Bullet fired at a Hinged door (rotational velocity problem)

  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. Doc Al

    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. Matterwave

    Matterwave 3,816
    Science Advisor
    Gold Member

    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. so how should i set this up then? as 2 rotational velocities with different moments of inertia?
  6. momentum would be conserved then. so do i set it up as initial linear momentum equals final rotational momentum?
  7. Matterwave

    Matterwave 3,816
    Science Advisor
    Gold Member

    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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