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Angular Motion Of A Bullet Striking A Door 
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#1
Dec1312, 07:31 AM

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The problem is:
"A 0.00600kg bullet traveling horizontally with a speed of 1.00*10^{3} m/s enters an 19.2kg door, imbedding itself 8.60 cm from the side opposite the hinges as in the figure below. The 1.00m wide door is free to swing on its frictionless hinges. (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Yes? No? (b) If so, evaluate this angular momentum. (If not, enter zero.) (c) Is mechanical energy of the bulletdoor system constant in this collision? Yes? No? (Answer without doing a calculation.) (d) At what angular speed does the door swing open immediately after the collision? (e) Calculate the energy of the bulletdoor system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision." I think I will be able to solve parts (d) and (e). For (a), the answer is yes, but I can't quite figure out why the answer is yes. For (c), the answer is no; but I am having trouble understanding how we can possibly know. Couldn't the kinetic energy of the bullet be completely absorbed as kinetic energy in the door? In this scenario, wouldn't mechanical energy be conserved? 


#2
Dec1312, 07:42 AM

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#3
Dec1312, 04:37 PM

P: 946

Angular momentum of a particle is found to be the product linear momentum of the particle and the particle's distance from the axis of rotation. Yes, I see now. I do have another question, however; for part (d), how do I calculate the momentum of inertia of the door?



#4
Dec1312, 04:44 PM

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Angular Motion Of A Bullet Striking A Door



#5
Dec1312, 04:47 PM

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I don't have the entire length of the door, though.



#7
Dec1312, 05:21 PM

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Okay, I got parts (b), (d), and (e) wrong.
Part (b): [itex]L=(0.00400~kg)(1.00\cdot 10^3~m/s)(8.10\cdot 10^{2}~m)=0.324~kg\cdot m^2/s[/itex] For (d), [itex]L_i=L_f \rightarrow (0.324~kg\cdot m^2/s)+0=[(0.00400~kg)((8.10\cdot 10^{2}~m/s)^2+1/3(15.8)(1.00~m)^2] \omega \rightarrow \omega = 0.0615~rad/s[/itex] For (e), I solved for the initial kinetic energy, and got it right; but I apparently didn't properly solve the final kinetic energy: [itex]K_f=1/2(0.00400~kg)((8.10\cdot 10^{2}~m/s)^2+1/3(15.8)(1.00~m)(0.0615~rad/s)^2= 0.00997 J[/itex] 


#8
Dec1312, 07:21 PM

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#9
Dec1312, 07:27 PM

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0.081 m, right? I accidentally put units of velocity with that number.



#11
Dec1312, 07:45 PM

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Well, in the problem it states that the bullet is embedded into the door at a distance of 8.60 cm from the hinges from the door, so I took that to be the distance...Oh, I calculated 8.1 cm...Was that the error?



#12
Dec1312, 07:54 PM

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#13
Dec1312, 08:00 PM

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Oh, 8.60 cm is the distance from where the bullet hits the door to the edge of the door not containing the hinges? If that's the case, I wouldn't know the other distance.



#15
Dec1312, 08:40 PM

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But how is that the distance from the axis of rotation (hinges) to the other end of the door?



#16
Dec1312, 08:44 PM

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#17
Dec1312, 08:46 PM

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1.0 m, and then it's length is 8.60 m plus the distance beyond the bullet.



#18
Dec1312, 09:32 PM

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The door is 1 meter wide. Therefore the far edge of the door is 1 meter from the edge with the hinges.
How far is the bullet from the hinges? (The figure is from a perspective looking down on the door.) 


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