Moment of Inertia of mud thrown on door

[PsychonautQQ]

Homework Statement

A solid wood door 1m wide and 2m high is hinged along one side and has a total mass of 40kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass .5kg.
What is the original and final moment of inertia for the door?
What is the final speed of the door if the mud hits the center of the door at 12 m/s?

Homework Equations

I=Cmr^2

The Attempt at a Solution

I'm really quite lost here, I have no idea how to figure out the constant for the door... am i supposed to have this memorized? Then mud is applied in the center of the door, which would add mass at a point r/2 if you take r to be the width of the door,, but I'm having problems putting it all together.

For solving for the final angular velocity of the door i used conservation of angular momentum

mivi x ri = mfvf x rf

mi = mass of mud
vi = velocity of mud initially
ri = .5 (the place where the mud hits the door)

mf = mass of mud + door combined
vf = final velocity of the door at it's furthest point from the hinges
rf = 1 (width of the door)
i solved this equation for vf and then divided it by the radius (1) to find angular speed. My answer was wrong ;-/ anyone know what went wrong in my brain?

 

[SteamKing]

You should be able to look up the MOI for a rectangle or, if all else fails, calculate it from first principles.

 

[mfb]

I have no idea how to figure out the constant for the door... am i supposed to have this memorized?

The door is equivalent to a very simple shape, where it is useful to know the moment of inertia - and if you do not, you should be able to derive it via integration.

mivi x ri = mfvf x rf
[...]
rf = 1 (width of the door)

That would require the whole door and mud to be 1m away from the hinge - they are not. You can use the moment of inertia here, as soon as you calculated it.