How Does Spring Mass Affect Its Behavior?

[Saitama]

The entire original length $dx$ becomes the entire extended length $dl$. The extension $\Delta f$ is the difference between the entire extended length and the entire original length.

Ah, I think I see why is it so. I somehow missed the word "relaxed" in #19 and got myself confused between $l$ and $x$ again. I am sorry for being careless.

So
$$dl-dx=\frac{\rho gx\,dx}{AE} \Rightarrow l(x)=x+\frac{\rho gx^2}{2AE}+C$$
I am a little confused about what should be the constant.

If I use the conditions $x=0$ and $l(0)=0$, I get $C=0$ but if I use the conditions $x=a$ and $l(a)=L$, I get a different $C$.

 

[voko]

No, you do not get a different $C$ when you use the conditions for the other end. Keep in mind that the only constants specified in the problem are $m$ and $L$ (and $g$ is specified by the nature), so the condition for $x = a$ is a constraint on the other constants.

 

[Saitama]

No, you do not get a different $C$ when you use the conditions for the other end. Keep in mind that the only constants specified in the problem are $m$ and $L$ (and $g$ is specified by the nature), so the condition for $x = a$ is a constraint on the other constants.

Do you mean to say $x=0$ and $l(0)=0$ is an invalid constraint?

 

[voko]

$l(0) = 0$ is most certainly correct. My point is that there is no contradiction between that and $l(a) = L$.

 

[Saitama]

$l(0) = 0$ is most certainly correct. My point is that there is no contradiction between that and $l(a) = L$.

From $l(0)=0$, I have $C=0$.

From $l(a)=L$, I have
$$C=L-a-\frac{\rho ga^2}{2AE}$$
I can replace $\rho a$ with $m$ but that still doesn't make it zero.

 

[voko]

$l(0) = 0$ means $C = 0$. The other condition gives you $$ L = {\rho g a^2 \over 2AE } + a $$ What is $\rho a$ and what is $AE \over a$?

 

[Saitama]

$l(0) = 0$ means $C = 0$. The other condition gives you $$ L = {\rho g a^2 \over 2AE } + a $$ What is $\rho a$ and what is $AE \over a$?

I think I understand, the constant $C$ should be same for both the conditions so this gives us a relation between AE/a and the specified constants.

$\rho a=m$, hence
$$L=\frac{mga}{2AE}+a \Rightarrow \frac{AE}{a}=\frac{mg}{2(L-a)}$$
Is this correct? What should be my next step?

 

[voko]

$AE \over a$ has a particular physical meaning. What is it? What happens as $a \to 0$?

 

[Saitama]

$AE \over a$ has a particular physical meaning.

Spring constant?

What happens as $a \to 0$?

It is equal to mg/(2L).

 

[voko]

Very well. What is the tensile energy stored in the tiny spring (=small element)? What is the tensile energy in the entire slinky? What is its potential energy?

 

[Saitama]

Very well. What is the tensile energy stored in the tiny spring (=small element)?

$$dU=\frac{1}{2}\frac{AE}{dx}(dl-dx)^2 \Rightarrow dU=\frac{1}{2}\frac{\rho^2 g^2x^2\,dx}{AE}$$

What is the tensile energy in the entire slinky? What is its potential energy?

Integrating the expression for dU and substituting $\rho a=m$ and $AE/a=mg/(2L)$ gives:
$$U=\frac{mgL}{3}$$
which I suppose is incorrect.

 

[voko]

That seems correct for the tensile potential energy. What about its gravitational potential energy?

 

[Saitama]

What about its gravitational potential energy?

Won't that be simply mgL/2? Adding it to U gives me an incorrect answer.

 

[voko]

$mgL/2$ assumes the slinky is distributed uniformly in equilibrium. $l(x)$ indicates quite plainly it is not.

 

[Saitama]

$mgL/2$ assumes the slinky is distributed uniformly in equilibrium. $l(x)$ indicates quite plainly it is not.

Yes, I am sorry about that.

The selected $dx$ part is at height $l(x)$ from the ground. Hence, its gravitational potential energy is:
$$dE=\rho g\left(x+\frac{\rho gx^2}{2AE}\right)\,dx$$
$$\Rightarrow E=\frac{m}{a}g\left(\frac{a^2}{2}+\frac{\rho ga^3}{6AE}\right)$$
Substituting $\rho=m/a$ and $AE/a=mg/(2L)$,
$$E=\frac{mgL}{3}$$
(in the limit $a\rightarrow 0$)
Adding this with the tensile potential energy gives me the correct answer.

For part b), I think I need to equate the sum of the above two energies with kinetic energy when slinky reaches its original relaxed length. This gives me the correct answer.

