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Bullet fired at a Hinged door (rotational velocity problem)

  1. Apr 17, 2010 #1
    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
     
  2. jcsd
  3. Apr 17, 2010 #2

    Doc Al

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    Staff: Mentor

    KE is not conserved--it's a perfectly inelastic collision. (If it were conserved, the second question would be rather trivial.)

    What is conserved?
     
  4. Apr 17, 2010 #3

    Matterwave

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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.
     
  5. Apr 17, 2010 #4
    so how should i set this up then? as 2 rotational velocities with different moments of inertia?
     
  6. Apr 17, 2010 #5
    momentum would be conserved then. so do i set it up as initial linear momentum equals final rotational momentum?
     
  7. Apr 17, 2010 #6

    Matterwave

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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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