Is this conservation of momentum?

AI Thread Summary
The discussion centers on a physics problem involving a rod and a bullet, where the bullet impacts the rod and the goal is to find the bullet's speed before impact. Participants clarify that conservation of angular momentum, rather than linear momentum, is the appropriate principle to apply in this scenario. The correct expression for the bullet's angular momentum before the collision is derived as mv(L/2)sin(theta). The initial calculations presented by the user were incorrect due to mixing linear and angular momentum concepts. Ultimately, the discussion emphasizes the importance of understanding angular momentum in solving the problem correctly.
Cyrad2
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The problem is:
A uniform thin rod of length = .5m and mass M = 4.0kg can rotate in a horizontal plane about a vertical axis through its center (I = ML^2/12). The rod is at rest when a bullet of mass m = 3.0g traveling in the horizontal plance of the rod is fired into one end of the rod. As viewed from above, the direction fo the bullets velocity makes an angle of theta=60 with the rod. If the bullet lodges into the rod and the angular velocity of the rod is 10rad/s immediately after the collision, what is the bullet's speed just before impact?
This is a review question for a test I have monday, the answer is1290m/s, but I can't get that.

Here's what I did (which is obviously incorrect):
I = (1/12)ML^2 + m(L/2)^2

conservation of momentum?
mv = Iw
v = Iw/m
= (1/12)ML^2 + m(L/2)^2 / m
= wrong.

Is conservation of momentum the right tool to be using to solve this? ...how does theta play into it? Any tips will be *greatly* appreciated.
 
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Cyrad2 said:
Here's what I did (which is obviously incorrect):
I = (1/12)ML^2 + m(L/2)^2
This is correct for the rotational inertia of the system (rod + bullet) after the collision.

conservation of momentum?
mv = Iw
v = Iw/m
= (1/12)ML^2 + m(L/2)^2 / m
= wrong.
Two problems here:
(1) You are mixing up linear and angular momentum; they aren't the same thing---the units don't even match.
(2) In this problem, angular momentum is conserved. What's the angular momentum of the bullet just before it hits the rod?

Is conservation of momentum the right tool to be using to solve this?
Conservation of angular momentum is.
...how does theta play into it?
It will allow you to determine the angular momentum of the bullet.
 
Cyrad2 said:
The problem is:
A uniform thin rod of length = .5m and mass M = 4.0kg can rotate in a horizontal plane about a vertical axis through its center (I = ML^2/12). The rod is at rest when a bullet of mass m = 3.0g traveling in the horizontal plance of the rod is fired into one end of the rod. As viewed from above, the direction fo the bullets velocity makes an angle of theta=60 with the rod. If the bullet lodges into the rod and the angular velocity of the rod is 10rad/s immediately after the collision, what is the bullet's speed just before impact?
...

Is conservation of momentum the right tool to be using to solve this?

No. Use conservation of angular momentum instead. The initial angular momentum is that of the bullet and it is mv(L/2)sin(theta) with respect to the rotation axis.

ehild
 
Great! Thanks guys. You helped a bunch...now I know what I need to go back and study :-) Have a great sunday!
 
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