Angular Momentum: mud hitting door

  • #1

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!
 

Answers and Replies

  • #2
I figured out the correct answer, but in order to get it, I needed to say that

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

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.
 

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