# Homework Help: Rotational Collision Door Problem

1. Jan 11, 2014

### megatyler30

1. The problem statement, all variables and given/known data
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).

2. Relevant equations
ΣLBefore=ΣLAfter
L=Iω=mvrsinϕ
IDoor swinging on hinges=(1/3)mr2
IOf point=mr2

3. The attempt at a solution
ΣLBefore=ΣLAfter
mvrsinϕ=(IDoor+IMud
(.5)(15)(.5)sin90=[(1/3)(49)(.5)2+(.5)(.5)2

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

Last edited: Jan 11, 2014
2. Jan 11, 2014

### haruspex

That looks right...
... but that seems much too big.

3. Jan 11, 2014

### megatyler30

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

4. Jan 12, 2014

### haruspex

Looks good.

5. Jan 12, 2014

### 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.

6. Jan 12, 2014

### haruspex

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.

7. Jan 12, 2014

### Tanya Sharma

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

Last edited: Jan 12, 2014
8. Jan 12, 2014

### megatyler30

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