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Is there a formula to calculate both the translational and rotational velocity? Does the bar always bend, and if so is there a formula on how it bends (maybe related to the velocity/force of the bullet)? What if the bar is very long, to the point where if the bar rotate it'd be at relativistic velocity?

What if I shoot at a string of cloth instead? Would the behavior of the bar/string be different if there's air drag?

Sorry for asking too many questions, but if you can answer any of them I'd be grateful.

Qmechanic
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Kim Dong
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  • The concepts you need are covered by basic mechanics. In particular moment of inertia, momentum and kinetic energy are terms to look up. But the question is off-topic (too broad, not enough effort at research). – StephenG - Help Ukraine Apr 24 '18 at 09:41
  • I think I forgot a lot of stuff from mechanics... I tried to look, but couldn't find anyone talk about how much the momentum of the bullet would be turned into translational motion, and how much rotational... Could you perhaps at least give me some more clues? I'd really appreiciated it. – Kim Dong Apr 24 '18 at 10:05

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Imagine a steel bar floating in space.

With you so far.

Assuming the bar wouldn't break or bend, if I shoot a bullet at one end

What do you mean "one end"? Be specific. Precisely describe the bar and where the bullet hits it.

would it rotate, fly away, or both?

That depends entirely on where the bullet hits it. If you hit it in the middle of the end, then it simply moves. If you hit it on the side of the end, it rotates too. Which is trivially calculated using basic trigonometry.

Maury Markowitz
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  • I was thinking of a 2D situation only, so its 3D equivalence would be the bullet hitting the exact middle of the end I assume...

    And the rotation I'm thinking about is one that goes perpendicular through the middle of the bar... Like how an egg would roll (rotate around its axis when push on the side similar to how a bullet pushing the bar) instead of sliding if the friction is high enough.

    I think I should ask this question differently with some illustration another time...

     – Kim Dong Apr 25 '18 at 12:03
  • Ok so then simplify: it's just a rectangle. If the bullet is aimed directly at the center of the rectangle, and hits directly at the middle of the short dimension, then it simply transfers its momentum. If it strikes somewhere else on the short axis, there will be a rotational component based on the triangle between that point, the center of the short side, and the center of the entire bar (well, it's CoG actually). This is basic trig, just draw a few examples and you'll see what I mean. – Maury Markowitz Apr 25 '18 at 13:11