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Coin Drop in Motion

  1. Nov 29, 2012 #1
    Hey. I've never formally taken a physics class but when I read the blurb on Brian Greene's The Fabric of the Cosmos it intrigued me and I bought it. I started reading it the other day and there's one particular idea that I'm confused by, and which I hope you all could clarify. Greene references how Galileo said that if you are in a boat and you hold a coin over your foot and drop it, it will land on your foot whether or not the boat is in motion. When I first read this it made sense because from both your frame of reference and the coins the only thing in motion is the water (when the boat is moving, that is). However, I later thought, when you let go of the coin, in its frame of reference the water is no longer moving towards the boat, but the boat is moving towards the water, correct? Obviously the boat couldn't move forward very much in the time it takes the penny to fall, but if you were to perform the same experiment dropping the penny from, say, 100 feet in the air, would it still land on your foot, or land behind you? Thanks!
     
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  3. Nov 29, 2012 #2

    Bandersnatch

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    Hi nsang.

    As you carry the coin in your hand, you impart it with the same velocity w/r to water that you have(and which is the same as the boat's). When you drop the coin, that velocity doesn't go away(Newton's 1st law of motion), so it keeps moving at the same constant velocity w/r to water as you do. There's just no force acting on it that would try to change its velocity in the horizontal plane. The only force acting is gravity, and it's only changing the coin's velocity vertically.

    Of course, that would be only true in vacuum.

    The atmosphere(assuming no wind) has got the same speed w/r to the boat as the water, so when you drop the coin it is slowed down by air friction as it falls, landing further back than it would without air drag.
    Unless it's windy in such a fashion that the wind exactly follows the boat, in which case we're back to the previous situation, where the coin lands exactly beneath the hand that dropped it.
     
  4. Nov 29, 2012 #3
    That makes sense, thanks!
     
  5. Nov 29, 2012 #4

    K^2

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    Easily verified on the airplane, by the way. An airliner can easily be traveling at up to 300m/s with respect to air outside. So in the time it takes something to drop even a foot, the airplane will travel more than it's own length. You can test this without getting out of your seat, and not have any question on whether the object continues traveling with an airplane or falls straight down with respect to earth.
     
  6. Nov 30, 2012 #5
    Do a quick search on inertial vs non-inertial frames of reference for more information on this. To clarify, your example holds true if the boat (or plane) is not accelerating. If it is travelling at constant velocity, and you drop the coin, it will appear to fall straight down.... it really doesn't since it has a forward velocity component when you release it, the same forward velocity that the boat or plane has, so it could be viewed as them cancelling out....the coin moves the same distance forward as the plane, so it appears to fall straight down. Try this riding in a car (don't do it if you're the driver!).... drop the coin while accelerating, and then drop the coin when travelling at constant speed.... use a vertical element for visual reference..... I've done this on a bus using the vertical grab bars to show my kids. Drop the coin next to the bar so it's easy to see if it appears to fall straight down, or not.
     
  7. Nov 30, 2012 #6
    Thanks for all of the responses.

    And just to be sure, this is because if the boat is accelerating your horizontal velocity will exceed that of the coins meaning there is now relative motion between the coin and you, causing it in the case of acceleration to appear to move backwards, correct?
     
  8. Nov 30, 2012 #7

    Bandersnatch

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