솔루션피아

$$ \begin{cases} L_i&=-600\ut{kg\cdot m^2/s}\\\tau&=+60\ut{N\cdot m}\\\end{cases} $$$$\vec L=I\omega,$$$$\omega_f=0\Rarr L_f=0$$$$\vec \tau_\net=\dyt{\vec L}={\Delta L\over \Delta t},$$$$ \begin{aligned}\Delta t&={\Delta L\over \tau}\\&={L_f-L_i\over \tau}\\&={-L_i\over \tau}\\&=10\ut{s}\end{aligned} $$

$$ \begin{cases} \vec F&=-8.0\i+4.0\j\ut{N}\\\vec r&=3.0\i+4.0\j\ut{m}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\vec \tau &= \vec r\times \vec F,\\&=44\k\ut{N\cdot m}\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\abs{\vec r\times \vec F} &= rF\sin\phi,\\\phi&=\sin^{-1}{\abs{\vec r\times \vec F}\over rF}\\&=\sin^{-1}{44\over 5\cdot4\sqrt5}\\&\approx 1.390942827002419\ut{rad}\\&\approx 1.4\ut{rad}..

$$ \begin{cases} R&=0.250\ut{m}\\v_i&=33.0\ut{m/s}\\S&=225\ut{m}\\v_f&=0\\I&=0.155\ut{kg\cdot m^2}\\\end{cases} $$ $$\ab{a}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} a&=\abs{v^2-{v_0}^2\over2S}\\&={121\over50}\ut{m/s^2}\\&=2.42\ut{m/s^2} \end{aligned} $$ $$\ab{b}$$ $$\alpha={a\over R},$$ $$ \begin{aligned} \alpha&={2.42\over0.250}\ut{rad/s^2}\\&=9.68\ut{rad/s^2}\\\end{aligned} $$ $$\ab{c}$$ $$\t..

$$ \begin{cases} \vec r&=3.0\i-2.0\j+4.0\k\ut{m}\\\vec F_1&=3.0\i-4.0\j+5.0\k\ut{N}\\\vec F_2&=-3.0\i-4.0\j+5.0\k\ut{N}\\\vec P&=3.0\i+2.0\j+4.0\k\ut{m}\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\vec \tau_a&=\vec r\times\vec F_1\\&=6.0\i-3.0\j-6.0\k\ut{N\cdot m}\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\vec \tau_b&=\vec r\times\vec F_2\\&=6.0\i-27\j-18\k\ut{N\cdot m}\end{aligned} $$$$\ab{c}$$$$..

(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/800 속이 채워진 구의 회전 관성$$\title{Rotational Inertia of Solid Sphere}$$$$\put \begin{cases} x&=\text{Latitude Line}\\ y&=\text{Longitude Line}\\ \end{cases} $$$$ \begin{cases} r_x&=r\sin\theta\\r_y&=r\end{cases} $$$$ \begin{cases} x&=r_x \phi=(r\sin\theta)\p..

$$ \begin{cases} m&=0.25\ut{kg}\\\vec r&=2.0i-2.0\k\ut{m}\\\vec v&=-5.0\i+8.0\k\ut{m/s}\\\vec F&=4.0\j\ut{N}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\vec L&=\vec r\times m\vec v\\&=-1.5\j\ut{kg\cdot m^2/s}\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\vec \tau&=\vec r\times \vec F\\&=8.0\i+8.0\k\ut{N\cdot m}\end{aligned} $$

$$ \begin{cases} \vec F&=2.0\i-6.0\k\ut{N}\\\vec r&=0.50\j-2.0\k\ut{m}\\\vec P&=2.0\i-3.0k\ut{m}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\tau_a&=\vec r\times \vec F\\&=-3.0\i-4.0\j-1.0\k\ut{N\cdot m}\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\tau_b&=(\vec r-\vec P)\times \vec F\\&=-3.0\i-1.0\times10\j-1.0\k\ut{N\cdot m}\end{aligned} $$

$$ \begin{cases} m&=2.0\ut{kg}\\r&=2.0\ut{m},\theta_r=45\degree\\v&=4.0\ut{m/s},\theta_2=30\degree\\F&=2.0\ut{N},\theta_3=30\degree\\\end{cases} $$$$ \begin{aligned}\vec r&=r\cos\theta_r\i+r\sin\theta_r\j\\&=\sqrt2\i+\sqrt2\j\taag1\\\end{aligned} $$$$\theta_v=\theta_r-180\degree-\theta_2,$$$$ \begin{aligned}\vec v&=v\cos\theta_v\i+r\sin\theta_v\j\\&=-(\sqrt6+\sqrt2)\i-(\sqrt6-\sqrt2)\j\taag2\\\e..

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