(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Bullet fired at a Hinged door (rotational velocity problem)

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