \theta}{t}-\omega_2\\&=5\ut{rad/s}\\&=5.00\ut{rad/s}\\\end{aligned} $$$$\ab{b}$$$$\Delta \theta=\..

$$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\\KE : \text{Kinetic Energy}\\\GE : \text{Gravitational Potential Energy}\\\end{cases} $$$$ \begin{cases} r&=0.20\ut{m}\\I&=0.40\ut{kg\cdot m^2}\\m&=6.0\ut{kg}\\\KE&=6.0\ut{J}\\g&=9.80665\ut{J}\\y_i&=0\\\KE_i&=0\\\RE_i&=0\end{cases} $$$$\ab{a}$$$$ \begin{cases} \KE&=\frac{1}{2}mv^2\\\RE&=\frac{1}{2}I\omega^2\\v&=r\omega\\\end{cases} $$..

$$ \begin{cases} r&=0.20\ut{m}\\m&=2.0\ut{kg}\\\theta&=20\degree\\a&=2.0\ut{m/s^2}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$ \begin{cases} \Sigma F&=ma\\\Sigma F&=mg\sin\theta-T\\a&=r\alpha\\\tau&=I\alpha\\\tau&=rT\end{cases} $$$$ \begin{aligned}I&=\frac{g\sin\theta-a}{a}mr^2\\&=\frac{g\sin20\degree -2}{25}\\&\approx 5.416287354178676\times10^{-2}\ut{kg\cdot m^2}\\&\approx 5.4\times10^{-2}\ut{kg\cdo..