Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 226: 73

Answer

$3×10^{−16}\%$

Work Step by Step

$L_i=L_f$ $L_e+L_a=L_f$ $I_e\omega_e+I_a\omega_a=I_f\omega_f$ $I_e\omega_e+I_a\frac{v_a}{r_e}=I_f\omega_f$ $\frac{2}{5}M_eR_e^2\omega_e+\frac{2}{5}M_aR_a^2\frac{v_a}{r_e}=I_f\omega_f$ $\frac{2}{5}(5.97\times10^{24}kg)(6.37\times10^3m)^2\omega_e+\frac{2}{5}(1.0\times10^5kg)(6.37\times10^3m)^2\frac{3.5\times10^4\frac{m}{s}}{6.37\times10^3m}=\frac{2}{5}(5.97\times10^{24}kg)(6.37\times10^3m)^2\omega_f$ $9.69\times10^{31}\omega_e+8.92\times10^{12}=9.69\times10^{31}\omega_f$ $\omega_f-\omega_e=9.21\times10^{-20}$ $\frac{\omega_f-\omega_e}{\omega_f}\times100\%=\frac{9.21\times10^{-20}}{\frac{2\pi}{24 hours}\times\frac{1 hour}{3600s}}\times100\%=3×10^{−16}\%$
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