11판/11. 굴림운동, 토크, 각운동량

11-53 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 6. 2. 16:02
{m=7.5[kg]v=2.0t3i^[m/s] \begin{cases} m&=7.5\ut{kg}\\ \vec v&=-2.0t^3\i\ut{m/s}\\ \end{cases} p=mv=15t3i^[kgm/s](1) \begin{aligned} \vec p &=m\vec v\\ &=-15t^3\i\ut{kg\cdot m/s}\taag1\\ \end{aligned} put X=xi^+yj^[m],\put \vec X=x\i+y\j\ut{m}, rX=rX=0tv ⁣dt(xi^+yj^)=(t42x)i^yj^[m](2) \begin{aligned} \vec r_X &=\vec r-\vec X\\ &=\int_0^t \vec v\dd t-\br{x\i+y\j}\\ &=\br{-{t^4\over2}-x}\i-y\j\ut{m}\taag2 \end{aligned} L=r×p,\vec L=\vec r\times\vec p, LX=rX×p=15t3yk^[kgm2/s](3) \begin{aligned} \vec L_X&=\vec r_X\times \vec p\\ &=-15t^3y\k\ut{kg\cdot m^2/s}\taag3 \end{aligned} τnet = ⁣dL ⁣dt,\vec \tau_\net=\dyt{\vec L}, τX= ⁣dLX ⁣dt=45t2yk^[Nm](4) \begin{aligned} \vec \tau_X&=\dyt{\vec L_X}\\ &=-45t^2y\k\ut{N\cdot m}\taag4 \end{aligned} {LX=15t3yk^[kgm2/s]τX=45t2yk^[Nm] \therefore \begin{cases} \vec L_X&=-15t^3y\k\ut{kg\cdot m^2/s}\\ \vec \tau_X&=-45t^2y\k\ut{N\cdot m}\\ \end{cases} (a,b)\ab{a,b} Xab=0,\vec X_{ab}=0, {La=0τb=0 \begin{cases} \vec L_a&=0\\ \vec \tau_b&=0\\ \end{cases} (c,d)\ab{c,d} Xcd=2.0i^+5.0j^[m],\vec X_{cd}=2.0\i+5.0\j\ut{m}, {Lc=75t3k^[kgm2/s]τd=2.25t2×102k^[Nm] \begin{cases} \vec L_c&=-75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_d&=-2.25t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} {Lc=75t3k^[kgm2/s]τd2.3t2×102k^[Nm] \begin{cases} \vec L_c&=-75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_d&\approx -2.3t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} (e,f)\ab{e,f} Xef=2.0i^5.0j^[m],\vec X_{ef}=2.0\i-5.0\j\ut{m}, {Le=75t3k^[kgm2/s]τf=2.25t2×102k^[Nm] \begin{cases} \vec L_e&=75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_f&=2.25t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} {Le=75t3k^[kgm2/s]τf2.3t2×102k^[Nm] \begin{cases} \vec L_e&=75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_f&\approx 2.3t^2\times10^2\k\ut{N\cdot m}\\ \end{cases}