How Does a Bullet Impact Affect the Angular Velocity of a Door?

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A 10 g bullet traveling at 370 m/s strikes an 8.0 kg door, causing it to swing open. The discussion revolves around calculating the door's angular velocity post-impact using conservation of angular momentum. The initial attempt incorrectly equated linear and angular momentum, leading to confusion. It is clarified that the bullet does possess angular momentum relative to the door's hinge, which must be considered in the calculations. The correct approach involves using the bullet's angular momentum before impact to find the door's angular velocity after the collision.
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Homework Statement



A 10 g bullet traveling at 370 m/s strikes a 8.0 kg , 0.80-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 just after impact?
Express your answer to two significant figures and include the appropriate units.

Homework Equations


use conservation of momentum for slab about edge: linear momentum = angular momentum
mv = Iw
I = (1/3)ML^2 (for a slab)
w = (3mv) / (ML^2)
(w = omega)

3. The Attempt at a Solution

w = (3mv) / (ML^2)

w = (3* 0.01kg * 370m/s) / (8.0kg * 0.80kg^2)
w = 2.167 = 2.2 rad/s
= wrong

cant see what I am doing wrong, thanks for any help
 
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sp3sp2sp said:
use conservation of momentum for slab about edge: linear momentum = angular momentum
o_O

Linear momentum and angular momentum are different quantities with different units. They cannot be set equal. In this case, only one of those is conserved.

Hint: Use conservation of angular momentum. (What's the angular momentum of the bullet just before impact?)
 
bullet doesn't have any angular momentum before impact So Iw = 0?
 
sp3sp2sp said:
bullet doesn't have any angular momentum before impact
Sure it does (about the door hinge axis). Read this: Angular Momentum of a Particle

sp3sp2sp said:
So Iw = 0?
No.
 
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