Answer
The final angular speed of the door is 0.223 rad/s.
Work Step by Step
We can find the moment of inertia of the door, which is related to the moment of inertia of a rod rotating about one end.
$I_{door} = \frac{1}{3}ML^2$
$I_{door} = \frac{1}{3}(40.0~kg)(1.00~m)^2$
$I_{door} = 13.33~kg~m^2$
We can find the moment of inertia of the door including the mud.
$I_{total} = I_{door}+I_{mud}$
$I_{total} = I_{door}+mr^2$
$I_{total} = 13.33~kg~m^2+(0.500~kg)(0.500~m)^2$
$_{total} = 13.46~kg~m^2$
We can use conservation of angular momentum to find the final angular speed of the door.
$L_2 = L_1$
$I\omega_f = mvr$
$\omega_f = \frac{mvr}{I}$
$\omega_f = \frac{(0.500~kg)(12.0~m/s)(0.500~m)}{13.46~kg~m^2}$
$\omega_f = 0.223~rad/s$
The final angular speed of the door is 0.223 rad/s.
