솔루션피아

$$ \begin{cases} m_A&=460\ut{g}=0.460\ut{kg}\\m_B&=500\ut{g}=0.500\ut{kg}\\R&=8.00\ut{cm}=8.00\times10^{-2}\ut{m}\\\Delta y_{0\rarr5}&=75.0\ut{cm}=0.750\ut{m}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$\ab{a}$$$$S=v_0t+\frac{1}{2}at^2,$$$$ \begin{aligned}a&=\frac{2\Delta y}{t^2}\\&=\frac{3}{50}\ut{m/s^2}\\&=6.00\times10^{-2}\ut{m/s^2}\\&=6.00\ut{cm/s^2}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\Si..

(풀이자주:실제 캔의 데이터를 문자로 두엇습니다.)$$ \begin{cases} F_{\text{in}}&=10\ut{N}\end{cases} $$$$ \begin{aligned}\tau_{\text{in}}&=\tau_{\text{out}}\\F_{\text{in}}\cdot r_1&=F_{\text{out}}\cdot r_2\\F_{\text{out}}&=\frac{r_1}{r_2}\cdot F_{\text{in}}\\\end{aligned} $$

$$ \begin{cases} \theta(t\ut{s})&=4.0t-5.0t^2+t^3\ut{rad}\\\end{cases} $$$$\ab{a,b}$$$$ \begin{aligned}\omega(t)&=\dyt{\theta}\\&=\dt(4.0t-5.0t^2+t^3)\\&=4 - 10 t + 3 t^2\\\end{aligned} $$$$\ab{a}$$$$ \begin{aligned}\omega(2)&=-4.0\ut{rad/s}\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\omega(5)&=29\ut{rad/s}\end{aligned} $$$$\ab{c}$$$$ \begin{aligned}\bar \alpha_{2\rarr5}&=\frac{\Sigma \Delta \om..

$$ \begin{cases} \omega_0&=0\\\alpha&=\Cons\\\Delta \theta_{0\rarr 5}&=50\ut{rad}\\\end{cases} $$$$\ab{a}$$$$\Delta \theta=\omega_0t+\frac{1}{2}\alpha t^2,$$$$ \begin{aligned}\alpha&=\frac{2\Delta \theta}{t^2}\\&=4.0\ut{m/s^2}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\bar \omega_{0\rarr5}&=\frac{\Sigma \Delta \theta}{\Sigma \Delta t}\\&=10\ut{rad/s}\end{aligned} $$$$\ab{c}$$$$\Delta \theta=\..

$$ \begin{cases} \theta&=20\degree\\ \omega&=70\ut{rad/s}=\frac{35}{\pi}\ut{rev/s}\\ v_0&=50\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$v_y=v_{y0}+at,$$ $$ \begin{aligned} t&=\frac{v-v_0}{a}\\ &=\frac{0-v_0\sin\theta}{-g}\\ &=\frac{v_0}{g}\sin\theta\\ \end{aligned} $$ $$ \begin{aligned} \theta&=\omega t\\ &=\frac{\omega v_0}{g}\sin\theta\\ &\approx 19.42760143020047\ut{rev}\\ &\approx 19\ut..

(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&=\d..

$$ \begin{cases} \omega_i&=160\ut{rad/s}\\\alpha&=-4.0\ut{rad/s^2}\\\omega_f&=0\\\end{cases} $$$$\omega=\omega_0+\alpha t,$$$$\ab{a}$$$$ \begin{aligned}t&=\frac{\omega_f-\omega_i}{\alpha}\\&=40\ut{s}\\&=4.0\times10\ut{s}\\\end{aligned} $$$$\ab{b}$$$$2\alpha \Delta \theta=\omega^2-{\omega_0}^2,$$$$ \begin{aligned}\Delta \theta&=\frac{{\omega_f}^2-{\omega_i}^2}{2\alpha}\\&=3200\ut{rad}\\&=3.2\time..

$$ \begin{cases} \alpha(t\ut{s})&=6.0t^4-5.0t^2\ut{rad/s^2}\\\omega_0&=+3.0\ut{rad/s}\\\theta_0&=+7.0\ut{rad}\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\omega(t\ut{s})&=\omega_0+\Delta \omega\\&=\omega_0+\int_0^t\alpha\dd t\\&=\frac{6 }{5}t^5-\frac{5 }{3}t^3+3\ut{rad/s}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\theta(t\ut{s})&=\theta_0+\Delta \theta\\&=\theta_0+\int_0^t\omega\dd t\\&=\frac{1}..

$$ \begin{cases} m&=0.0200\ut{kg}\\L&=3d\\L&=3.40\ut{cm}=3.4\times10^{-2}\ut{m}\\\end{cases} $$$$I=\Sigma mr^2,$$$$ \begin{aligned}I(k)&=m{r_k}^2\\&=m(kd)^2\\&=md^2k^2\\\end{aligned} $$$$ \begin{aligned}\Sigma I &=\sum_{k=1}^3\(md^2k^2\)\\&=md^2\sum_{k=1}^3 k^2\\&=14md^2\\\end{aligned} $$$$ \begin{aligned}\Ans&=\frac{I_k}{\Sigma I}\\&=\frac{md^2k^2}{14md^2}\\&=\frac{k^2}{14}\\\end{aligned} $$$$\..

(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$$\title{Rotational Inertia of Rod}$$$$ \lambda=\frac{\dd m}{\dd l}=\frac{M}{L}\taag1$$$$ \dd m=\lambda\cdot\dd l $$$$ \begin{aligned} \dd I&={r}^2\cdot \dd m\\&={l}^2\cdot (\lambda\cdot\dd l)\\&=\lambda{l}^2\cdot \dd l\taag2\\\end{alig..