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## Homework Statement

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.

## Homework Equations

[itex]L_{initial}=L_{final}[/itex]

[itex]L=Iw=rp[/itex]

## The Attempt at a Solution

[itex]L_{initial}=L_{door}+L_{mud}=0+mvr=(0.5)(14)(0.5)=3.5[/itex]

[itex]L_{final}=I_{total}+w_{f}[/itex]

[itex]I_{total}=I_{door}+I_{mud}[/itex]

I'm not sure how to calculate [itex]I_{door}[/itex], which gets me stuck here.

I had read online that it is:

[itex]I_{door}=\frac{1}{3}(42)(0.5)^2=3.5[/itex]

Mud I know:

[itex]I_{mud}=Iw=(0.5)(0.5)^2=1.125[/itex]

Assuming I calculated the door correctly,

[itex]L_{initial}=L_{final}[/itex]

[itex]3.5=(3.5+0.125)w_{f}[/itex]

[itex]w_{f}=0.966[/itex]

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