11판/11. 굴림운동, 토크, 각운동량

11-48 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 5. 29. 21:33
{IA=3.30[kgm2]ωAi=450[rev/min]IB=6.60[kgm2]=2IAωBi=900[rev/min]=2ωAi \begin{cases} I_A&=3.30\ut{kg\cdot m^2}\\ \omega_{Ai}&=450\ut{rev/min}\\ I_B&=6.60\ut{kg\cdot m^2}=2I_A\\ \omega_{Bi}&=900\ut{rev/min}=2\omega_{Ai}\\ \end{cases} ΔΣL=0,\Delta \Sigma \vec L=0, 0=ΔΣ(Iω)=Σ(Iωf)Σ(Iωi)=ωfΣIΣ(Iωi) \begin{aligned} 0&=\Delta \Sigma \br{I\omega}\\ &=\Sigma \br{I\omega_f}-\Sigma \br{I\omega_i}\\ &=\omega_f\Sigma {I}-\Sigma \br{I\omega_i}\\ \end{aligned} ωf=Σ(Iωi)ΣI\omega_f={\Sigma \br{I\omega_i}\over\Sigma {I} } (a)\ab{a} ωf=IAωAi+IBωBiIA+IB=IAωAi+(2IA)(2ωAi)IA+(2IA)=53ωAi=750[rev/min] \begin{aligned} \omega_f&={{I_A\omega_{Ai}+I_B\omega_{Bi}}\over I_A+I_B }\\ &={{I_A\omega_{Ai}+(2I_A)(2\omega_{Ai})}\over I_A+(2I_A) }\\ &={5\over 3 }\omega_{Ai}\\ &=750\ut{rev/min}\\ \end{aligned} (b,c)\ab{b,c} {IC=6.60[kgm2]=2IAωCi=800[rev/min] \begin{cases} I_C&=6.60\ut{kg\cdot m^2}=2I_A\\ \omega_{Ci}&=-800\ut{rev/min} \end{cases} ωf=IAωAi+ICωCiIA+IC=11503[rev/min] \begin{aligned} \vec \omega_f &={{I_A\omega_{Ai}+I_C\omega_{Ci}}\over I_A+I_C }\\ &=-{1150\over 3}\ut{rev/min}\\ \end{aligned} (b)\ab{b} ωf383.3333333333333[rev/min]383[rev/min] \begin{aligned} \omega_f &\approx 383.3333333333333\ut{rev/min}\\ &\approx 383\ut{rev/min}\\ \end{aligned} (c)\ab{c} ω<0Clockwise\vec \omega \lt 0 \Harr \text{Clockwise}