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Coriolis Force On A Bullet

  1. Feb 21, 2010 #1
    Hey guys, I wondered if you could give me a hand with this?

    1.A marksman follows a rapidly moving target by swinging his rifle from left to right about a vertical axis whilst holding the barrel horizontal. Just as the bullet is about to leave the barrel it has a muzzle velocity of 800 ms-1.
    1. If, at this instant, the rifle is rotating at 1 rad/s, what is the magnitude of the Coriolis force on the 20g bullet?
    2. Find the direction of force the rifle barrel exerts on the bullet.




    As far as I can see, the only relevant equation is that for the Coriolis pseudoforce:

    [tex]\vec{F}=-2m\vec{\omega}\times\vec{v}[/tex]

    Where F is the coriolis force, m is the mass of the bullet (0.02 kg), omega is the angular frequency of the rifle and v is the muzzle velocity.



    I assumed I don't have to take into account the coriolis effect due to the rotation of the earth (negligible compared to the rotation of the rifle) and plugged in the numbers for the first part to get -320N. This seems like an awfully large number and I just want to check that I have it right.

    As for the second part, the muzzle is exerting a force on the bullet in the direction of rotation up until the point the bullet leaves the muzzle, after which there is only the Coriolis pseudoforce acting perpendicular to both the axis of rotation (vertically down) and the direction of motion (radial in x-y plane) such that the bullet is deflected to the marksman's left (I think...). Which of these forces is the one they're asking for?

    Thanks in advance. :)
     
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2
    hey im doing this question as well and just noticed a small error that would make things clearer. You said you got -320N as the force, well I get -32N and I think you may have just put an extra 0 somewhere. About the second part im not really sure..
     
  4. Feb 21, 2010 #3
    Ah yeah thanks, T- I did make a silly decimal place error on the arithmetic. Just shoved it through the calculator again and realised what happened. Annoying how ambiguous the second part of the question is. -_- I don't know if they want me to say "the coriolis force opposes the direction of rotation" or "the coriolis force doesn't really exist, but the rifle barrel exerts in the direction of rotation"....
     
  5. Feb 21, 2010 #4

    Cleonis

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    Gold Member

    In general, in any problem if you use the Coriolis vector you must use the centrifugal vector too.

    Let me first discuss the idealized case of a frictionless barrel.
    The barrel is pointing in horizontal direction, spinning around a vertical axis.
    As the bullet slides down the barrel the force that the barrel exerts upon the bullets has two effects:
    - It makes the bullet co-spin with the barrel.
    - It causes the bullet to speed up down the barrel even more.

    (In fact, there is an abandoned design for http://mythbustersresults.com/episode93" [Broken]. Mythbusters Jamie and Adam recreated that design, to find out if it would work.)

    The symmetry of the problem suggests the following:
    - half the force goes into making the bullet co-spin
    - half the force goes into speeding the bullet up.

    However, in the real world there is friction between the bullet and the barrel, which complicates matters. If the bullet does not accelerate down the barrel then the force on the bullet is not as strong as the magnitude of the Coriolis vector, but half that.


    I wonder whether this physics is relevant in skeet shooting.
    I expect not. I think the bullet travels down the barrel so fast that there is no time for the bullet to disrupt the aim.
     
    Last edited by a moderator: May 4, 2017
  6. Feb 22, 2010 #5
    Hmm ok thanks for the information- that Mythbusters link was really interesting! I tried it again, using the equation for effective force in a rotating system:

    [tex]\vec{F}_eff=m\vec{a}-m\vec{R''}-m\vec{\omega'}\times\vec{r}-m\vec{\omega}(\vec{\omega}\times\vec{r})-2m\vec{\omega}\times\vec{v}[/tex]

    Where [tex]m\vec{a}[/tex] is the inertial force (0), [tex]m\vec{R''}[/tex] is due to acceleration of the rotating frame of reference (assume 0 as man is not accelerating linearly), [tex]m\vec{\omega'}\times\vec{r}[/tex] is due to the angular acceleration of the rotating frame (0 as man is rotating at constant rate), and the other two terms are the centrifugal and coriolis force respectively.

    Considering only directions, I ended up with the centrifugal force acting radially outwards (as expected, obviously) and the coriolis force acting opposite to the direction of rotation of conthe barrel (again, as expected). So I assume that's the best way to solve the second part of the problem?
     
    Last edited: Feb 22, 2010
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