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Spring Though Question

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WilliamWJ
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.


No. We're talking about the very bottom where there really is no weight (it seems counter-intuitive but I promise I'm right).

You're right about the lighter and heavier thing, but the tension is throughout the slinky and so it closes up as a whole faster when it is heavier. A heavy slinky and a light slinky have the same amount of weight at the bottom (none).
Reply 21
Avatar for aKarma
aKarma
11
WilliamWJ
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.


The idea of no mass refers to the boundary plane as it were along the lowest point of the spring. It's not referring to part of the spring having no mass.
Reply 22
Avatar for WilliamWJ
WilliamWJ
OP
9
Audrey Hepburn
No. We're talking about the very bottom where there really is no weight (it seems counter-intuitive but I promise I'm right).

You're right about the lighter and heavier thing, but the tension is throughout the slinky and so it closes up as a whole faster when it is heavier. A heavy slinky and a light slinky have the same amount of weight at the bottom (none).


mmmm I swear at Bham I was told what I said, I wrote down notes on it!

But seeing as you have a years undergrad experience I will beleive you :P

What University are you at?
WilliamWJ
mmmm I swear at Bham I was told what I said, I wrote down notes on it!

But seeing as you have a years undergrad experience I will beleive you :P

What University are you at?


They might have simplified it because they didn't want to bring non-rigid body dynamics into it or something, not sure why they explained it like that really. Every year they tell you one thing and then the next year they explain why it's wrong and tell you something else, yay Physics! aKarma explained it much better and more succinctly than I did though. I'm at Oxford myself.
Reply 24
Avatar for WilliamWJ
WilliamWJ
OP
9
Audrey Hepburn
They might have simplified it because they didn't want to bring non-rigid body dynamics into it or something, not sure why they explained it like that really. Every year they tell you one thing and then the next year they explain why it's wrong and tell you something else, yay Physics! aKarma explained it much better and more succinctly than I did though. I'm at Oxford myself.


ahh i dont like that! GCSE says they are right, A level says GCSE is wrong, University says A level is wrong. And then University says it doesnt know what is right, what they 'know' is just a very good approximation :P
Isn't a nicer way to look at it to just say from the perspective of the bottom of the slinky, it can't know that you've let go of the top until the signal has travelled all the way along it. So it must initially remain still.

This also neatly explains why we wouldn't normally observe stiffer objects doing this, since the speed of sound in them is much faster.
Its a famous question.

I did it in the lab and got the students to video it then watch the playback in slow motion.

The bottom stays still. The top moves down. When the slinky is no longer extended , the whole thing falls.
Reply 27
Avatar for Scipio90
Scipio90
teachercol
Its a famous question.

I did it in the lab and got the students to video it then watch the playback in slow motion.

The bottom stays still. The top moves down. When the slinky is no longer extended , the whole thing falls.


That's pretty cool. I would imagine it's less impressive for an ideal spring though, since it will continue to oscillate as it falls.

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