11판/11. 굴림운동, 토크, 각운동량

11-43 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 5. 23. 17:10
{ω=1.5[rev/s]Ii=8.0[kgm2]If=2.0[kgm2] \begin{cases} \omega&=1.5\ut{rev/s}\\ I_i&=8.0\ut{kg\cdot m^2}\\ I_f&=2.0\ut{kg\cdot m^2}\\ \end{cases} (a)\ab{a} ΔΣL=0,\Delta \Sigma \vec L=0, 0=LfLi=IfωfIiωi \begin{aligned} 0&=L_f-L_i\\ &=I_f\omega_f-I_i\omega_i\\ \end{aligned} ωf=IiIfωi=6[rev/s] \begin{aligned} \omega_f&={I_i\over I_f}\omega_i\\ &=6\ut{rev/s}\\ \end{aligned} (b)\ab{b} put {RE:Rotational Kinetic Energy \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\ \end{cases} EfEi=REfREi=12Ifωf212Iiωi2=IfIi(ωfωi)2=IfIi(1ωiIiIfωi)2=IiIf=4.0 \begin{aligned} {E_f\over E_i}&={\RE_f\over \RE_i}\\ &={{1\over2}I_f{\omega_f}^2\over {1\over2}I_i{\omega_i}^2}\\ &={I_f\over I_i}\cdot\br{{\omega_f}\over {\omega_i}}^2\\ &={I_f\over I_i}\cdot\br{{{1}\over {\omega_i}}\cdot{I_i\over I_f}\omega_i}^2\\ &={I_i\over I_f}\\ &=4.0\\ \end{aligned} (c)\ab{c} Moving Stone by Man \begin{aligned} \text{Moving Stone by Man} \end{aligned}