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Ball dropping and frame of reference

  1. Jul 16, 2010 #1
    Consider I am in the train with a ball traveling at normal speed of a train. I drop the ball, it follows a straight path of descent and falls in front of my feet.
    For an observer on the platform, the ball follows a parabolic path and falls in front of my feet.
    (Please go to this http://www.phys.unsw.edu.au/einsteinlight/jw/module1_Galileo_and_Newton.htm" [Broken] and play the flash. It is similar to the setup I have mentioned above)

    The doubt -
    The moment I, standing on the train, release the ball, the ball is in free space. And the only force acting on it is gravitational. Then why doesnt it fall straight with respect to the Earth [as that is the one which is pulling it] from the position in space it is released? That is, when the ball is released, it should fall few meters behind my feet as I would have moved with the train; whereas the ball was in free space.

    I know that the above description is a fact. But I just don't know why. Please do explain the reason.

    I think it is the same thing when we get down from a bus which is motion with respect to the road - you experience a force which pushes you for few meters and then you come to a halt on the road. Am I correct?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 16, 2010 #2
    You are correct at the end of your post. The ball when released has forward momentum. It keeps that as it falls.

    Think of it this way: if your train is in space away from gravity and you let go of the ball does it stop or stay next to your hand?

    Adding gravity acting perpendicularly to your line of motion won't change that.

    What may change is the measured value of the gravity in the train. Is it still 9.8 m/ss?
     
  4. Jul 16, 2010 #3
    Thanks for answering.
    But my question was with respect to the train reference frame and you answered for platform frame.
    From the platform's observer [Op] point of view, the train, the observer [Ot] inside that with the ball, all have relative velocity and thus a momentum. So because of which, the Op would observe that the ball would have momentum and gradual loose it, thus having a parabolic path.

    But I am asking the question from Ot point of view. If Ot releases the ball, the ball is in the free space.
    And from this point, I am completely confused. because, if it is free space, just being released from the clutches of Ot, then it should have the same parabolic path [of gradual losing of momentum] as it is NO more in contact with a body (either train or Ot) which is traveling with some velocity.
    But at the same time, I know that the ball doesn't have relative velocity with the Ot or train.
    I cannot pinpoint which point in the above thought flow is the wrong one. Please clarify.

    Compare the above scenario of leaving the ball inside the train to one exactly opposite - of leaving the ball outside the train. That is, Ot is standing near the edge of bogie [or a hand outside the window of the train] and releases the ball. Then in this case, the ball will have a parabolic path.
    What is the difference between the two points in space - one being inside the train where the ball was released and the other being outside where the ball was released?

    And regarding your statement "if your train is in space away from gravity and you let go of the ball does it stop or stay next to your hand?" - my answer would be that it gradually comes to halt. Am i correct in this?
     
  5. Jul 17, 2010 #4
    In the train frame, every body moves with same velocity in forward direction. As the relative velocity among the man, train, ball is zero before releasing, they appear to be at rest w.r.t each other. When the ball is just released, it continues to possess the forward velocity, which doesn't sees any resistance in the horizontal direction. Due to this undiminishing horizontal velocity, it is still at relative rest in the horizontal direction. But, due to force of gravity it falls down.

    (Obsrvr moving fwd with const v + ball moving fwd with const v + falling downward with const g) = falling downward with const g i.e first two terms cancel each other's effect leaving the last.

    In the Earth frame, when the ball is seen to be released, it possesses the same forward velocity as in the above case, which also doesn't sees any resistance in the horizontal direction. As the observer is at rest here, the ball is not at relative rest in the horizontal direction- it carries on forward with the velocity it possessed when it was in the train-man's hand.

    (Obsrvr at rest + ball moving fwd with const v + falling downward with const g) = falling downward with parabolic trajectory.

    There's no gradual loss of horizontal momentum as you are thinking of. The key concept is the independence between horizontal and vertical motions. The superposition of an horizontal arithmetic sequence with an independent vertical geometric sequence produces a parabola - Thats the beauty!
     
  6. Jul 17, 2010 #5
    Thanks for the answer superkan619.
    It helped!
     
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