솔루션피아

$$ \begin{cases} \theta_i&=102\degree\\ w&=-\frac{360}{24}\ut{\degree/h}=-15\ut{\degree/h}\\ \end{cases} $$$$\ab{a}$$$$\theta_f=25\degree$$$$\Delta \theta=\omega t$$$$ \begin{aligned} t&=\frac{\Delta \theta}{\omega}\\ &=\frac{25\degree-102\degree}{-15\ut{\degree/h}}\\ &=\frac{77}{15}\ut{h}\\ &\approx 5.133333333333334\ut{h}\\ &\approx 5.1\ut{h} \end{aligned} $$$$\ab{b}$$$$\theta_f=-20\degree$$$$..

(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/801 두꺼운 원형 고리의 회전 관성$$\title{Rotational Inertia of Ring}$$$$ \put \begin{cases} R_i:\text{Inner Radius}\\R_o:\text{Outer Radius} \end{cases} $$$$ \sigma=\frac{\dd m}{\dd A}=\frac{M}{A}=\frac{M}{\pi {R_o}^2-\pi {R_i}^2} \taag1$$$$ \begin{aligned} L&=r\t..

$$ \begin{cases} m_1&=70.0\ut{kg}\\ I_1&=12.0\ut{kg\cdot m^2}\\ x_1 &= 3.0\ut{cm}=3.0\times10^{-2}\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \tau_1 &= r_\perp F,\\ &= +x_1 m_1g\\ &=2.1g \end{aligned} $$ $$\Sigma \tau =I\alpha,$$ $$ \begin{aligned} \alpha &=\frac{\tau_1}{I}\\ &=\frac{7}{40}g\\ &=1.71616375\ut{rad/s^2}\\ &\approx 1.7\ut{rad/s^2}\\ \end{aligned} $$ $..

$$ \put \begin{cases} s:\text{ship}\\g:\text{gas}\\\end{cases} $$$$\put \dot y=\dyt{y},$$$$ \begin{cases} v_i&=6.0\times10^3\ut{m/s}\\v_{g\larr s}&=-3.0\times10^3\ut{m/s}\\m_s&=2.5\times10^4\ut{kg}\\a_s&=2.0\ut{m/s^2}\\M&=m_s+m_g\\\end{cases} $$$$\Delta \Sigma \vec p=0,$$$$ \begin{cases} Mv_i&=m_gv_{g}+m_sv_s\\v_{g\larr s}&=v_g-v_s\\\Delta v_s&=v_s-v_i\end{cases} $$$$ \begin{aligned}-m_gv_{g\lar..

$$ \begin{cases} v_i&=2.0\ut{m/s}\\m&=4.0\ut{kg}\\\Delta \Sigma \KE &=16\ut{J}\\\end{cases} $$$$ \begin{cases} \Delta \Sigma \vec p&=0,\\\Delta \Sigma \KE &=16\ut{J}\\\end{cases} $$$$ \begin{cases} 2mv_i&=mv_A+mv_B\\\frac{1}{2}(2m) {v_i}^2+16&=\frac{1}{2}m {v_A}^2+\frac{1}{2}m {v_B}^2\end{cases} $$$$ \begin{cases} v_A&=v_i+\frac{4}{\sqrt m}\\v_B&=v_i-\frac{4}{\sqrt m}\\\end{cases} $$$$ \begin{ca..

$$ \begin{cases} m_2&=1.0\ut{kg}\\v_{2i}&=0\\k&=490\ut{N/m}\\m_1&=2.0\ut{kg}\\M&=m_1+m_2=3\ut{kg}\\v_{1i}&=4.0\ut{m/s}\\\end{cases} $$$$\Delta \Sigma \vec p=0,$$$$ \begin{aligned}m_1v_{1i}&=Mv_f,\\v_f&=\frac{m_1}{M}v_{1i}\end{aligned} $$$$\Sigma \Delta E=0,$$$$\frac{1}{2}M{v_f}^2=\frac{1}{2}kx^2$$$$ \begin{aligned}x&=v_f\sqrt\frac{M}{k}\\&=\frac{m_1v_{1i}}{\sqrt{Mk}}\\&=\frac{4}{7}\sqrt{\frac{2}..

$$ \put \begin{cases} 0:\text{Start}\\1:\text{After Burst}\\2:\text{Stop}\end{cases} $$$$ \begin{cases} m_L&=3.0\ut{kg}\\\mu_L&=0.40\\\mu_R&=0.50\\d_L&=0.15\ut{m}\\d_R&=0.25\ut{m}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$\Delta \Sigma \vec p=0,$$$$0=m_Lv_{L1}+m_Rv_{R1}$$$$W=\Delta \KE,$$$$f d=\frac{1}{2}mv^2,$$$$ \begin{aligned}{v_1}^2&=2\mu gd\\\end{aligned} $$$$ \begin{case..

$$ \begin{cases} v_i&=v\\M&=m_A+m_B=m\\m_A&=4m_B\\v_B&=0\\\end{cases} $$$$ \begin{cases} m_A&=\frac{4}{5}M\\m_B&=\frac{1}{5}M\\\end{cases} $$$$\Delta \Sigma \vec p=0,$$$$ \begin{aligned}Mv_i&=m_Av_A+m_Bv_B\\&=m_Av_A\end{aligned} $$$$ \begin{aligned}v_A&=\frac{M}{m_A}v_i\\&=\frac{5m_A}{m_A}v_i\\&=5v_i\\\end{aligned} $$$$ \begin{aligned}\Delta \Sigma \KE&=\Delta \sum \(\frac{1}{2}mv^2\)\\&=\(\frac..

$$ \begin{cases} m&=140\ut{g}=0.140\ut{kg}\\v_i&=9.30\ut{m/s}\\v_f&=-v_i=-9.30\ut{m/s}\\\Delta t&=3.80\ut{ms}=3.80\times10^{-3}\ut{s}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\vec J&=\Delta \vec p\\&=\Delta m\vec v\\&=m\Delta \vec v\\&=m (\vec v_f-\vec v_i)\\&=m (-2\vec v_i)\\\end{aligned} $$$$ \begin{aligned}J&=2mv_i\\&=\frac{651}{250}\ut{N\cdot s}\\&=2.604\ut{N\cdot s}\\&\approx 2.60\ut{N\cd..

$$ \begin{cases} v_i&=7600\ut{m/s}\\m_A&=290.0\ut{kg}\\m_B&=150.0\ut{kg}\\M&=m_A+m_B=440\ut{kg}\\v_{A\larr B}&=910.0\ut{m/s}\end{cases} $$$$ \begin{cases} v_{A\larr B}&=v_A-v_B\\\Delta \Sigma \vec p&=0\end{cases} $$$$ \begin{cases} v_{A\larr B}&=v_A-v_B\\Mv_i&=m_Av_A+m_Bv_B\end{cases} $$$$ \begin{cases} v_A&=v_i+\cfrac{m_B}{M}v_{A\larr B}\\v_B&=v_i-\cfrac{m_A}{M}v_{A\larr B}\\\end{cases} $$$$\ab..