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What is the angular momentum of the bullet

  • Thread starter rasgar
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  • #1
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1. Homework Statement
A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length L and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

What is the angular momentum of the bullet- block system? (For the following answers, use M for the mass M, m for the mass m, and L for the length .)


2. Homework Equations

L=Iw=rP

3. The Attempt at a Solution
Well I easily got the solution with the known v (m*v*l), but the question asks for the solution with respect to only the mass of the bullet and the block, and the length of the rod. Since angular momentum has time as a part of its unit, I don't see how I'm supposed to do it. I thought I could find the torque and take the integral with respect to time, but the force of the system isn't known, so I'm completely lost. I think the physics teacher solved this, but I missed class, and there's a test in two days (I can't make it to his office hours either).
 

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Answers and Replies

  • #2
cepheid
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I'm not sure what your issue is, but here's my take on how to approach the problem. Using conservation of momentum, you can figure out the momentum of the system the at very instant of collision. This is the linear momentum with which the bullet+block would move off in a straight line if it weren't attached to the rod. The rod provides centripetal force, so the momentum does change, but only in direction. It's magnitude remains the same as it was at the instant of collision. So, by conservation of momentum:

mv = (m+M)v'

where v' is the velocity of the bullet + block after collision.

From the definition of angular momentum, the magnitude of the angular momentum is given by the the length of the rod times the linear momentum. Since the letter L is being used for the length of the rod, I'll just use J for angular momentum:

J = L*p = L*(M+m)v' = L*(m+M)*((m/(M+m))v = Lmv
 
  • #3
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"J...=Lmv"
Yea I actually got that
"Well I easily got the solution with the known v (m*v*l)"

The problem is I have to do the problem solely with M, m, and L (I accidentally put lower case L there sorry). Thanks for your response though.
 
  • #4
Doc Al
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The problem is I have to do the problem solely with M, m, and L (I accidentally put lower case L there sorry).
Your answer is correct--and the answer must incorporate the speed of the bullet. I suspect you're misinterpreting the instructions. (Where does it say that your answer can only have M, m, & L? Did you leave out part of the problem statement?)
 
  • #5
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Actually I just realized what my mistake was after the second post I made. The homework i had was online and the response checker was case-sensitive. I made the L a caps and got it right, sorry for the trouble guys.
 

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