## Essential University Physics: Volume 1 (3rd Edition)

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