How Does Spring Mass Affect Its Behavior?

[TSny]

I am not sure if I understood the problem statement correctly. Is the compressed position same as the initial relaxed position?

Yes, I probably should have used the phrase "initial relaxed position". However, a slinky is never in a truly relaxed state, it is still under tension even when it's back to its shortest length.

 

[Saitama]

As I said in #55, the different answers imply that the "collapsed slinky" must have some internal energy when it reaches the bottom. It is either oscillations (waves) or heat (which is also oscillations, just at the atomic level).

Yes, I probably should have used the phrase "initial relaxed position". However, a slinky is never in a truly relaxed state, it is still under tension even when it's back to it's shortest length.

Thanks a lot both of you! :)

 

[ehild]

Here are some amazing videos about slinky drop experiment.


https://www.youtube.com/watch?v=oKb2tCtpvNU&NR=1
https://www.youtube.com/watch?featu...v=JsytnJ_pSf8&annotation_id=annotation_314765

At the end, when the slinky is collapsed, its centre of mass has speed that corresponds to the gravitational potential energy of the fully extended slinky. The internal forces and motions do not count.

Again, that problem was for high-school students. Here is a high-school slinky model. The mass of the slinky is M , its spring constant is D. Imagine to cut the slinky to N equivalent parts, and make a chain of beads with masses m=M/N connected by equivalent massless springs, with spring constant k=DN. The unstretched length of all springs can be taken zero.

In equilibrium, The springs have lengths L1, L2, ...Ln, and tensions T1, T2...Tn. The positions of the beds are y1, y2, ...yn.

According to Hook's Law, T1=kL1...Tn=kLn.

From the equilibrium condition, T1=mg, T2-T1-mg=0 --> T2=2mg, .. Tn=nmg.

The length are: L1=mg/k, L2=2mg/k, Ln=nm/k.
The positions of the beads:
y1=0, y2=L1, y3=L1+L2,...$y_n=\sum_1^{n-1}L_i =\frac{mg}{k}\sum_1^{n-1}i=\frac{mg}{2k}n(n-1)$

The gravitational potential energy is $\sum_1^{N}mgy_n$, the elastic potential energy is $\sum_1^{N}0.5kL_n^2$.

The total length of the slinky is $L=\sum_1^{N}L_n$
Edit: [STRIKE]For the whole extended slinky, Mg=DL[/STRIKE].

Substitite m and k in the energy expression with m=M/N and k=ND=N (Mg/L).

ehild

 

[Saitama]

Hi ehild! :)

Here are some amazing videos about slinky drop experiment.


https://www.youtube.com/watch?v=oKb2tCtpvNU&NR=1
https://www.youtube.com/watch?featu...v=JsytnJ_pSf8&annotation_id=annotation_314765

Yes, those videos are truly amazing!

BTW, I just noticed that you and TSny linked to the websites and videos of the same person.

In equilibrium, The springs have lengths L1, L2, ...Ln, and tensions T1, T2...Tn. The positions of the beds are y1, y2, ...yn.

According to Hook's Law, T1=kL1...Tn=kLn.

From the equilibrium condition, T1=mg, T2-T1-mg=0 --> T2=2mg, .. Tn=nmg.

The length are: L1=mg/k, L2=2mg/k, Ln=nm/k.
The positions of the beads:
y1=0, y2=L1, y3=L1+L2,...$y_n=\sum_1^{n-1}L_i =\frac{mg}{k}\sum_1^{n-1}i=\frac{mg}{2k}n(n-1)$

The gravitational potential energy is $\sum_1^{N}mgy_n$, the elastic potential energy is $\sum_1^{N}0.5kL_n^2$.

The total length of the slinky is $L=\sum_1^{N}L_n$
For the whole extended slinky, Mg=DL.

Substitite m and k in the energy expression with m=M/N and k=ND=N (Mg/L).

ehild

I evaluate the summations. First, for the potential energy:
$$\begin{aligned}
\sum_{n=1}^{N}mgy_n & =\frac{Mg}{N}\sum_{n=1}^{N} \frac{mg}{2k}n(n-1)\\
&= \frac{MgL}{2N^3}\sum_{n=1}^{N} \left(n^2-n\right) \,\,\,\,\,\,\,\,\left(\because \frac{mg}{2k}=\frac{L}{2N^2}\right)\\
& = \frac{MgL}{2N^3}\left(\frac{N(N+1)(2N+1)}{6}-\frac{N(N+1)}{2}\right)\\
& = \frac{MgL}{2N^3}\frac{N(N+1)}{2}\frac{2(N-1)}{3}\\
& = \frac{MgL}{6}\left(1-\frac{1}{N^2}\right)\\
\end{aligned}$$
I could be wrong but the above doesn't look right to me. If I take the limit $N\rightarrow \infty$, I get $MgL/6$ instead of $MgL/3$.

