Rotational Collision Door Problem

[megatyler30]

Homework Statement

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 49.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud of mass 0.500 kg, traveling perpendicular to the door at 15.0 m/s just before impact. Find the final angular speed of the door (in rad/sec).

Homework Equations

ΣL_Before=ΣL_After
L=Iω=mvrsinϕ
I_{Door swinging on hinges}=(1/3)mr²
I_{Of point}=mr²

The Attempt at a Solution

ΣL_Before=ΣL_After
mvrsinϕ=(I_Door+I_Mud)ω
(.5)(15)(.5)sin90=[(1/3)(49)(.5)²+(.5)(.5)²]ω
w=.8911 rad/sec

Sorry, I did do .8911 on the question (webassign), I just typed it in wrong

 

[haruspex]

(.5)(15)(.5)sin90=[(1/3)(49)(.5)²+(.5)(.5)²]ω

That looks right...

w=89.11 rad/sec

... but that seems much too big.

 

[megatyler30]

That looks right...

... but that seems much too big.

I fixed it in OP, I meant to put .8911

 

[haruspex]

I fixed it in OP, I meant to put .8911

Looks good.

 

[megatyler30]

It's wrong though... (Or at least I used exact same method for practice another version and got it wrong but can't check since I'll be locked out if I get it wrong again.

 

[haruspex]

It's wrong though... (Or at least I used exact same method for practice another version and got it wrong but can't check since I'll be locked out if I get it wrong again.

OK, I see it. It's an easy confusion.
The moment of inertia of a rod length 2L about its centre is mL²/3.
The moment of inertia of a rod length 2L about one end is 4mL²/3.
The moment of inertia of a rod length L about one end is mL²/3.
You halved the door width, but only used 1/3 instead of 4/3.

 

[Tanya Sharma]

(.5)(15)(.5)sin90=[(1/3)(49)(.5)²+(.5)(.5)²]ω

The item in red should be width of the door i.e 1.That should give you ω = .23 rad/sec .

 

[megatyler30]

Okay, thank you very much, I ended up getting ω = .2278 rad/sec which was right.