11판/11. 굴림운동, 토크, 각운동량

11-31 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 5. 18. 13:22
{m=2.0[kg]r=2.0[m],θr=45°v=4.0[m/s],θ2=30°F=2.0[N],θ3=30° \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} r=rcosθri^+rsinθrj^=2i^+2j^(1) \begin{aligned} \vec r&=r\cos\theta_r\i+r\sin\theta_r\j\\ &=\sqrt2\i+\sqrt2\j\taag1\\ \end{aligned} θv=θr180°θ2,\theta_v=\theta_r-180\degree-\theta_2, v=vcosθvi^+rsinθvj^=(6+2)i^(62)j^(2) \begin{aligned} \vec v&=v\cos\theta_v\i+r\sin\theta_v\j\\ &=-(\sqrt6+\sqrt2)\i-(\sqrt6-\sqrt2)\j\taag2\\ \end{aligned} θF=θr+θ3,\theta_F=\theta_r+\theta_3, F=FcosθFi^+FsinθFj^=622i^+6+22j^(3) \begin{aligned} \vec F&=F\cos\theta_F\i+F\sin\theta_F\j\\ &={\sqrt6-\sqrt2\over2}\i+{\sqrt6+\sqrt2\over2}\j\taag3\\ \end{aligned} (a,b)\ab{a,b} L=m(r×v)=8k^[kgm2/s] \vec L=m(\vec r\times \vec v)=8\k\ut{kg \cdot m^2/s} (a)\ab{a} L=8.0[kgm2/s]L=8.0\ut{kg \cdot m^2/s} (b)\ab{b} L>0CounterClockwise\vec L\gt0\Rarr \text{CounterClockwise} (c,d)\ab{c,d} τ=r×F=2k^[Nm] \vec \tau=\vec r\times \vec F=2\k\ut{N\cdot m} (c)\ab{c} τ=8.0[Nm]\tau=8.0\ut{N\cdot m} (d)\ab{d} τ>0CounterClockwise\vec \tau\gt0\Rarr \text{CounterClockwise}