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Bullet fired at a Hinged door (rotational velocity problem) 
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#1
Apr1710, 06:20 PM

P: 5

1. The problem statement, all variables and given/known data
A 0.005kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 17.2kg door, imbedding itself 9.8 cm from the side opposite the hinges as in the figure below. The 1.00mwide door is free to swing on its hinges. relevant img: http://www.webassign.net/sercp8/p856.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 doorbullet 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 


#2
Apr1710, 06:27 PM

Mentor
P: 41,436

What is conserved? 


#3
Apr1710, 06:28 PM

Sci Advisor
P: 2,793

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. 


#4
Apr1710, 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?



#5
Apr1710, 06:40 PM

P: 5




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