Linear Momentum to Angular Momentum

[brad hars]

Homework Statement

A 10 g bullet traveling at 400 m/s strikes a 10 kg , 1.2-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door immediately after impact?

Homework Equations

p[/B]= mv
L = Iω

The Attempt at a Solution

For the bullet:
p = (0.010) * (400) = 4
For the door:
I = (1/3)(10.01)(1.2)^2 = 4.8048

Momentum is conserved.
p = Iω
4 = 4.8048 * ω
ω = 0.833
This seems right, and I've traced it over again and still seems right, but it's not and I can't seem to find out why.

 

[kuruman]

p = Iω

Here is your problem. You are setting linear momentum p equal to angular momentum Iω.

 

[brad hars]

Well, yea. Because the bullet goes in the door and all that linear momentum p goes into angular momentum. I'm not sure what else I could do with it.

 

[vbrasic]

Well, yea. Because the bullet goes in the door and all that linear momentum p goes into angular momentum. I'm not sure what else I could do with it.

Try to calculate the angular momentum of the bullet. Angular momentum and linear momentum are two separate quantities.

 

[kuruman]

Linear momentum is not conserved because the hinges prevent the door's center of mass from moving in a straight line. You have to set the angular momentum of the bullet (relative to the hinges) equal to the angular momentum of the door (also relative to the hinges). Keep apples with apples and oranges with oranges.

 

[brad hars]

But doesn't the bullet have 0 angular momentum?

 

[kuruman]

Not with respect to the hinges. $\vec{L}=\vec r \times \vec p$.