What is the Angular Momentum of a Bullet Before Hitting a Door?

[MacDougal87]

Hey guys, I've been a member of this site for a while now, looking at other people's homework problems and attempting them as extra practice. I find that reading about what other people have trouble with when attempting physics problems helps me avoid them myself, and it has worked pretty well so far.
Most of the time when I make a mistake in physics problems, and have the correct answer on hand, I am eventually able to figure out where I went wrong and apply the correct steps to get there.
However, this situation truly baffles me:

Homework Statement

A .005-kg bullet traveling horizontally with speed 1,000-m/s strikes an 18-kg door, embedding itself 10 cm from the side opposite the hinges. The 1-m wide door is free to swing on its frictionless hinges.
Before it hits the door, what is the bullet's angular momentum relative to the door's axis of rotation?

(to distinguish between the two, m will be the mass of the bullet and M will be the mass of the door, the distance from the axis of rotation of the door is d)

Homework Equations

The way I approached this problem I figured we'd need:
L=Iω
To get I; I modeled the bullet as a particle, and used I=md²
Using v=dω; ω=v/d
Substituting, I could use L=(md²)(v/d)
Which simplifies to L=mdv

The Attempt at a Solution

Plugging in the given values, I calculated (.005kg)(.1m)(1,000m/s) = .5 kgm²/s
The answer in the back of the book however, gives the solution as 4.5 kgm²/s. The only way I can figure they might have gotten this solution was to factor in the mass of the door somewhere, maybe using the door's moment of inertia, but that doesn't make any sense to me since the problem specifically asks for the bullet's angular momentum before it hits the door. I can't seem to think of any other way to take this problem since the bullet is not interacting with anything else in the system at this moment.
Any suggestions would be great, I have a big exam coming up on Wednesday, and need all the practice I can get =)

 

[gneill]

The mass of the door is extraneous information. All you need to do is recall that angular momentum is given by r x mv. You have the mass and velocity of the bullet, so that gives you mv. You should be able to figure out the perpendicular distance from the axis of rotation from the given impact location.

 

[MacDougal87]

Right, but since the bullet embeds itself in the door 10 cm from the hinges (aka its axis of rotation) that gives a distance of .1m, which is what I used in the calculations. Also, taking the cross product of r x mv gives (0i + (-.1)j) X .005(1000i + 0j) = (-.5)j*i = -(-.5) = .5 which is what my previous method gave me also. Unless I did cross product wrong, I suppose it is a little rusty...

 

[gneill]

"...embedding itself 10 cm from the side opposite the hinges"

 

[MacDougal87]

*facepalm* well, that solves that mystery. Not sure if I should be glad or ashamed that it wasn't my physics that was off as much as it was just me not realizing that. I must have re-read that problem at least five or ten times, and missed it every time.
Thanks for the help =P

 

[gneill]

No worries. Happens to everyone.