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Accelerate a ball bearing through a 90 degree turn

  1. Dec 15, 2013 #1
    I am looking to accelerate a ball bearing through a 90 degree turn using an impulse from an electromagnet wrapped around a bent tube. What I want to know is what is the minimum radius turn you could use, obviously if the radius of the turn is too small the ball will simply impact the wall at the turn and not make it around the bend or there will be excessive deceleration due to friction at the bend. There must be a minimum ball radius to turn radius ratio
     
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  3. Dec 15, 2013 #2

    sophiecentaur

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    If you look up this link, it will tell you the force needed to keep a mass moving on a given radius.
    There is no 'answer' to the question about the actual force unless the radius of the turn is specified. In the end, you need an unspecified force for an infinitesimal time to produce the required impulse - which is why collisions and deflections like your example, usually involve momentum change and Impulse rather than Force. I'm saying there is no "minimum" radius.
     
  4. Dec 16, 2013 #3

    This problem is similar to curved barrel guns designed to shoot around corners, you would assume that if the curve at the bend was too sharp the bullet would simply blow the barrel in two?
     
  5. Dec 16, 2013 #4

    sophiecentaur

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    If the radius were too small then a (cylindrical) bullet could jam but I can't think of a reason why the pressure should change inside the barrel, as the bullet turns the corner. There is no fundamental reason why the bullet would slow up at the bend, imo. There would, of course, be a force against the outside of the barrel - the value, given by mv2/r. Plus, there would be some more friction as the bullet dragged against the outside of the curve.
     
  6. Dec 16, 2013 #5
    I just tried it with a bendy straw and a plastic bead, I put the bead in and blew it like a pea shooter. I found if I made the bend at the corner too tight it impacted just like I thought it would, when the bend was more gradual it went around the corner.
     
  7. Dec 16, 2013 #6

    sophiecentaur

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    Bending a straw like that would not ensure that the bore was constant all the way along. A very flawed experiment, I'm afraid. If you actually 'bored' out a tube in a serious experiment then there is no reason to think that there would be any slowing up of the bullet.
     
  8. Dec 17, 2013 #7

    So if I had a cannon with a sudden 90 degree bend (radius equal to ball diameter) you would fully expect that upon firing that cannon the ball would make a sudden 90 degree turn and exit the other end without loss in velocity?

    You wouldn't expect the momentum of the cannon ball to keep it on a straight path and rupture the wall at the bend?

    And if I had a cannon with a gradual bend and a cannon with a sudden bend you would fully expect the ball velocity exiting both cannons to be the same?
     
    Last edited: Dec 17, 2013
  9. Dec 17, 2013 #8

    sophiecentaur

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    When you say "A sudden 90 degree bend", you would need to define it in more detail. What sort of radius did you have in mind? It's obviously true that, if you cut the barrel and made a mitre joint, the ball would stop dead or bounce back. But I don't think you mean that. You imply that you mean the ball is constrained to move on a circular path. If we assume that there is no significant friction then there is no mechanism for the ball to lose energy. No energy loss means the speed would't change.

    I have now to bounce the question back to you and ask what mechanism you had in mind that would cause a loss of energy. It isn't sufficient that you have an intuitive feel that it 'must' lose some energy.

    Clearly, there will be a radius, below which the thing can no longer be treated as 'ideal' because the forces would be larges enough to distort the materials involved and the friction would become significant. But we must assume this is an 'ideal' case until we god into more detail about the extra parameters involved in a practical situation.
     
  10. Dec 17, 2013 #9
    I consider a sudden bend being a radius equal to ball diameter as I stated, and yes my intuitive feeling is that it will rupture the wall at the bend due to the momentum imparted to the ball. Momentum is the tendency to keep an object moving in the same direction unless acted upon by an external force. In this case the external force is the contact with the wall at the bend. If the curve is gradual then that force is exerted on the ball over a longer time duration but if the curve is small that force is exerted on the ball suddenly over a shorter time period. I fully expect that the ball if imparted with enough impulse velocity will simply rupture the wall at the bend. If the ball is not imparted with enough velocity to rupture the wall I expect it to lose most its momentum at the bend and exit with a fraction of the velocity were it a straight cannon.
     
