(a) Before it hits the door, why does the bullet have angular momentum relative

Question

(a) Before it hits the door, why does the bullet have angular momentum relative the door's axis of rotation?

(b) 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.)
= rad/s

(d) 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. (Give the kinetic energy of the door-bullet system just after collision.)
= J

Explanation / Answer

a) the bullet has angular momentum because it is traveling at some perpendicular radius (8.7cm in this case)

b) Now we have to assume momentum is conserved so:
The bullets prior to the collision (mvr) will equal the door and bullet after
mvr=[m(r)2+(1/3)M(L)2]

.005(1000)(.087)=[.005(.087)2+(1/3)(17.9)12]

=.0729 rad/sec

d)So first I'll find the kinetic energy of the bullet before

KE=(1/2)mv2=(1/2)(.005)(1000)2

KE=2500 joules

Now for the door-bullet after collision:

KE=(1/2)[m(r)2+(1/3)M(L)2]2=(1/2)[.005(.087)2+(1/3)(17.9)12](.0729)2

KE=.0159 joules

This means the door-bullet system after the collision is less than the bullet before the collision.

Hope that helps

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