11-39 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} A : \text{Bug}\\ B : \text{Disk} \end{cases} $$ $$ \begin{cases} m_A&=0.25\ut{kg}\\ R&=15\ut{cm}=0.15\ut{m}\\ I_B&=5.0\times10^{-3}\ut{kg\cdot m^2}\\ v_{Ai}&=2.0\ut{m/s}\\ \omega_{Bi}&=2.8\ut{rad/s}\\ v_{A\larr B f}&=0\\ \end{cases} $$ $$\ab{a}$$ $$\Delta \Sigma \vec L=0,$$ $$ \begin{aligned} 0 &=\Delta\Sigma L\\ &=\Sigma L_f-\Sigma L_i\\ &=\Sigma (I \omega_f)-\br{L_{Ai}+L_{Bi}}\\ &= \omega_f\Sigma I- (I_A\omega_{Ai}+I_B\omega_{Bi})\\ \end{aligned} $$ $$I=mR^2,$$ $$ \begin{aligned} \omega_f &={I_A\omega_{Ai}+I_B\omega_{Bi}\over I_A+I_B}\\ &={Rm_Av_{Ai}+I_B\omega_{Bi}\over m_AR^2+I_B}\\ &={712\over85}\ut{rad/s}\\ &\approx 8.376470588235295\ut{rad/s}\\ &\approx 8.4\ut{rad/s}\\ \end{aligned} $$ $$ \put \begin{cases} \ME : \text{Mechanical Energy}\\ \RE : \text{Rotational Kinetic Energy}\\ \KE : \text{Translational Kinetic Energy}\\ \end{cases} $$ $$\ab{b}$$ $$v=\omega R,$$ $$I=mR^2,$$ $$ \begin{aligned} \Sigma \Delta \ME &=\Delta \ME_{A}+\Delta \ME_{B}\\ &=\Delta \KE_{A}+\Delta \RE_{B}\\ &=\Delta \br{{1\over2}m_A{v_{A}}^2}+\Delta \br{{1\over2}{I_B}{\omega_{B}}^2}\\ \end{aligned} $$ $$ \begin{aligned} 2\cdot\Sigma \Delta \ME &=\Delta \bra{m_A\br{\omega_{A}R}^2}+\Delta \br{{I_B}{\omega_{B}}^2}\\ &=\Delta \bra{I_A{\omega_{A}}^2}+\Delta \br{{I_B}{\omega_{B}}^2}\\ &=\br{I_A{\omega_{f}}^2-I_A{\omega_{Ai}}^2}+\br{{I_B}{\omega_{f}}^2-{I_B}{\omega_{Bi}}^2}\\ &={\omega_{f}}^2\br{I_A+{I_B}}-\br{I_A{\omega_{Ai}}^2+{I_B}{\omega_{Bi}}^2}\\ &=\br{{I_A\omega_{Ai}+I_B\omega_{Bi}\over I_A+I_B}}^2\br{I_A+{I_B}}-\br{I_A{\omega_{Ai}}^2+{I_B}{\omega_{Bi}}^2}\\ &=-{I_AI_B\br{\omega_{Ai}-\omega_{Bi}}^2\over {I_A+I_B}}\\ \end{aligned} $$ $$ \begin{aligned} \therefore \Sigma \Delta \ME &=-{1\over2}\cdot \br{I_AI_B\over I_A+I_B}\cdot \br{\omega_{A\larr Bi}}^2,\\ \end{aligned} $$ $$ \begin{cases} I_A=m_AR^2\ne0\\ I_B\ne0\\ \omega_{A\larr Bi}\ne0 \end{cases} $$ $$\text{Non-Conservation of Mechanical Energy}$$