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I What is the final velocity of a ball rolling through a tube?

  1. Dec 22, 2017 #1
    I am making a physics simulator and this problem is tricky. You have two freely moving objects with known masses and velocities in a frictionless environment. The first object is a ball and the second is a pipe. They are moving toward each other. I need to know what formulas are used to simulate the changes in their vectors. I attached an image for this problem.
     

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  3. Dec 22, 2017 #2

    berkeman

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    You are writing a physics simulator, but are not familiar with the kinematic equations of motion? That doesn't sound like a good match on the surface of it.

    What is your background in math and physics? Have you had calculus yet? Calculus-based physics?
     
  4. Dec 22, 2017 #3
    This is no ordinary physics simulation. It is for a very specific action that I could not find any information on. I have already taken undergrad courses in math and physics. I have already solved and programmed the initial components to this simulation and understand the math behind kinetic energy transfer, angular momentum and just about everything else I need for this problem. However, I have not posted how I am thinking about solving this problem because I don't want to taint anyone's thought process. Even though I do not have this one solved (yet), I want to make sure my thinking is headed the right direction. Also for verification purposes. Care to take a crack at it? The graphic I attached should contain all the necessary information.
     
  5. Dec 30, 2017 #4
    [​IMG]
    I am making a simulator in java to show what happens when a ball rounds an 180 degree arc. This sounds easy at first but the ball and arc can move freely in space with zero friction. My simulator breaks down each frame by a specific distance traveled then each time-frame by the length of time it takes for a specific action. I need to find the formulas to determine the final velocities and travel directions for the ball and the arc per frame (distance). I attached a picture with the specifics. I know at first glance this should be a simple two-dimensional elastic collision problem.

    https://drive.google.com/open?id=1cExtz4_7ZpVE55HPSk-5ZpR8xKE8n1Rf

    [​IMG]
     
  6. Dec 30, 2017 #5

    berkeman

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    (Two threads on the same question merged...)
     
  7. Dec 30, 2017 #6

    kuruman

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    I assume the ball is given some initial velocity, goes around inside the tube and comes out the other end. How do you calculate the new velocity vector given the old velocity vector at each step in your simulation? The conservation laws apply here but they can only give you the final quantities.

    To get the final quantities using a step by step method, you need to figure out how much velocity to add to the old velocity in order to get the new velocity at each step. The amount of velocity ##\Delta \vec{v}## that you add per step is the acceleration of the ball. Whatever ##\Delta \vec{v}## you add to the ball must be accompanied by another ##\Delta \vec{v}## added to the tube. Of course, this addition must be consistent with momentum conservation in two dimensions at each step by ensuring that ##m_{ball}\Delta \vec{v}_{ball}+m_{tube}\Delta \vec{v}_{tube}=0##.

    Don't forget that that the acceleration of the ball is centripetal, i.e. it points towards the center of the arc, wherever that is at each step, and has magnitude ##v_{ball}^2/R##.
    This is a neat problem!
     
  8. Dec 30, 2017 #7
    The ball and arc both have opposing initial velocities. If the ball where to only hit the apex of the arc then this problem would be much easier. Each frame of animation in my simulation deals with an exact distance that is traveled between frames. So a total travel would be 'X' frames of 'S' distances with time between frames being the only thing that changes based on final velocity. I simplified this part of the simulation to deal with length segments around the arc instead of a continuous curve. I have been testing a set of formulas for the changes in X and Y velocities of the ball and the arc at each step of the simulation. I haven't coded them yet though, just put them in a spreadsheet. I haven't quite solved this one yet.
     
  9. Dec 30, 2017 #8

    kuruman

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    It's still easy if all you care about is the final velocities of the masses. The only thing you have to keep in mind is that the center of mass travels in a straight line which means that the ball moves gradually to the right while the arc moves gradually to the left. In the end though the final velocities would be the same as in the case where the ball hits the apex of the arc.
    Are you saying that you have preset the sequential positions of the ball or did I misunderstand you?
     
  10. Dec 31, 2017 #9
    When the ball is at the apex then the centrifugal force of the ball is in direct opposition to the movement of the arc PY=mv⋅SIN(Θ), PX=mv⋅COS(Θ). Unless I am wrong, the total kinetic energy transferable would be KT=K/√2 since the two are traveling toward each other.
     
  11. Dec 31, 2017 #10

    Nidum

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    pipeandball-png.png

    Surely the condition shown in far rhs image can never be ? The arc shaped pipe will have been tipping over in the clockwise direction continuously since the ball first entered it ?
     
  12. Dec 31, 2017 #11
    I suppose it would depend on the difference in kinetic energy. If the pipe was extremely massive then the ball would round the bend and not lose much of any speed. If the pipe had almost no mass then the ball would never round the bend and send pipe spinning off into oblivion. At any rate, the formulas are the important part. I just added numbers to give people something to work with.
     
