Linear and Angular Momentum (conservation of)

  • Thread starter patm95
  • Start date
  • #1
31
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Hi all, I have been working on this problem all night and although it should be easy, I can not seem to get the correct answer as in the back of the book.


Homework Statement



A uniform, thin rod of length .5m and a mass of 4kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle 60 deg with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/sec immediately after the collision, what is the bullet's speed just before impact?



Homework Equations



(1/12 M L^2)* Omega = this is momentum of rod after impact

m*v = this is momentum of bullit right before impact



The Attempt at a Solution



Using conservation of momentum, I would like to set the two equations equal to each other.

1/12*4kg*(.5^2)*(10) = .003kg*v*Cos 60 (using cos 60 because only the velocity in the y direction should contribute to the rods angular momentum)

The answer in the back of the book is 1300m/s

Any help is greatly appreciated.

Thanks!
 

Answers and Replies

  • #2
46
0
You need to use the angular momentum of the bullet. (You are trying to equate momentum with angular momentum, which don't even have the same units).
 
  • #3
114
0
use conservation of angular momentum:

for the bullet the angular momentum right before it hits is :
[itex]
\vec{L} = \vec{r} \times \vec{p} = rmv \sin (\theta )
[/itex]

give it a shot from here
 
  • #4
31
0
Got it! Thanks!
 

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