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Angular Motion Of A Bullet Striking A Door

by Bashyboy
Tags: angular, bullet, door, motion, striking
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Bashyboy
#1
Dec13-12, 07:31 AM
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The problem is:

"A 0.00600-kg bullet traveling horizontally with a speed of 1.00*103 m/s enters an 19.2-kg door, imbedding itself 8.60 cm from the side opposite the hinges as in the figure below. The 1.00-m 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 bullet-door 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 bullet-door 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?
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Doc Al
#2
Dec13-12, 07:42 AM
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Quote Quote by Bashyboy View Post
For (a), the answer is yes, but I can't quite figure out why the answer is yes.
How is the angular momentum of a particle defined?

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?
When the colliding objects end up merged and traveling together, that is a perfectly inelastic collision. You can show (using conservation of momentum) that mechanical energy cannot be conserved. Only when they bounce off each other can energy be conserved.
Bashyboy
#3
Dec13-12, 04:37 PM
P: 936
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?

Doc Al
#4
Dec13-12, 04:44 PM
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Angular Motion Of A Bullet Striking A Door

Quote Quote by Bashyboy View Post
I do have another question, however; for part (d), how do I calculate the momentum of inertia of the door?
You can treat it like a thin rod.
Bashyboy
#5
Dec13-12, 04:47 PM
P: 936
I don't have the entire length of the door, though.
Doc Al
#6
Dec13-12, 04:59 PM
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Quote Quote by Bashyboy View Post
I don't have the entire length of the door, though.
Use the 'width' of the door, which is given.
Bashyboy
#7
Dec13-12, 05:21 PM
P: 936
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]
Doc Al
#8
Dec13-12, 07:21 PM
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Quote Quote by Bashyboy View Post
Part (b): [itex]L=(0.00400~kg)(1.00\cdot 10^3~m/s)(8.10\cdot 10^{-2}~m/s)=0.324~kg\cdot m^2/s[/itex]
I don't understand what are doing here. What's the momentum of the bullet? How far is it from the axis?
Bashyboy
#9
Dec13-12, 07:27 PM
P: 936
0.081 m, right? I accidentally put units of velocity with that number.
Doc Al
#10
Dec13-12, 07:41 PM
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0.081 m, right? I accidentally put units of velocity with that number.
That's not right. How did you determine that distance?
Bashyboy
#11
Dec13-12, 07:45 PM
P: 936
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?
Doc Al
#12
Dec13-12, 07:54 PM
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Quote Quote by Bashyboy View Post
Well, in the problem it states that the bullet is embedded into the at at a distance 8.60 cm from the door, so I took that to be the distance...Oh, I calculated 8.1 cm...Was that the error?
The bullet hits 8.60 cm from the side opposite the hinges (see the figure). What's the distance to the axis?
Bashyboy
#13
Dec13-12, 08:00 PM
P: 936
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.
Doc Al
#14
Dec13-12, 08:27 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.
You have the width of the door.
Bashyboy
#15
Dec13-12, 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?
SammyS
#16
Dec13-12, 08:44 PM
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Quote Quote by Bashyboy View Post
But how is that the distance from the axis of rotation (hinges) to the other end of the door?
How wide is the door ?
Bashyboy
#17
Dec13-12, 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.
SammyS
#18
Dec13-12, 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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