솔루션피아

$$ \put \begin{cases} A:\text{Disk}\\B:\text{Man}\\C:\text{Stone}\\\end{cases} $$$$\ab{a}$$$$\Delta \Sigma \vec L=0,$$$$ \begin{aligned}0&=\Sigma \vec L_i=\Sigma \vec L_f\\&=\vec L_{Af}+\vec L_{Bf}+\vec L_{Cf}\\&=I_A\vec \omega_f+I_B\vec \omega_f+\vec r_C\times m_C\vec v_C\\&=\vec \omega_f\br{I_A+m_BR^2}+Rm_C\vec v_C\\\end{aligned} $$$$ \vec \omega_f=-{Rm_C\vec v_C\over I_A+m_BR^2},$$$$\omega_{A..

$$\vec \tau_\net=\dyt{\vec L},$$$$\ab{a}$$$$\vec L=-6.5\k\ut{kg\cdot m^2/s},$$$$ \begin{aligned}\vec \tau_\net&=\dyt{\vec L}=0\end{aligned} $$$$\ab{b}$$$$\vec L=-6.5t^2\k\ut{kg\cdot m^2/s},$$$$ \begin{aligned}\vec \tau_\net&=\dyt{\vec L}=-13t\k\ut{N\cdot m}\end{aligned} $$$$\ab{c}$$$$\vec L=-6.5\sqrt t\ut{kg\cdot m^2/s},$$$$ \begin{aligned}\vec \tau_\net&=\dyt{\vec L}=-{13\over4\sqrt t}\k\ut{N\c..

$$ \begin{cases} m&=2.90\times10^{-4}\ut{kg}\\v&=3.03\ut{m/s}\\2d&=4.20\ut{cm}=4.20\times10^{-2}\ut{m}\\\end{cases} $$$$ \begin{cases} \vec v&=v\i\\\vec S&=vt\i+d\j\\\vec O&=x\i+y\j+z\k\\\vec r&=\vec S - \vec O=(vt-x)\i+(d-y)\j-z\k\\\end{cases} $$$$ \begin{aligned}\vec L&=m(\vec r\times\vec v)\\&=mv\bra{-z\j+\br{y-d}\k}\\\end{aligned} $$$$ L=\abs{mv}\sqrt{\br{y-d}^2+{z}^2}$$$$ \begin{cases} d_1&..

(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/799 속이 빈 얇은 구 껍질의 회전 관성$$\title{Rotational Inertia of Hollow Sphere}$$ $$\put \begin{cases} x&=\text{Latitude Line}\\ y&=\text{Longitude Line}\\ \end{cases} $$ $$ \begin{cases} r_x&=R\sin\theta\\ r_y&=R \end{cases} $$ $$ \begin{cases} x&=r_x \phi=(R\si..

$$ \begin{cases} m_A&=3.00\ut{kg}\\\vec v_{Ai}&=-3.00\j\ut{m/s}\\m_B&=4.00\ut{kg}\\M&=m_A+m_B=7.00\ut{kg}\\\vec v_{Bi}&=4.50\i\ut{m/s}\\\vec r_P&=-0.500\i-0.100\j\ut{m}\\\end{cases} $$$$\Delta \Sigma \vec p=0,$$$$ \begin{aligned}\vec p_f&=\Sigma \vec p_i\\&=m_A\vec v_{Ai}+m_B\vec v_{Bi}\\\end{aligned} $$$$ \begin{aligned}\vec L_f&=\vec r\times\vec p_{f},\\&=\vec r\times(\Sigma \vec p_i),\\&=\vec..

$$ \begin{cases} \omega_{Ai}&=900\ut{rev/min}=30\pi\ut{rad/s}\\I_B&=2I_A\\\omega_{Bi}&=0\\\end{cases} $$$$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\\end{cases} $$$$\ab{a}$$$$\Delta \Sigma \vec L=0,$$$$ \begin{aligned} 0&=\Delta L_A+\Delta L_B\\&=I_A\Delta \omega_A+I_B\Delta \omega_B\\&=I_A (\omega_f-\omega_{Ai})+2I_A \omega_f\\&=3\omega_f -\omega_{Ai}\\\end{aligned} $$$$ \begin..

$$ \put \begin{cases} A : \text{Disk}\\B : \text{Man}\\\end{cases} $$$$ \begin{cases} R&=1.20\ut{m}\\m_A&=200\ut{kg}\\k_A&=91.0\ut{cm}=0.910\ut{m}\\m_B&=44.0\ut{kg}\\v_{Ai}&=0\\v_{Bi}&=3.00\ut{m/s}\\\end{cases} $$$$\ab{a}$$$$k_A=\sqrt\frac{I_A}{m_A},$$$$ \begin{aligned} I_A&=m_A{k_A}^2\\&=\frac{8281}{50}\ut{kg\cdot m^2}\\&=165.62\ut{kg\cdot m^2}\\&\approx 166\ut{kg\cdot m^2}\\\end{aligned} $$$$\..

$$ \put \begin{cases} A : \text{Disk}\\B : \text{Something}\\\end{cases} $$$$ \begin{cases} m_A&=0.10\ut{kg}\\R&=0.10\ut{m}\\\omega_{Ai} &= 4.7\ut{rad/s}\\I_A&=5.0\times10^{-4}\ut{kg\cdot m^2}\\m_B&=0.036\ut{kg} \end{cases} $$$$I_B=m_BR^2,$$$$\Delta \Sigma \vec L=0,$$$$ \begin{aligned} 0&=\Delta \vec L_A+\Delta \vec L_B\\&=\vec L_{Af}-\vec L_{Ai}+\vec L_{Bf}\\&=\vec L_{f}-\vec L_{Ai}\\&=\vec \om..

$$ \put \begin{cases} A : \text{Disk}\\B : \text{Man}\\C : \text{Ball}\\\end{cases} $$$$ \begin{cases} \omega_i&=0\\m_B&=20\ut{kg}\\R&=2.0\ut{m}\\I_A&=150\ut{kg\cdot m^2}\\m_C&=1.0\ut{kg}\\v_{Ci}&=12\ut{m/s}\\\phi&=37\degree\\\end{cases} $$$$ \begin{aligned} \omega&=\frac{v_t}{R},\\&=\frac{v\cos\phi}{R}\taag1\\\end{aligned} $$$$ \begin{cases} I_B&=m_BR^2\\I_C&=m_CR^2 \end{cases} $$$$\Delta \Sigm..

$$ \begin{cases} m&=0.500\ut{kg}\\R&=9.00\ut{cm}=9.00\times10^{-2}\ut{m}\\\theta&=30\degree\\v_0&=0\\v_1&=3.5\ut{m/s}\\\end{cases} $$$$v=v_0+at,$$$$ \begin{aligned}a&=\frac{v_1-v_0}{t}\\&=\frac{7}{2}\ut{m/s^2}\\\end{aligned} $$$$ \begin{cases} \Sigma F&=m a\\\Sigma \tau&=I\alpha\\a&=r\alpha\\\end{cases} $$$$ \begin{cases} f-mg\sin\theta&=m (-a)\\Rf&=I\alpha\\a&=R\alpha\\\end{cases} $$$$ \begin{a..