# Angular Momentum: mud hitting door

1. Nov 19, 2011

### PirateFan308

1. The problem statement, all variables and given/known data
A solid wood door 1.00m wide and 2.00m high is hinged along one side and has a total mass of 42.0kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500kg, traveling perpendicular to the door at 14.0m/s just before impact.
Find the angular speed of the door.

2. Relevant equations
$L_{initial}=L_{final}$

$L=Iw=rp$

3. The attempt at a solution
$L_{initial}=L_{door}+L_{mud}=0+mvr=(0.5)(14)(0.5)=3.5$

$L_{final}=I_{total}+w_{f}$

$I_{total}=I_{door}+I_{mud}$

I'm not sure how to calculate $I_{door}$, which gets me stuck here.

$I_{door}=\frac{1}{3}(42)(0.5)^2=3.5$

Mud I know:

$I_{mud}=Iw=(0.5)(0.5)^2=1.125$

Assuming I calculated the door correctly,

$L_{initial}=L_{final}$

$3.5=(3.5+0.125)w_{f}$

$w_{f}=0.966$

But this is wrong, and I'm not sure where I messed up. Thanks in advance!

2. Nov 19, 2011

### PirateFan308

I figured out the correct answer, but in order to get it, I needed to say that

$I_{door}=\frac{1}{3}(42)(1)^{2}$

I don't understand why I should use 1 for the radius of the door, when the force was applied to the centre of the door.