How Does Spring Mass Affect Its Behavior?

[Saitama]

Homework Statement

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?

Homework Equations

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!

 

[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?

 

[Saitama]

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

Even if the slinky is massive, consider what happens when it is stretched horizontally: what energy is stored in it?

The potential energy due to elongation of spring.

And what changes energy-wise when it is stretched vertically upward?

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.

 

[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?

 

[Saitama]

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?

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

 

[voko]

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

 

[Saitama]

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

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. :(

 

[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.

 

[Saitama]

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.

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

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

Please take your time. :)

 

[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?

 

[Saitama]

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?

$$E=\frac{\rho xg\,dx}{A\,dl}$$
$$\Rightarrow dl=\frac{\rho xg\,dx}{AE}$$
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!

 

[voko]

But how do I introduce the spring constant?

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.

 

[Saitama]

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.

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.

 

[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.

 

[Saitama]

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.

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.

 

[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.

 

[Saitama]

It is really a way to mark some material point in the slinky. $l(x)$ is the position of that material point in equilibrium.

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. :(

 

[voko]

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 $x+dx$.

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.

 

[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 $dx$.

 

[Saitama]

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 $dx$.

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.

 

[voko]

Pranav, I do suggest that you try the approach outlined in #19 instead. It eliminates the confusion about what $dx$ and $dl$ mean.

 

[Saitama]

Pranav, I do suggest that you try the approach outlined in #19 instead. It eliminates the confusion about what $dx$ and $dl$ mean.

In equilibrium, the part contained within $x$ and $x+dx$ is acted upon by three forces. The part above it exerts a force $F'$ in the upward direction. The part below exerts a force $F$ downward. The third force is its weight i.e
$$\rho g\,dx+F=F'$$
Is the above equation correct?

Is $F=\rho xg$ and $F'=mg-\rho (a-x)g$?

I haven't used $l(x)$ anywhere in my equations so I feel this is wrong.

 

[voko]

There is a principle in statics, that is sometimes referred to as the "principle of solidification". It states that any part of a system in equilibrium can be considered solid (and rigid). Applying this principle here, you can consider the "upper" and the "lower" parts of the slinky solid, while the "middle part" (that between $x$ and $x + dx$) is a tiny spring. Note also that the solid upper part is held in place by an external force. What is the FBD of the "solidified" system?

 

[Saitama]

Note also that the solid upper part is held in place by an external force. What is the FBD of the "solidified" system?

I have described the FBD in my previous post. Is it incorrect?

 

[voko]

Are there any forces due to the springy nature of the spring?

 

[Saitama]

Are there any forces due to the springy nature of the spring?

Doesn't that cancel out? I mean the spring force acts on both the points of middle part where it is attached with the upper and lower parts of slinky so it should cancel out.

 

[voko]

Doesn't that cancel out? I mean the spring force acts on both the points of middle part where it is attached with the upper and lower parts of slinky so it should cancel out.

Reasoning like this, you will conclude that the spring force has nothing to do with the equilibrium of the slinky. Which is obviously wrong.

Let's simplify this further. The upper and lower interfaces of the tiny spring with the upper and lower solidified parts are themselves immobile. That means the forces at each interface must be balanced.

 

[Saitama]

Reasoning like this, you will conclude that the spring force has nothing to do with the equilibrium of the slinky. Which is obviously wrong.

Let's simplify this further. The upper and lower interfaces of the tiny spring with the upper and lower solidified parts are themselves immobile. That means the forces at each interface must be balanced.

If F is the spring force on the lower solidified part, then at equilibrium, $F=\rho gx$.

Similarly, if F' is the spring force on the upper solidified part, then $F'=mg-\rho (a-x)g$, is this correct?

 

[voko]

Yes, that is correct. Observe also that $\rho a = m$, so $F' = -F$. So what is $F$ as the spring force?

 

[Saitama]

So what is $F$ as the spring force?

I don't understand what you are asking. Do you ask me to substitute $F$ and $F'$ in $\rho gdx+F=F'$?

 

[voko]

You need to express $F$ in terms of the elastic properties of the slinky. Remember, we are dealing with a tiny spring (small element of the slinky) of the original length $dx$ stretched to $dl$.

 

[Saitama]

You need to express $F$ in terms of the elastic properties of the slinky. Remember, we are dealing with a tiny spring (small element of the slinky) of the original length $dx$ stretched to $dl$.

$$dl=\frac{F\,dx}{AE}=\frac{\rho gx\,dx}{AE}$$
$$\Rightarrow l(x)=\frac{\rho gx^2}{2AE}$$

Is this correct?

 

[voko]

Yes and no. You will see the reason for "yes" later in this thread, and now I will address the reason for "no".

$F \over A$ is $\sigma$, the stress. $E$ is the elastic modulus, so you must have ${\sigma \over E} = \epsilon$, which is the strain. For a deformation from length $f$ into length $f + \Delta f$, we have, basically by definition, $\epsilon = {\Delta f \over f }$. Putting this all together, you must have $$ {F \over AE} = {\Delta f \over f} $$ so $$ F = {AE \Delta f \over f} $$ You have $$ F = {AE dl \over dx} $$ which must mean $\Delta f = dl$ and $f = dx$. But is this so? It is true that the original length of the small element is $dx$, so $f = dx$. But is $dl$ the extension $\Delta f$, or is it the entire length $f + \Delta f$?

 

[Saitama]

But is $dl$ the extension $\Delta f$, or is it the entire length $f + \Delta f$?

I am unable to think about a reason why $dl$ shouldn't be $\Delta f$. Since you commented on this, I guess $dl$ should br $f+\Delta f$ but this would mean that the initial length of the selected part is zero. This doesn't make sense.

 

[voko]

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.

 

[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!