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Bullet fired at a Hinged door (rotational velocity problem)

by Number47
Tags: angular velocity, bullet, door, momentum
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Number47
#1
Apr17-10, 06:20 PM
P: 5
1. The problem statement, all variables and given/known data
A 0.005-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 17.2-kg door, imbedding itself 9.8 cm from the side opposite the hinges as in the figure below. The 1.00-m-wide door is free to swing on its hinges.

relevant img: http://www.webassign.net/sercp8/p8-56.gif

At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.)

Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.


2. Relevant equations
moment of inertia=(1/3)mL^2

KErotational= (1/2)*I*Wf^2


3. The attempt at a solution

I calculated the moment of inertia for the door and bullet system I=(1/3)(17.205kg)(1.0m-.098m)^2=4.66

Then i set the KE of the bullet equal to the KE of the door bullet system equal to each other and solved for Wf

Wf=sqrt((.005kg*(1000m/s)^2)/4.66) and got the answer to be 32.76rad/s which sounds way to big to be true. This is due just before midnight tonight, so thanks in advance for any help
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Doc Al
#2
Apr17-10, 06:27 PM
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Quote Quote by Number47 View Post
Then i set the KE of the bullet equal to the KE of the door bullet system equal to each other and solved for Wf
KE is not conserved--it's a perfectly inelastic collision. (If it were conserved, the second question would be rather trivial.)

What is conserved?
Matterwave
#3
Apr17-10, 06:28 PM
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The KE is not conserved since this collision is inelastic. Linear momentum is also not conserved due to external (tension) forces keeping the door on its hinge.

Angular momentum; however, should be conserved as there are no external torques. The tension forces are radial in direction.

Number47
#4
Apr17-10, 06:34 PM
P: 5
Bullet fired at a Hinged door (rotational velocity problem)

so how should i set this up then? as 2 rotational velocities with different moments of inertia?
Number47
#5
Apr17-10, 06:40 PM
P: 5
Quote Quote by Doc Al View Post
KE is not conserved--it's a perfectly inelastic collision. (If it were conserved, the second question would be rather trivial.)

What is conserved?
momentum would be conserved then. so do i set it up as initial linear momentum equals final rotational momentum?
Matterwave
#6
Apr17-10, 06:48 PM
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You can't have a linear momentum equaling an angular momentum, they are different phenomena which are measured in different units.

You want initial angular momentum equal to final angular momentum.


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