Angular momentum of a bullet

  • Thread starter randyp92
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  • #1
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Homework Statement


A uniform stick 1.3 m long with a total mass of 260 g is pivoted at its center. A 3.6 g bullet is shot through the stick midway between the pivot and one end. The bullet approaches at 250 m/s and leaves at 140 m/s. With what angular speed is the stick spinning after the collision?



Homework Equations


Iw(initial)=Iw(final)


The Attempt at a Solution


I dont really know what intial condition you would use. Obviously the final condition is the bullet left and the thing is spinning. Also, what would you use for the R of the bullet in calculating moment of inertia? One thought i had was the moment just before the bullet enters the stick the R approaches 1.3/4. So Iw(initial) would be (.0036)(1.3/4)^2(250/(1.3/4)). Then my guess for Iw(final) is (1/12)(.26)(1.3)^2(W) + (.0036)(1.3/4)^2(140/(1.3/4)) where W is what we want.
 
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Answers and Replies

  • #2
Filip Larsen
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The equation you have in 2) states that the angular momentum of just the stick is preserved, which leaves out the angular momentum of the bullet. You need to equate the total angular momentum (angular momentum of the stick plus angular momentum of the bullet) before and after collision around the pivot axis (note that choosing this axis for calculating the angular momentum allows us to ignore reaction forces from the pivot).

Also note, that the angular momentum of the bullet can be calculated directly from the definition of angular momentum, that is, L = mv x r = mvr sin(theta), where theta is the angle between the v and r vector.
 
  • #3
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Ok, so theta is 90 for the initial and final conditions. Now r. If you consider the instant before the bullet hits and the instant after the bullet leaves could you use (1.3/4)?
 
  • #4
Filip Larsen
Gold Member
1,286
213
Ok, so theta is 90 for the initial and final conditions. Now r. If you consider the instant before the bullet hits and the instant after the bullet leaves could you use (1.3/4)?
Yes, that is correct.
 

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