솔루션피아

$$ \begin{cases} 0:\text{Start}\\1:\text{Some Time}\\2:\text{1 After 50 rev}\\\end{cases} $$$$ \begin{cases} \alpha&=\Cons\\\omega_0&=0\\\omega_1&=10\ut{rev/s}\\\omega_2&=20\ut{rev/s}\\\Delta \theta_{1\rarr2}&=50\ut{rev}\\\end{cases} $$$$\ab{a}$$$$2\alpha \Delta \theta=\omega^2-{\omega_0}^2,$$$$ \begin{aligned}\alpha&=\frac{{\omega_2}^2-{\omega_1}^2}{2 \Delta \theta}\\&=3.0\ut{rev/s^2}\\\end{ali..

$$ \begin{cases} r&=12\ut{m}\\\theta&=0.20t^2\\t&=5.0\ut{s}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\omega&=\dyt{\theta}\\&=\dt\(0.20t^2\)\\&=\frac{2}{5}t\\\end{aligned} $$$$ \omega(5)=2.0\ut{rad/s} $$$$\ab{b}$$$$ \begin{aligned}v&=r\omega\\&=r\(\frac{2}{5}t\)\\&=\frac{24}{5}t\\\end{aligned} $$$$ v(5)=24\ut{m/s} $$$$\ab{c}$$$$ \begin{aligned}a_t&=r\alpha\\&=r\cdot\dyt{\omega}\\&=r\cdot\dt\(\f..

$$ \begin{cases} r_A&=8.0\ut{m}\\F_A&=16\ut{N}\\\theta_A&=135\degree\\\end{cases} $$$$ \begin{cases} r_B&=4.0\ut{m}\\F_B&=16\ut{N}\\\theta_B&=160\degree\\\end{cases} $$$$ \begin{cases} r_C&=3.0\ut{m}\\F_C&=24\ut{N}\\\theta_C&=90\degree\\\end{cases} $$$$ \begin{cases} \tau_A&=+r_AF_A\sin\theta_A\\\tau_B&=+r_BF_B\sin\theta_B\\\tau_C&=-r_CF_C\sin\theta_C\\\end{cases} $$$$ \begin{cases} \tau_A&=+64\..

$$ \begin{cases} 2R&=76\ut{m}\\T&=2\ut{min}=120\ut{s}\\N&=60\\\end{cases} $$$$\text{Symmetry}\Harr\Sigma \tau=0$$$$ \begin{aligned}W&=\int_{\theta_i}^{\theta_f}\Sigma \tau\dd\theta\\&=\Sigma \tau\int_{\theta_i}^{\theta_f}\dd\theta\\&=0\end{aligned} $$

$$ \begin{cases} 2R&=8.0\ut{cm}\\R&=4.0\times10^{-2}\ut{m}\\L&=8.2\ut{m}\\\omega_0&=0\\\alpha&=0.86\ut{rad/s^2}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\theta&=\frac{L}{r},\\&=205\ut{rad}\\&\approx 2.1\times10^2\ut{rad}\\\end{aligned} $$$$\ab{b}$$$$\Delta \theta=\omega_0t+\frac{1}{2}\alpha t^2,$$$$ \begin{aligned}t&=\sqrt\frac{2\Delta \theta}{\alpha}\\&=10\sqrt\frac{205}{43}\ut{s}\\&\approx 2..

$$ \begin{cases} m&=30\ut{kg}\\R&=0.50\ut{m}\\r&=0.20\ut{m}\\F&=140\ut{N}\\a&=0.80\ut{m/s^2}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$ \begin{cases} \Sigma F&=ma\\\Sigma \tau&=I\alpha\\a&=r\alpha\\\end{cases} $$$$ \begin{cases} T-mg&=ma\\FR-Tr&=I\alpha\\a&=r\alpha\\\end{cases} $$$$ \begin{aligned}I&=\frac{r }{a}\bra{F R-m r (a+g)}\\&=\frac{163}{10}-\frac{3 g}{2}\\&=1.590025\ut{kg\cdot m^2}\\&\approx..

$$ \begin{cases} \alpha&=2.00\ut{rad/s^2}\\\Delta t_{1\rarr2}&=3.00\ut{s}\\\Delta \theta_{1\rarr2} &= 90.0\ut{rad}\\\end{cases} $$$$\ab{a}$$$$\Delta \theta=\omega_0t+\frac{1}{2}\alpha t^2,$$$$ \begin{aligned}\omega_1&=\frac{\Delta \theta}{t}-\frac{\alpha t}{2}\\&=27.0\ut{rad/s}\\\end{aligned} $$$$\ab{b}$$$$\omega=\omega_0+\alpha t,$$$$ \begin{aligned}\Delta t_{0\rarr1}&=\frac{\omega_1-\omega_0}{..

(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/795  속이 채워진 원기둥의 회전 관성$$\title{Rotational Inertia of Solid Cylinder}$$ $$ \rho=\frac{\dd m}{\dd v}=\frac{M}{V}=\frac{M}{AH}=\frac{M}{\pi R^2 H} \taag1$$ $$ \begin{aligned} l&=r\theta,\\ \dd l&=r \dd \theta\\ \end{aligned} $$ $$ \begin{aligned} \dd a&=\..

$$\title{Rotational Inertia of Ring Cylinder}$$ $$ \put \begin{cases} R_i:\text{Inner Radius}\\ R_o:\text{Outer Radius}\\ \end{cases} $$ $$ \rho=\frac{\dd m}{\dd v}=\frac{M}{V}=\frac{M}{AH}=\frac{M}{\pi H( {R_o}^2- {R_i}^2) } \taag1$$ $$ \begin{aligned} l&=r\theta,\\ \dd l&=r \dd \theta\\ \end{aligned} $$ $$ \begin{aligned} \dd a&=\dd l\cdot\dd r \\ &=(r \dd \theta)\cdot\dd r \\ &=r \cdot \dd r\..

$$\title{Rotational Inertia of Ring}$$$$ \put \begin{cases} R_i:\text{Inner Radius}\\ R_o:\text{Outer Radius} \end{cases} $$$$ \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{\pi {R_o}^2-\pi {R_i}^2} \taag1$$$$ \begin{aligned} l&=r\theta,\\ \dd l&=r \dd \theta\\ \end{aligned} $$$$ \begin{aligned} \dd a&=\dd r \cdot \dd l\\ &=\dd r \cdot (r \dd\theta)\\ &=r\cdot\dd r \dd\theta\taag2\\ \end{aligne..