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Angular Momentum -Opening a door with a bullet

  1. Apr 14, 2004 #1
    Sorry for so many questions, just coming back here after going through all the review questions I could get. Here we go:

    A cowboy wants to open a saloon door with a revolver shot. The swinging door (Mass M, width b) is hit on the very edge and the bullet (Mass m, velocity v) lodges into the door.
    a) Derive an expression for the moment of inertia of the door. - I got this to be J = Mb^2/3, and not too worried about this question.
    b) Derive an expression for the angular velocity w, with which the door swings open after being hit. -I'm supposed to use the conservation of angular moment here, right? I ended up getting w = 3mv/((M+m)b^2), but I don't think this is right.
    c) How many degrees will the door open at most, with D* as the "angular benchmark"? Use the numbers M = 10 kg, b = .6 m, m = 10 g, v = 500 m/s, D* = 1.2 Nm. - I don't even know if "angular benchmark" is the right translation for D* (it's Winkelrichtgroesse in German if that helps anyone). We were also given the formula Moment of Force = -D*phi. Am I going to want to plug the info I'm given into the formula I created in b?

    Thank you again.
  2. jcsd
  3. Apr 14, 2004 #2

    Doc Al

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    Yes, use conservation of angular momentum. But you need the rotational inertia of the "door + bullet", not just the door. (It's a tiny correction.)
    Assuming by "moment of force" is meant a restoring torque acting to close the door (it's on a spring), then find energy stored in it as a function of angle. (Integrate; or just compare to a spring.) Then find the KE that the system has after impact, using the results of step b. Then consider that KE is transformed into spring potential energy.
  4. Apr 14, 2004 #3
    I used mv = (M + m)b^2w/3, which is including both the door and the bullet, isn't it? Or do you mean in the initial momentum of the bullet as well? The door has no momentum to start off with though.
  5. Apr 14, 2004 #4


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    In this problem all you care about is the angular momentum of the bullet (rotating about the door hinge). The balance of the linear momentum will be taken up by the door hinge.

    Part b:
    The angular momentum of the bullet should be [tex]|\vec{p} \times \vec{r} | = mvb[/tex] (angular speed should be in per second - you had per meter second). And, since [tex]M>>m[/tex] you can just use [tex]\omega\approx\frac{3mv}{Mb}[/tex]

    Part d: The saloon door is on a spring that's working against the bullet. D is effectively the spring constant for the door. You should be able to solve this using energy.
  6. Apr 14, 2004 #5

    Doc Al

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    You are leaving out a "b" (in the initial angular momentum of the bullet) and are not capturing the rotational inertia of the bullet properly:
    mvb = (1/3M + m)b^2ω, so ω = 3mv/(M+3m)b.
    But as NateG says, you can ignore the bullet's small contribution to the rotational inertia (3m compared to M), so ω ≈ 3mv/Mb.
    Last edited: Apr 14, 2004
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