# Angular Motion Of A Bullet Striking A Door

by Bashyboy
Tags: angular, bullet, door, motion, striking
 P: 798 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? Attached Thumbnails
Mentor
P: 40,243
 Quote by Bashyboy 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.
 P: 798 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?
Mentor
P: 40,243

## Angular Motion Of A Bullet Striking A Door

 Quote by Bashyboy 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.
 P: 798 I don't have the entire length of the door, though.
Mentor
P: 40,243
 Quote by Bashyboy I don't have the entire length of the door, though.
Use the 'width' of the door, which is given.
 P: 798 Okay, I got parts (b), (d), and (e) wrong. Part (b): $L=(0.00400~kg)(1.00\cdot 10^3~m/s)(8.10\cdot 10^{-2}~m)=0.324~kg\cdot m^2/s$ For (d), $L_i=L_f \rightarrow (0.324~kg\cdot m^2/s)+0=[(0.00400~kg)((8.10\cdot 10^{-2}~m/s)^2+1/3(15.8)(1.00~m)^2] \omega \rightarrow \omega = 0.0615~rad/s$ For (e), I solved for the initial kinetic energy, and got it right; but I apparently didn't properly solve the final kinetic energy: $K_f=1/2(0.00400~kg)((8.10\cdot 10^{-2}~m/s)^2+1/3(15.8)(1.00~m)(0.0615~rad/s)^2= 0.00997 J$
Mentor
P: 40,243
 Quote by Bashyboy Part (b): $L=(0.00400~kg)(1.00\cdot 10^3~m/s)(8.10\cdot 10^{-2}~m/s)=0.324~kg\cdot m^2/s$
I don't understand what are doing here. What's the momentum of the bullet? How far is it from the axis?
 P: 798 0.081 m, right? I accidentally put units of velocity with that number.
Mentor
P: 40,243
 Quote by Bashyboy 0.081 m, right? I accidentally put units of velocity with that number.
That's not right. How did you determine that distance?
 P: 798 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?
Mentor
P: 40,243
 Quote by Bashyboy 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?
 P: 798 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.
Mentor
P: 40,243
 Quote by Bashyboy 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.
 P: 798 But how is that the distance from the axis of rotation (hinges) to the other end of the door?
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