Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 323: 45b

Answer

The ratio of the new kinetic energy of the system to the original kinetic energy is $~~3$.

Work Step by Step

Note that $~~I_f = \frac{I_i}{3}$ We can use conservation of angular momentum to find an expression for the new angular speed: $L_f = L_i$ $I_f~\omega_f = I_i~\omega_i$ $\frac{I_i}{3}~\omega_f = I_i~\omega_i$ $\omega_f = 3~\omega_i$ We can write an expression for the initial kinetic energy: $K_i = \frac{1}{2}I_i\omega_i^2$ We can write an expression for the final kinetic energy: $K_f = \frac{1}{2}I_f\omega_f^2$ $K_f = \frac{1}{2}(\frac{I_i}{3})(3~\omega_i)^2$ $K_f = 3\times \frac{1}{2}I_i\omega_i^2$ $K_f = 3~K_i$ Therefore: $\frac{K_f}{K_i} = 3$ The ratio of the new kinetic energy of the system to the original kinetic energy is $~~3$
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