  13. Dec 31, 2017 #12
    Yes, the positions are the constant in the simulation. It is only the time between frames that changes. I was thinking about reversing that but then I am afraid I would lose accuracy. Although it is a real pain using a variable sleep timer.
     
  14. Dec 31, 2017 #13

    kuruman

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    That's a big problem! What do you think the observer of the simulation would make of it? Our brains are programmed to think that time flows linearly, frame by frame, each frame being separated by the other by the same amount Δt. Suppose you drive on the highway and the driver in front of you hits the brakes but his brake lights are defective and don't light up. You can deduce that he is slowing down because the distance between cars in successive time intervals gets shorter. If your brain interpreted the visual information according to your scheme, the distance between you and the car ahead would stay constant until you crash into it. It is safe to surmise that the proto-humans who interpreted motion in terms of equal space intervals got caught and were eaten by predators before they could pass their genes down to us.
    What is better? A simulation that looks unrealistic but is accurate, or a simulation that looks realistic but is inaccurate? If the two were sold side by side in a store, which one would you pay money for?
    Yes it is. I hope I have persuaded you to abandon it.
     
  15. Dec 31, 2017 #14
    I am just not looking forward to rewriting the whole thing. Although there is a benefit to using a fixed timer, I could better handle the thread that the simulation runs in.
     
  16. Dec 31, 2017 #15

    kuruman

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    OK.
     
  17. Jan 1, 2018 #16

    Khashishi

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    Often, when constraint forces are very difficult to calculate, it's easier to use a Lagrangian approach with generalized coordinates. Consider a parameter ##\theta## which is the angle of the ball relative to the middle of the bottom face of the pipe block while the ball is in the pipe are in contact. It goes from 0 to pi.

    Energy, momentum, and angular momentum are conserved. Assuming there is no deformation, all the energy is kinetic energy, so you have:
    ##T=(1/2)*m_b*v_b^2 + (1/2)*m_p*v_{cp}^2 + (1/2)*I_p*w_p^2##
    The second and third terms are the translational and angular kinetic energy of the pipe block. We neglect the angular kinetic energy of the ball.
    ##v_{cp}## is the velocity of the center of mass of the pipe block.
    The system momentum is ##p=m_b v_b + m_p v_p##

    Let ##\alpha## be the orientation of the pipe block (which varies in time). Let ##(x_p, y_p)## be the position of the center of the bottom face of the pipe block.
    Let ##(x_1,y_1)## be the position of the center of mass of the pipe block relative to ##(x_p, y_p)##. So the center of mass of the pipe block is
    ##(x_{cp},y_{cp}) = (x_p + x_1 \cos(\alpha) + y_1 \sin(\alpha), y_p + y_1 \cos(\alpha) - x_1 \sin(\alpha))##
    The center of mass of the system is
    ##pt/(m_b+m_p) \equiv C_m##
    (moving slightly upward)
    so we can write the coordinates of the ball in terms of the generalized coordinates using the center of mass.
    ##m_b x_b + m_p (x_p + x_1 \cos(\alpha) + y_1 \sin(\alpha)) = C_{mx}##
    ##m_b y_b + m_p (y_p + y_1 \cos(\alpha) - x_1 \sin(\alpha)) = C_{my}##

    We need to write the angular velocity of the block
    ##\omega_p = \dot{\alpha}##
    Now, we need to write the Lagrangian in terms of ##\theta, \alpha, x_p, y_p##
    This part is messy, but I think the equations are all above.
    Using the Euler-Lagrange equation, you should be able to calculate the equations of motion for ##\alpha, \theta, x_1, y_1##. Then you know how to get ##x_b, y_b##.
     
  18. Jan 1, 2018 #17

    kuruman

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    I agree. The equations of motion I got are not pretty. I figured that they can be converted to difference equations and solved using a step-by-step procedure that is good enough for simulation purposes.
    How is this neglect justified? Admittedly, ##m_p## is two orders of magnitude larger than ##m_b##, but shouldn't one introduce approximations after the equations of motion are obtained? I learned by experience that if I make an approximation too early, I end up throwing out the baby with the bath water.
     
  19. Jan 2, 2018 #18

    A.T.

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    If there is no friction between pipe and ball, and the pipe cannot rotate, then this is equivalent to an elastic collision:
    https://en.wikipedia.org/wiki/Elastic_collision
     
    Last edited: Jan 2, 2018
  20. Jan 8, 2018 #19
    I thought it would be an elastic collision except for the fact that it is continuous with time being a factor. Is not an elastic collision void of the time variable?
     
  21. Jan 9, 2018 #20

    A.T.

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    Any real collision has some non-zero duration, but for many applications only the final velocities are of interest, not what happens during the collision.
     
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