11-53 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=7.5\ut{kg}\\ \vec v&=-2.0t^3\i\ut{m/s}\\ \end{cases} $$ $$ \begin{aligned} \vec p &=m\vec v\\ &=-15t^3\i\ut{kg\cdot m/s}\taag1\\ \end{aligned} $$ $$\put \vec X=x\i+y\j\ut{m},$$ $$ \begin{aligned} \vec r_X &=\vec r-\vec X\\ &=\int_0^t \vec v\dd t-\br{x\i+y\j}\\ &=\br{-{t^4\over2}-x}\i-y\j\ut{m}\taag2 \end{aligned} $$ $$\vec L=\vec r\times\vec p,$$ $$ \begin{aligned} \vec L_X&=\vec r_X\times \vec p\\ &=-15t^3y\k\ut{kg\cdot m^2/s}\taag3 \end{aligned} $$ $$\vec \tau_\net=\dyt{\vec L},$$ $$ \begin{aligned} \vec \tau_X&=\dyt{\vec L_X}\\ &=-45t^2y\k\ut{N\cdot m}\taag4 \end{aligned} $$ $$ \therefore \begin{cases} \vec L_X&=-15t^3y\k\ut{kg\cdot m^2/s}\\ \vec \tau_X&=-45t^2y\k\ut{N\cdot m}\\ \end{cases} $$ $$\ab{a,b}$$ $$\vec X_{ab}=0,$$ $$ \begin{cases} \vec L_a&=0\\ \vec \tau_b&=0\\ \end{cases} $$ $$\ab{c,d}$$ $$\vec X_{cd}=2.0\i+5.0\j\ut{m},$$ $$ \begin{cases} \vec L_c&=-75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_d&=-2.25t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} $$ $$ \begin{cases} \vec L_c&=-75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_d&\approx -2.3t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} $$ $$\ab{e,f}$$ $$\vec X_{ef}=2.0\i-5.0\j\ut{m},$$ $$ \begin{cases} \vec L_e&=75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_f&=2.25t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} $$ $$ \begin{cases} \vec L_e&=75t^3\k\ut{kg\cdot m^2/s}\\ \vec \tau_f&\approx 2.3t^2\times10^2\k\ut{N\cdot m}\\ \end{cases} $$