  11. Dec 17, 2013 #10

    jbriggs444

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    As long as the radius of the bend is less than the radius of the cannonball and as long as certain other idealizations are assumed then yes, I would expect the exit velocities to be the same. The idealizations that spring to mind are:

    The frictional force is directly proportional to normal force over the range of normal forces that will be encountered.

    The barrel is perfectly rigid and is held motionless so that it dissipates no energy due to the short duration and localized forces that it experiences. It can still dissipate heat energy from friction.

    The ball is perfectly rigid so that it dissipates no energy due to the short duration and localized forces that it experiences.

    The ball fits tightly in the barrel. It does not bounce from side to side.

    We are not considering the pressure of the propellant continuing to pushing the ball down the barrel. The pressure in a gas flowing down a curved tube is a different problem.

    If the ball is considered to roll down the barrel under the effect of unbalanced frictional force, the resulting roll rate is assumed to be independent of turn radius. [I suspect that the correctness of this assumption follows from the others but am not certain]

    In practice, my suspicion is that increasingly large instantaneous forces encountered as the turn radius approaches the ball radius will challenge the correctness of these assumptions and that energy will be dissipated in vibration in the barrel and the cannonball or in the permanent deformation/destruction of either or both.

    There is an analogy that could be drawn between this scenario and a series of polarized sunglass lenses, each rotated slightly from the one before. With just two lenses at 90 degrees you get no light to pass. But with a series of lenses, each rotated only slightly but with a total rotation of 90 degrees between first and last, you do get light to pass. In the limit with a series of sufficiently thin, completely transparent, polarizing lenses, I believe that you get 100% transmission [of properly polarized input light] regardless of twist rate per centimeter.
     
  12. Dec 17, 2013 #11

    Good so we agree :smile:

    So back to my original question, how do I find what is the minimum radius which will not result in dissipation of energy at the bend?
     
  13. Dec 17, 2013 #12

    sophiecentaur

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    We have to be careful to define the level of practicality we describe this sort of problem with. If we are talking about an ideal case then start introducing small degrees of 'realism' the answers will change.

    Momentum has a more precise definition than that. Classically, it is Mass times Velocity (ignoring relativistic bolt-ons, for now). There is a principle that Momentum is Conserved, meaning that the total momentum in an isolated system is the same before and after any collisions or other interactions. When the ball is deflected by the barrel, there will be a change of the ball's momentum and a corresponding equal and opposite change in the momentum of the gun (gun plus shooter and possibly to the Earth too, if the gun is held rigidly. If the gun / shooter etc are massive enough, the resulting kinetic energy, imparted to them is vanishingly small because their velocity is so near zero. (Any movement in a practical case will affect the result of our thought experiment)
    The change in the ball's momentum is described as an Impulse and will be a force times a time (not a uniform force all the time round the curve, of course). The total value of this impulse will be the same for all curvatures - except in the case you have now introduced. If, as you now suggest, the barrel has finite strength, there will be a time for deflection, below which the force is enough to break the barrel. But you would have to specify a lot more information about it and, so far, you have been talking ideally.,
     
  14. Dec 17, 2013 #13

    sophiecentaur

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    Apart from the obvious requirement that the ball must fit all the way round (dictated by geometry and not by Physics) there is no limit. There are practical issues like friction against the sides but, as you haven't specified them, we have to assume friction is near enough zero.
    If you can specify a force that the barrel will stand then the radius can be calculated.
     
  15. Dec 17, 2013 #14

    Nugatory

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    If the wall ruptures, we can always try again with the same curve and propellant charge, but a thicker stronger cannon barrel that can handle the forces. No matter how great the forces are, we can always imagine an even stronger cannon barrel. However, as the forces become larger, they will start to deform the cannon ball and the curve in the wall of the barrel so that the geometry changes as the ball is forced through the curve. At this point the situation becomes very difficult to analyze, but one way or another the ball will jam in the barrel instead of making the sharp turn that a perfect sphere in a perfectly curved curved runway would make.

