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

Question

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

yes/no    

Explain your answer.

(b) Is mechanical energy conserved in this collision? Answer without doing a calculation.

yes/no    

(c) 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)
Yes. Any object which has a velocity vector which does not pass directly through the chosen axis has angular momentum about that axis.

b)
No, mechanical energy is not conserved. Angular momentum is conserved, and any excess energy is dissipated within the door.

c)
W.before = W.after
Solve for both, then solve for final w. (note that W is angular momentum, w is angular velocity)

W.before = I.bullet * w.bullet + I.door * w.door
I.bullet = m.bullet * (0.086 m)^2
w.bullet = V.bullet / (.914m)
W.before = m.bullet * V.bullet * (.914 m) + 0

W.after = I.bullet * w.bullet + I.door + w.door
W.after = m.bullet * (.914 m)^2 * w.bullet + 1/3 * m.door * (1m)^2 * w.door
since w.bullet = w.door, we'll just call that w

W.after = w * (m.bullet * (.914 m)^2 + 1/3 * m.door * (1m)^2)

Set W.before = W.after and solve for w

m.bullet * V.bullet * (.914 m) + 0 = w * (m.bullet * (.914 m)^2 + 1/3 * m.door * (1m)^2)
.004kg * 1000 m/s * .914 m = w * (.004 kg * (.914 m)^2 + 1/3 * 16.6 kg * (1 m)^2)
3.65kgm^2/s = w * 5.531 kgm^2

w = 0.66 radians/second

d)

Before:
Ek = 1/2 * m.bullet * V.bullet^2
Ek = 1/2 * .004 kg * 1000000 m^2/s^2
Ek = 2000 joules

After:
Ek = I.bullet * w^2 + I.door * w^2
Ek = (.66 rad/sec)^2 * (.004 kg * (.914 m)^2 + 1/3 * 16.6 kg * (1m)^2)
Ek = 2.40 joules

Note that kinetic energy after is significantly less than kinetic energy before

there might be calculation errors but on the overall procedure is correct. :-)

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