I have one question, how do you get Mg=DL? I got a different result when voko helped me to calculate the spring constant in post #42.

 

[ehild]

I evaluate the summations. First, for the potential energy:...
because $\frac{mg}{2k}=\frac{L}{2N^2}$

Are you sure? $L=\sum_1^{N}L_n=\sum_1^{N}\frac{mg}{k}n=\frac{mg}{2k}N(N+1)$
$\frac{mg}{2k}=\frac{L}{(N+1)N}$

ehild

 

[Saitama]

Are you sure? $L=\sum_1^{N}L_n=\sum_1^{N}\frac{mg}{k}n=\frac{mg}{2k}N(N+1)$
$\frac{mg}{2k}=\frac{L}{(N+1)N}$

ehild

But then what's wrong with this:

As you stated before, $m=M/N$ and $k=N(Mg/L)$ i.e
$$\frac{mg}{2k}=\frac{MgL}{2N^2Mg}=\frac{L}{2N^2}$$


Can you please answer my question about the spring constant? :)

 

[ehild]

But then what's wrong with this:

As you stated before, $m=M/N$ and $k=N(Mg/L)$ i.e
$$\frac{mg}{2k}=\frac{MgL}{2N^2Mg}=\frac{L}{2N^2}$$


Can you please answer my question about the spring constant? :)

It must be wrong. L is the sum of length of the individual springs. D is not needed really.

ehild

 

[Saitama]

It must be wrong. L is the sum of length of the individual springs. D is not needed really.

ehild

Why is it wrong? I can't see anything wrong with using the relations you posted.

 

[ehild]

Why is it wrong? I can't see anything wrong with using the relations you posted.

The hanging slinky does not extend uniformly because its own weight. Voko explained it already. F=DL would be true for the horizontal slinky . If you pull the horizontal slinky with force F=Mg, it length changes by Mg/D.
You can hold the slinky by force F=Mg. The tension in the hanging slinky is zero at the bottom and Mg at the top, and changes linearly along the length. You can apply Hook's law with the average tension: LD=Mg/2.
Using the sum of the individual lengths you get the same L at the limit of infinite N.

Also it might be useful to see http://www.urch.com/forums/gre-physics/7891-massive-spring.html

ehild

 

[Saitama]

The hanging slinky does not extend uniformly because its own weight. Voko explained it already. F=DL would be true for the horizontal slinky . If you pull the horizontal slinky with force F=Mg, it length changes by Mg/D.
You can hold the slinky by force F=Mg. The tension in the hanging slinky is zero at the bottom and Mg at the top, and changes linearly along the length. You can apply Hook's law with the average tension: LD=Mg/2.
Using the sum of the individual lengths you get the same L at the limit of infinite N.

Also it might be useful to see http://www.urch.com/forums/gre-physics/7891-massive-spring.html

ehild

Thanks ehild! I reached the answer with your method too.

$$\begin{aligned}
\sum_{n=1}^N mgy_n & = \sum_{n=1}^N mg\frac{mg}{2k}n(n-1)\\
& = \frac{mgL}{N(N+1)}\sum_{n=1}^N \left(n^2-n\right)\\
& = \frac{mgL}{N(N+1)}\left(\frac{N(N+1)(2N+1)}{6}-\frac{N(N+1)}{2}\right)\\
& = \frac{mgL}{3}(N-1)\\
& = \frac{MgL}{3}\left(1-\frac{1}{N}\right) \\
\end{aligned}$$

$$\begin{aligned}
\sum_{n=1}^N \frac{1}{2} k L_n^2 & = \sum_{n=1}^N \frac{1}{2} k \frac{n^2m^2g^2}{k^2}\\
& = \frac{mgL}{N(N+1)}\sum_{n=1}^N n^2\\
& = \frac{mgL}{N(N+1)}\frac{N(N+1)(2N+1)}{6} \\
& = \frac{MgL}{6}\left(2+\frac{1}{N}\right) \\
\end{aligned}$$
Adding both the energies and taking the limit $N\rightarrow \infty$ gives the right answer.

Thanks!

 

[ehild]

Congratulation! You are very smart.

And it was pure high-school Physics...

ehild