    The forces involved in your ball bearing experiment obviously aren't great enough to distort anything, but you still have a problem with a tight radius: the inside radius is less than the outside, so some parts of the ball bearing will have to change direction more quickly than others. There's no way of doing that smoothly without transferring some energy from the ball to the tube, slowing or stopping the ball.

    As an aside... You can always get a sharp 90-degree turn just by firing the ball at a piece of armor plate at a 45-degree angle. Although that's not going to be useful in your application, it's worth thinking about the physics of that situation for a moment -it's an interesting extreme case in which the direction change happens without any mechanical jamming.
     
  16. Dec 17, 2013 #15
    I assumed it was obvious the barrel would have finite strength as to assume otherwise would be to assume it had infinite strength.

    You say the force experienced by the barrel at the bend is force X time, don't you mean force divided by time ie shorter the time period ^ the force on the wall at the bend (which equates to smaller radius turn = larger force on wall at bend)
     
  17. Dec 17, 2013 #16
    Is it not true that if the curve was more gradual the force exerted on the barrel would be less and therefore the same material would experience less deformation at the bend were it a small radius turn?
     
  18. Dec 17, 2013 #17
    I am not concerned with the ball rupturing the wall at the bend, what I am concerned with is the force acted upon the ball at the turn slowing it down. The track is going to be made from copper or aluminum and once the ball (12mm dia) turns the 90 degree bend it will return to the electromagnet to be accelerated again. This will happen repeatedly with increasing velocity (only if the ball would not experience massive deceleration at the bend). I want to use the smallest radius bend to maximize centripetal acceleration of the ball at the bend.
     
    Last edited: Dec 17, 2013
  19. Dec 17, 2013 #18

    sophiecentaur

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    Of course. So you need to specify the max force that the barrel can handle before you can know what the minimum radius is, for a given speed. Otherwise you have to assume the limit is at (or an infinitessimal amount less) than the radius of the ball.
    As usual, you need to know all the variables except the one you are trying to find out.

    There is no force on the ball "acting to slow it down", if you are in an ideal theoretical situation (which is what I assumed, initially). The only force is at right angles to its motion (basics of circular motion) so the is no force 'against' the instantaneous direction of motion. You cannot treat things intuitively when the theory is so rock solidly correct. Newton sussed this out a long time ago and it explains why the planets keep going round and round without a motor.

    If you are trying to design an accelerator, you are in the practical world and friction will be an important factor which you would need to quantify by tests and measurements. This makes things very different and you need to avoid sharp turns, where friction will be relevant. But there is no 'minimum'; friction effects will just be worse as the radius reduces.
     
  20. Dec 17, 2013 #19
    Intuition is probably the wrong word, experience would be a better word. I knew from seeing similar situations in the past what could be reasonably expected. In simulating this in my mind several times I think I now understand how the ball loses energy at the bend and thus experiences deceleration. Its not friction as is commonly thought ie sliding of materials at the surface. It loses energy because it imparts a short but intense mechanical impulse to the wall at the sharp bend which will be experienced by an observer as sound radiating as the ball passes the bend, thus kinetic energy is lost and the ball decelerates. This will happen even in a theoretical situation with a ball and track of high strength, the mechanical impulse will still be transferred at the bend.
     
  21. Dec 17, 2013 #20

    sophiecentaur

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    You asked a theoretical question and I gave you the right theory in reply. As you have never given any help about actual orders of magnitude for the variables, the theory is all we can go on. I repeat that the only mechanisms for loss are entirely practical ones. Your experience is very likely correct but your system is far too complicated for a simple answer, based on ideal conditions.
    Have you an idea about what velocities, masses, etc. are involved here? I think, if you want an answer to this, you will have to experiment and measure things.

    I see from your last sentence that you choose to disagree 'even with the theory'. If you do that, you will not learn anything from this exercise. I may have been wasting my time.
     
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