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Coriolis Force on a Playground

  1. Jun 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Four children are playing toss on a merry-go-round which has a radius of r=2m. The merry-go-round turns counterclockwise and completes one revolution in 2 seconds. The child who has the ball wants to toss it to its right neighbour. It tosses the ball towards the center of the merry-go-roung.

    2. Relevant equations
    a) How fast does the child have to throw the ball?
    b) How fast does the merry-go-round have to turn to let the child catch its own ball, if it was thrown at the same speed as in a)?
    Solve the problem in the (x,y) plane of the carousel, from an outside, resting frame of reference, neglecting gravity and friction.

    3. The attempt at a solution
    I saw at the beginning of someone's solution that:
    [itex]x(t) = x_0 + v_x t + \frac{1}{2} a_x t^2[/itex]
    [itex]y(t) = -r + v_y t [/itex]
    I don't understand why it should be correct. However I would know if someone could directly explain me a nice approach to this problem and solve it to make it clear to me. Thanks a lot.
     
  2. jcsd
  3. Jun 30, 2015 #2

    jbriggs444

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    As I read the problem, we do not need to worry about the vertical dimension. We can work in two dimensions. The proposed equation for x(t) involving ##\frac{1}{2}a_xt^2## is not appropriate. We do not have a constant acceleration situation.

    You are told to solve the problem using the non-rotating frame of reference. In that frame the you should be able to compute launch velocity based on the speed with which the thrower tosses the ball. In the non-rotating frame, will the path taken by the thrown ball be straight or curved? Can you figure out where its path will intersect with the rim of the merry-go-round?
     
  4. Jun 30, 2015 #3
    I think the path of the ball from outside will appear only along the y-axis (back and forth).
    And we can imagine the angle between the two children to be given.
     
  5. Jun 30, 2015 #4

    jbriggs444

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    The ball is launched with a velocity relative to the moving child that is purely in the y direction.
     
  6. Jun 30, 2015 #5
    Yes, this is what I meant..
     
  7. Jun 30, 2015 #6

    jbriggs444

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    Given a launch speed and the fact that the launch angle relative to the throwing child is along the y axis, can you compute the speed and direction of the throw in the motionless reference frame?
     
  8. Jun 30, 2015 #7
    We said that the direction of the throw in the motionless reference frame is along the y-axis, right?
    The speed depends on where is the other child on the circle. But I don't get the relation between the speed and the other child's position..
     
  9. Jun 30, 2015 #8

    jbriggs444

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    It does simplify the problem greatly if you make the assumption that the direction is along the y axis of the motionless frame. My reading was that the direction is [initially] along the y axis of the rotating frame.
     
  10. Jun 30, 2015 #9
    Is there along y-direction some kind of acceleration to be considered?
     
  11. Jun 30, 2015 #10

    jbriggs444

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    From your original post [emphasis mine]
     
  12. Jun 30, 2015 #11
    I was not talking about that quantities... I mean, if from outside point of view the ball moves toward positive y-axis, and then toward negative y-axis, maybe some kind of acceleration which modifies our velocity could be considered. However, if the speed is constant, we will see the ball reaching a max value S along y-axis and then again toward zero. It means the ball travel a distance of 2S, right? So Speed = 2S / time..
     
  13. Jun 30, 2015 #12

    SammyS

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    No.
     
  14. Jun 30, 2015 #13

    jbriggs444

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    Where do you get the idea that the ball moves back and forth?

    What is S supposed to be? No S is defined in the problem. No S is defined anywhere in anything you have written.
     
  15. Jun 30, 2015 #14
    If the ball go toward the center of the circle, then it has to come back to reach the other child. That's why I thought the ball moves toward positive y-axis first, and then toward negative y-axis.
    S is not defined.. I just set it as the half "right distance" to walk in order to reach the other child.
    How would you solve the task?
     
  16. Jun 30, 2015 #15

    SammyS

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    As briggs pointed out, that "someone's" solution is wrong. -- no need to involve acceleration . This assumes that we don't need to consider the vertical direction. This allows the problem to be worked out in two dimensions. If the vertical direction is to be included, that can be worked out after the two dimensional solution is completed.

    Some suggestions:
    Pick a location in the resting frame to place the center of the merry-go-round. -- maybe at the origin?

    Pick a location in the resting frame at which the child tosses the ball. Since the discussion so far has suggested that the child attempts a toss along the y-axis, I suggest a location of (0, -2) in the resting frame. I also suggest that when the child is at this location, the rotating reference frame (fixed to the merry-go-round) is aligned with the resting frame.

    Now, for some questions:

    What is the velocity of the child (in the resting frame) when at location (0, -2) ?
     
  17. Jun 30, 2015 #16
    According to the given information the velocity of the child (which is tangential to the circle, due to its motion) should be [itex]2 \pi[/itex]
     
  18. Jun 30, 2015 #17

    SammyS

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    (in units of m/s.) Right?

    If the child attempts to toss the ball to the center, he/she will give the ball a velocity that only has a component in the y-direction (in the rotating frame). But, in the resting frame, what is the x-component of that velocity?
     
  19. Jun 30, 2015 #18
    Yes m/s.
    This point is not very clear to me... why should the velocity from the resting frame have a x-component?
     
  20. Jun 30, 2015 #19

    SammyS

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    In the rotating frame, neither the child, nor the ball have an x-component of velocity at the moment it's tossed.

    But, in the resting frame the child does an x-component of velocity at that moment. (You just told me what it is.) So the ball must have that same component with respect to the resting frame. Right?
     
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