Thanks a lot voko for all the help.

I never thought dealing with springs having mass would be so difficult. I could have never solved it without your help. Thank you again!

 

[voko]

Very well. A couple of observations.

The gravitational potential energy in the equilibrium state has the term $\frac 1 2 mga$, which becomes zero in the limit. Note, however, that without taking the limit, this term is the gravitational potential energy of the slinky in its natural state, so the change in the gravitational potential energy and thus the change in the full potential energy are independent of the natural length.

Secondly, the full energy integral, which you never wrote, is $$ \int \limits_0^a \left( {AE \over 2}(y' - 1)^2 + \rho g y \right) dx $$ where I replaced $l$ with $y$ for readability. In the equilibrium state, this integral is minimal, which, using calculus of variations (Euler-Lagrange equation) leads to the differential equation $$ AEy'' = \rho g $$ Compare this with the equation for $l(x)$ that you obtained by balancing forces in infinitesimal elements.

 

[Saitama]

In the equilibrium state, this integral is minimal, which, using calculus of variations (Euler-Lagrange equation) leads to the differential equation $$ AEy'' = \rho g $$

I have never used Euler-Lagrange equations before. I don't see how you get that differential equation.

Compare this with the equation for $l(x)$ that you obtained by balancing forces in infinitesimal elements.

$l''(x)$ comes out to be same as the D.E you wrote.

 

[voko]

I have never used Euler-Lagrange equations before.

You will learn one day. You may recall this problem then :)

 

[TSny]

For part b), I think I need to equate the sum of the above two energies with kinetic energy when slinky reaches its original relaxed length. This gives me the correct answer.

I'm surprised that this leads to the correct answer for the speed of the slinky at total collapse. Do you mind stating the correct answer for the problem? It could be that I am not thinking correctly about part (b) of the problem. I get a final speed of $\small \sqrt{2gL/3}$.

If you want to see some interesting videos and animations of the falling slinky, see
http://www.physics.usyd.edu.au/~wheat/slinky/

This site also has a link to a journal article that analyzes the falling slinky.

 

[voko]

I'm surprised that this leads to the correct answer for the speed of the slinky at total collapse. Do you mind stating the correct answer for the problem? It could be that I am not thinking correctly about part (b) of the problem. I get a final speed of $\small \sqrt{2gL/3}$.

Your comment opens an interesting can of worms. Your result is based on the centre of mass analysis. Conserving energy, as seems to be required by the problem, gives a different answer. Which means energy is lost in the process, which cannot happen with a Hookean spring. The problem did say the slinky was Hookean, but apparently the requirement that it reaches the fully collapsed state, where it behaves essentially as a rigid body, at the bottom of the fall is not compatible with that.

 

[TSny]

Yes, it seems to me that the speed of the spring when it is fully collapsed would be the speed of the center of mass. The center of mass speed is determined by just the gravitational force, not the internal elastic forces. If mechanical energy is conserved, then I think there would be elastic waves set up such that it would not attain a fully collapsed state. To be fully collapsed requires conversion of mechanical energy into internal heat. At least that's how I see it.

 

[Saitama]

I'm surprised that this leads to the correct answer for the speed of the slinky at total collapse. Do you mind stating the correct answer for the problem? It could be that I am not thinking correctly about part (b) of the problem. I get a final speed of $\small \sqrt{2gL/3}$.

If you want to see some interesting videos and animations of the falling slinky, see
http://www.physics.usyd.edu.au/~wheat/slinky/

This site also has a link to a journal article that analyzes the falling slinky.

Thank you TSny for bringing this up, you are indeed correct. I am sorry for being hasty.

Here are the correct answers: http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P4615&l=en

It looks like I misread part b). How do I find the compressed position? Do I have to do a similar analysis as I did for the case when the slinky is stretched to length L?

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

 

[voko]

I understand "fully compressed" is the same as "initially relaxed", of negligible length as stated. #56 gives more than a hint on how to proceed.

 

[Saitama]

I understand "fully compressed" is the same as "initially relaxed", of negligible length as stated. #56 gives more than a hint on how to proceed.

As per TSny's post #56, gravitational potential energy is equal to final kinetic energy i.e
$$\frac{mgL}{3}=\frac{1}{2}mv^2 \Rightarrow v=\sqrt{\frac{2}{3}gL}$$

TSny said that there would be elastic waves set up if there is no loss of the spring potential energy into heat. I find this little difficult to understand. This reasoning isn't very much obvious to me, am I missing something? How am I supposed to know that there would be waves if no loss of energy?

 

[voko]

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).

 

[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!