# Spring with mass problem

1. Apr 17, 2014

### Pranav-Arora

1. The problem statement, all variables and given/known data
A slinky, a helical spring (used as a toy which is able to travel down a flight of steps) whose unstretched length is negligibly small, obeys Hooke's law with a good approximation, and it has a considerable elongation due to its own weight.
a) The slinky of mass m resting on the table is slowly raised at its top end until its lower end just raised from the table. The length of the spring at this position is L. How much work was done during lifting the spring?

b) If the slinky is released from this position, interestingly its lowermost turn does not move until the whole spring reaches its totally compressed length, (see the figure). What is the initial speed of the slinky at which it begins to fall right after reaching its totally compressed position?

2. Relevant equations

3. The attempt at a solution
I have never dealt with springs having mass so I am not sure how to start with the problems.

If the spring would have been massless, the work done is given by $\frac{1}{2}kL^2$ but it isn't and even if this formula applies to the current situation, I don't have the value of $k$. How to proceed?

Any help is appreciated. Thanks!

2. Apr 17, 2014

### voko

Even if the slinky is massive, consider what happens when it is stretched horizontally: what energy is stored in it? And what changes energy-wise when it is stretched vertically upward?

3. Apr 17, 2014

### Pranav-Arora

Hi voko! You were missed. Its great to see you again.

The potential energy due to elongation of spring.
In this case, I will have to consider the change in gravitational potential energy too. For part a), the change in gravitational potential energy is mgL/2. I don't see how to find the potential energy due to elongation.

4. Apr 17, 2014

### voko

Well, if the spring obeys Hooke's law as stated, then the potential energy of elongation must be equal to the work of the restoring force over the elongation. Right?

5. Apr 17, 2014

### Pranav-Arora

Yes but how do I find the restoring force? I don't have the spring constant. :(

6. Apr 17, 2014

### voko

Perhaps you can obtain some useful relationship among the mass, stiffness and length from the equilibrium condition?

7. Apr 17, 2014

### Pranav-Arora

I think I am missing something. The problem statement mentions nothing about the equilibrium position.

Let the force be $F$ which raises the upper end of spring. At any time,
$$F=mg-N$$
where $N$ is the normal reaction from ground.

When the spring is stretched by length L, then as per the problem statement, the lower end rises up which means $N=0$, hence $F=mg$ at that instant.

But I don't think this helps.

Do I analyse the forces acting on the uppermost end of the spring? If so, the forces acting on it are $F$ and $kL$ when the spring is about to leave the ground. Equating them gives $k=mg/L$. I think this is also wrong because it doesn't lead to the right answer. :(

8. Apr 17, 2014

### voko

The end state of "slowly raised at its top end until its lower end just raised from the table" is the equilibrium position. You are right assuming that in that position there is no reaction force from the table. Only the restoring force and gravity are at play.

P.S. I have to go now and will be back a few hours later.

9. Apr 17, 2014

### Pranav-Arora

I can't understand how the restoring force comes into play when I do the force balance on the complete spring.

10. Apr 17, 2014

### haruspex

Even though the relaxed spring has no length, you can have a notional distance x from the lower end of the relaxed spring. I.e. the spring from 0 to x has mass ρx.
If the modulus of elasticity is E, what is the extension of the section (x, x+dx) when the whole spring is clear of the table?

11. Apr 17, 2014

### Pranav-Arora

$$E=\frac{\rho xg\,dx}{A\,dl}$$
$$\Rightarrow dl=\frac{\rho xg\,dx}{AE}$$
But how do I introduce the spring constant?

haruspex, can you please look at the posts I recently posted in the "Ring rolling inside a cone" thread? I have got some doubts regarding the solution. Thanks!

12. Apr 17, 2014

### voko

You don't. The slinky does not extend uniformly in this case, so you have to look at "small elements", which means dealing with the modulus.

13. Apr 17, 2014

### Pranav-Arora

Is my expression for $dl$ correct then? If so, the potential energy stored in this small part is given by:
$$dE=\frac{1}{2}\frac{EA}{dx}dl^2$$
Do I have to integrate the above? But still it would depend on E which is not mentioned in the problem statement.

14. Apr 17, 2014

### voko

I would first integrate the equation you got in #11. That should you give a useful relationship among $L$, $\rho$ and $AE$, and it will give the distribution of mass in the equilibrium state.

15. Apr 17, 2014

### Pranav-Arora

Integrating the equation from #11 gives:
$$L=\frac{\rho g}{AE}\frac{L^2}{2}$$
$$\Rightarrow \rho=\frac{2AE}{gL}$$
The potential energy stored is given by integrating the following equation:
$$dU=\frac{1}{2}\frac{EA}{dx}dl^2=\frac{1}{2}\frac{\rho^2 g^2x^2 \,dx}{AE}$$
Substituting for AE gives:
$$dU=\frac{\rho^2 g^2x^2\,dx}{\rho gL}=\frac{\rho g x^2\,dx}{L}$$
$$U=\frac{\rho gL^2}{3}$$
Since $\rho=m/L$,
$$U=\frac{mgL}{3}$$
This is definitely wrong because if I add this with the gravitational potential energy, I get a wrong answer.

16. Apr 17, 2014

### voko

It looks like you are confusing $x$ with $l$. $x$ is, as haruspex said, a "notional distance". It is really a way to mark some material point in the slinky. $l(x)$ is the position of that material point in equilibrium. The part of the slinky between $0$ and $x$ has mass $\rho x$. The entire slinky has mass $m$, which gives you a condition for the maximum $x_1$: $\rho x_1 = m$. You can renormalise things so that $\rho = m$ and then $x_1 = 1$, so $x \in [0, 1]$. Of course, you can choose some other interval for $x$, but $[0, 1]$ seems simplest.

17. Apr 17, 2014

### Pranav-Arora

Now I am even more confused. I was using $x$ as the distance of a point on spring from the ground. $dl$ is the extension of the part of the spring contained within $x+dx$. So what I have done above is completely obsolete?

I don't see how to proceed and what's wrong with what I have done in my previous post. :(

18. Apr 17, 2014

### voko

If $x$ is the distance of a point on spring from the ground, then "the extension of the part of the spring contained within $x+dx$" has to be $x + dx$, because the initial length is zero (negligibly small as stated). So you get $dl = x + dx$, which obviously does not make any sense.

Instead, you should treat $x$ as something proportional to the "amount" of the slinky. Imagine that you paint some marks on the slinky and label them with values from the $[0, 1]$ interval. As the slinky extends, those marks move, but they still label the "amount" of the slinky.

$l(x)$ is the actual position of mark $x$ in the equilibrium state. #11 has a differential equation for $l(x)$. With an initial condition, you can determine $l(x)$, which gives you the distribution of the mass in the equilibrium state, as well as some other useful relationships.

19. Apr 18, 2014

### voko

It might be easier to approach this problem by assuming that the slinky's relaxed length is finite, say $a$, then $x$ may be the coordinate of a point in the relaxed string, in $[0, a]$, and $l(x)$ is the coordinate of the point in the equilibrium state. Then consider what the equilibrium state implies for the small element between $x$ and $dx$.

20. Apr 18, 2014

### Pranav-Arora

I am sorry for a typo in my previous post. I meant $dl$ is the extension of part contained within $x$ and $x+dx$ but I guess that doesn't make sense too.

You say my equation in #11 is correct but now I can't make sense of it. If $l$ is the position of a point on the spring in equilibrium, what is $dl$ supposed to represent then?

I wrote that equation in #11 considering the $dl$ as the extension of part contained within $x$ and $x+dx$ but then you say it doesn't make sense. :uhh: