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

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Homework Help Overview

The problem involves a bullet impacting a door and the subsequent effect on the door's angular velocity. It is situated within the context of rotational dynamics and conservation laws, specifically focusing on angular momentum.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and angular momentum, with some confusion regarding the definitions and conservation principles. There is an attempt to apply a formula for angular velocity, but discrepancies in understanding lead to questions about the initial conditions of the bullet's angular momentum.

Discussion Status

The discussion is ongoing, with participants exploring the correct application of conservation principles. Some guidance has been offered regarding the need to consider the angular momentum of the bullet before impact, but there is no consensus on the correct approach yet.

Contextual Notes

Participants are navigating the complexities of angular versus linear momentum, with some expressing uncertainty about the initial conditions and assumptions related to the bullet's motion and its impact on the door.

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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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