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