10-40 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=460\ut{g}=0.460\ut{kg}\\ m_B&=500\ut{g}=0.500\ut{kg}\\ R&=8.00\ut{cm}=8.00\times10^{-2}\ut{m}\\ \Delta y_{0\rarr5}&=75.0\ut{cm}=0.750\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} a&=\frac{2\Delta y}{t^2}\\ &=\frac{3}{50}\ut{m/s^2}\\ &=6.00\times10^{-2}\ut{m/s^2}\\ &=6.00\ut{cm/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma F_B&=m_Ba_B\\ T_2-m_Bg&=m_B(-a)\\ \end{aligned} $$ $$ \begin{aligned} T_2&=m_B(g-a) \\ &=\frac{1}{2} \(g-\frac{3}{50}\) \\ &=4.873325\ut{N}\\ &\approx 4.87\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \Sigma F_A&=m_Aa_A\\ T_1-m_Ag&=m_Aa\\ \end{aligned} $$ $$ \begin{aligned} T_1&=m_A(g+a)\\ &=\frac{23}{50} \(g-\frac{3}{50}\)\\ &=4.483459\ut{N}\\ &\approx 4.48\ut{N}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \alpha&=\frac{a}{R}\\ &=\frac{3}{4}\ut{rad/s^2}\\ &=0.750\ut{rad/s^2}\\ \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \Sigma \tau_C&=I_C\cdot\alpha_C\\ T_1R-T_2R&=I_C\cdot(-\alpha)\\ \end{aligned} $$ $$ \begin{aligned} I_C&=\frac{(T_1-T_2)R}{-\alpha}\\ &=\frac{4}{75} \(g-\frac{3}{50}\)\\ &\approx 0.5198213333333334\ut{kg\cdot m^2}\\ &\approx 0.520\ut{kg\cdot m^2}\\ \end{aligned} $$