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    Oh ********, 'Spring ThoughT Question' haha.


    I was at Birmingham Uni last week and they set us a few problems to work out. This was one that I found rather straightforward/simple to think about, but not many people managed to get it right.

    Imagine you are holding a spring/slinky from its highest point. It is hanging freely under its own weight, and is motionless.
    You let go of the spring. What does the lowest point of the spring do, the instant you let go.

    Does it
    a) Move up
    b) Move down
    c) Stay still

    and why?

    PS do not go and get a slinky to try it out!
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    I would say it moves up slightly. According to Hooke's law the force on a stretched spring is toward the equilibrium position. As the center of mass of the slinky falls, this force will cause the slinky to shorten slightly. Once the spring force is overcome by the force of gravity then it will fall as one object. I could be wrong. It's early, I'm tired and brain-dead.
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    ^

    i guess it retracks first before it starts to fall
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    I'd say stay still, potential force would equal gravity temporarily blablabla. But then again I'm not great at elasticity.
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    OK does nobody else know?! Am I asking for too much :P
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    Surely it depends how tight the spring is?
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    (Original post by The Muon)
    Surely it depends how tight the spring is?
    What do you mean how tight? Lots of tension?
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    (Original post by WilliamWJ)
    Oh ********, 'Spring ThoughT Question' haha


    I was at Birmingham Uni last week and they set us a few problems to work out. This was one that I found rather straightforward/simple to think about, but not many people managed to get it right.

    Imagine you are holding a spring/slinky from its highest point. It is hanging freely under its own weight, and is motionless.
    You let go of the spring. What does the lowest point of the spring do, the instant you let go.

    Does it
    a) Move up
    b) Move down
    c) Stay still

    and why?

    PS do not go and get a slinky to try it out!
    It will move up .according to newtons 1st law " a object will continue to move in a straight line unless acted upon by an external resultant force"
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    (Original post by rbnphlp)
    It will move up .according to newtons 1st law " a object will continue to move in a straight line unless acted upon by an external resultant force"
    I'll leave you to think a little more about that.
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    (Original post by WilliamWJ)
    OK does nobody else know?! Am I asking for too much :P
    Why would anyone else answer? - it's obviously not b, so someone's already got it right - either me or everyone else...

    anyways I'm pretty sure it's me - if it was going to move upwards then it would have moved upwards before you release it because the elastic tension would be greater than the gravity keeping it down. The bottom of the slinky would remain stationary until the rest of it has caught up because there is no resultant force acting on it yet and so nothing inducing it to move. It's in a state of rest until the slinky closes up. (the force is zero at the bottom because the tension is due to the underlying weight, of which there is none)
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    (Original post by Audrey Hepburn)
    Why would anyone else answer? - it's obviously not b, so someone's already got it right - either me or everyone else...

    anyways I'm pretty sure it's me - if it was going to move upwards then it would have moved upwards before you release it because the elastic tension would be greater than the gravity keeping it down. The bottom of the slinky would remain stationary until the rest of it has caught up because there is no resultant force acting on it yet and so nothing inducing it to move. It's in a state of rest until the slinky closes up. (the force is zero at the bottom because the tension is due to the underlying weight, of which there is none)
    Tell me why its not b
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    (Original post by WilliamWJ)
    Tell me why its not b
    1. Because it's c.
    2. Because it's the intuitive answer, and if there's one thing that I've leaned in my first year of a physics degree it's that it is NEVER the intuitive answer.
    3. It would mean that the spring is acting as a rigid body, which it's not.
    4. I'm pretty sure my previous posts explains...
    5. Because I say so.
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    Assuming no air resistance it would surely be b)
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    (Original post by aKarma)
    Assuming no air resistance it would surely be b)
    What about the tension in the spring, surely that would pull the bottom up?
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    (Original post by WilliamWJ)
    What about the tension in the spring, surely that would pull the bottom up?
    Well at the very instant of the drop the bottom is stationary where forces are in equi but then it falls. It's only temporarily stationary for a very short time in the same way that technically a pencil would react in the same way (but the time the bottom is still would be infinitesimal). I suppose if you used a large spring the pause would be noticeable
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    (Original post by Audrey Hepburn)
    1. Because it's c.
    2. Because it's the intuitive answer, and if there's one thing that I've leaned in my first year of a physics degree it's that it is NEVER the intuitive answer.
    3. It would mean that the spring is acting as a rigid body, which it's not.
    4. I'm pretty sure my previous posts explains...
    5. Because I say so.
    It was the intuitive answer for me

    It is c yes. If its stationary when its let go, the tension in the spring will try and pull the bottom up, but g tries to pull it down with the same force. The forces cancel out and the bottom remains where it is.

    Then think of it as a longitual wave, the bottom will only move down once the energry transferred from the top of the spring reaches the bottom. It will stay still untill then.
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    (Original post by WilliamWJ)
    It was the intuitive answer for me

    It is c yes. If its stationary when its let go, the tension in the spring will try and pull the bottom up, but g tries to pull it down with the same force. The forces cancel out and the bottom remains where it is.
    No.

    The hanging Slinky is in tension due to its self-weight. Because the weight of the Slinky is distributed, so is the tension. At the top, with the entire weight of the Slinky suspended below, the tension equals the weight. At the bottom there is no underslung weight, and so the tension there is zero.

    The bottom of the Slinky is not, as you seem to suggest, in equilibrium between a downward force due to its weight and an upward force due to tension. The bottom of the Slinky experiences no forces at all, either before or after the moment of release. Then according to newton's law blablabla. In this case, a state of rest applies. There is nothing going on to cause the bottom of the Slinky to move.

    You were right but for the wrong reasons I'm afraid.
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    (Original post by Audrey Hepburn)
    The hanging Slinky is in tension due to its self-weight. Because the weight of the Slinky is distributed, so is the tension. At the top, with the entire weight of the Slinky suspended below, the tension equals the weight. At the bottom there is no underslung weight, and so the tension there is zero.

    The bottom of the Slinky is not, as you seem to suggest, in equilibrium between a downward force due to its weight and an upward force due to tension. The bottom of the Slinky experiences no forces at all, either before or after the moment of release. Then according to newton's law blablabla. In this case, a state of rest applies. There is nothing going on to cause the bottom of the Slinky to move.

    You were right but for the wrong reasons I'm afraid.
    And the exact same will happen to any object iirc. It's just too small to perceive for most things
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    (Original post by aKarma)
    And the exact same will happen to any object iirc. It's just too small to perceive for most things
    Yep indeedy. (unless you work assuming that rigid bodies exist - which they don't)

    That's some sweet physics skills right there :five:
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    (Original post by Audrey Hepburn)
    No.

    The hanging Slinky is in tension due to its self-weight. Because the weight of the Slinky is distributed, so is the tension. At the top, with the entire weight of the Slinky suspended below, the tension equals the weight. At the bottom there is no underslung weight, and so the tension there is zero.

    The bottom of the Slinky is not, as you seem to suggest, in equilibrium between a downward force due to its weight and an upward force due to tension. The bottom of the Slinky experiences no forces at all, either before or after the moment of release. Then according to newton's law blablabla. In this case, a state of rest applies. There is nothing going on to cause the bottom of the Slinky to move.

    You were right but for the wrong reasons I'm afraid.
    The bottom part of the slinky has mass. The whole slinky is connected together. The bottom is being held where it is, due to the tension (because of the weight) of the slinky as a whole.

    If the slinky was lighter, there would be less tension, and the bottom of the slinky would stay still for longer.
    If the slinky was heavier, there would be more tension, and the bottom would stay still for a shorter amount of time.
 
 
 
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