Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 11 - Exercises and Problems - Page 202: 54


$m=2.8 \times 10^7 \ kg$

Work Step by Step

We add the two angular momentums to find: $L_i=I\omega_0^2+mvdsin\theta$ $L_i=\frac{2}{5}MR^2\omega_0^2+mvRsin\theta$ Where $\omega_0=.684$ We set this equal to the final angular momentum: $\frac{2}{5}MR^2\omega_0^2+mvRsin\theta=(mR^2+\frac{2}{5}MR^2)\omega_f^2$ Where $\omega_f=.621\ rads/s$ Plugging in the known values, we find: $m=2.8 \times 10^7 \ kg$
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