Subfield: Mechanics

Zero-length springs and slinky coils

A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$.

<img_4535>

Figure 1: the relation between the force $F$ and the spring length $L$

A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as "slinky" may be (but not necessarily) such a ZLS.

Part A: Statics
Context question:
A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring?
Context answer:
$\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$


Context question:
A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$.
Context answer:
\boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$}


Extra Supplementary Reading Materials:

Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches.
Context question:
A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$.
Context answer:
\boxed{$H=\frac{L_{0}}{2}(\alpha+\frac{1}{\alpha})$}


Extra Supplementary Reading Materials:

Part B: Dynamics

Experiments show that when the spring is hung at rest and then released, it gradually contracts from the top, while the lower part remains stationary (see Figure 2). As time advances, the contracting part moves as a solid chunk and accumulates additional turns of the spring, while the stationary part becomes shorter. Every point on the spring begins to move only when the moving part reaches it. The bottom end of the spring starts moving only when the spring is fully collapsed and reaches its unstretched length $L_{0}$. After that, the contracted spring continues falling straight downwards, without tumbling, as a rigid body under the influence of gravity.
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Figure 2: Left: a sequence of pictures taken during the free fall of slinky. Right: the moving part I and the stationary part II during the free fall of the spring.

In the remaining parts of the question, you are asked to base your solution on this described model. You may neglect air resistance, but you are not allowed to neglect $L_{0}$.
Context question:
B.1 Calculate the time $t_{c}$ it takes from the moment the spring is released, until it fully collapses back to its minimal length $L_{0}$. Express your answer in terms of $L_{0}, g$ and $\alpha$.

Compute the numerical value of $t_{c}$ for a spring with $k=1.02 \mathrm{~N} / \mathrm{m}, L_{0}=0.055 \mathrm{~m}$ and $M=0.201 \mathrm{~kg}$, while taking $g$ to be $9.80 \mathrm{~m} / \mathrm{s}^{2}$.
Context answer:
\boxed{$t_{c}=\sqrt{\frac{L_{0}}{3 g \alpha}(1-\alpha)^{3}}$ , $t_{c}=0.245$}


Question: B.2 In this task $\ell$ is used to denote the coordinate of the boundary between parts I (in figure 2, the moving part) and II (the stationary part). At a certain moment, while a stationary part still exists its mass is $m(\ell)=\frac{\ell}{L_{0}} M$, and the moving part moves with uniform instantaneous velocity $v_{I}(\ell)$. Show that at this moment (while there exists a stationary part) the velocity of the moving part is $v_{I}(\ell)=$ $\sqrt{A \ell+B}$. Express the constants $A$ and $B$ in terms of $L_{0}, g$ and $\alpha$.

Solution:
The moving top section of the spring is pulled down by its own weight, $m_{t o p} g=M g \frac{\left(L_{0}-l\right)}{L_{0}}$ and also by the tension in the spring below, which is equal to the weight $M g l / L_{0}$ of the stationary section of the spring. Thus, the moving top section experiences a constant force $F=$ $M g$ throughout its whole fall. Another way to see that, is that a total force of $M g$ is exerted on the spring, but only the moving part experiences it. Let's calculate the position of the center of mass at equilibrium of the upper part, i.e., all points with $l^{\prime}>l$ for some $l>l_{0}$. From part A, the position of a small portion $\Delta l^{\prime}$ with coordinate $l^{\prime}$ is: $H\left(l^{\prime}\right)=\frac{l^{\prime 2}}{2 l_{0}}+\frac{l_{0}}{2}$ and the center of mass of this part is:

$$
\begin{aligned}
H_{\text {cm-upper }-i} & =\frac{L_{0}}{M\left(L_{0}-l\right)} \int_{l}^{L_{0}} H\left(l^{\prime}\right) d m=\frac{L_{0}}{M\left(L_{0}-l\right)} \int_{l}^{L_{0}}\left(\frac{l^{\prime 2}}{2 l_{0}}+\frac{l_{0}}{2}\right) d m \\
& =\frac{L_{0}}{M\left(L_{0}-l\right)} \int_{l}^{L_{0}}\left(\frac{l^{\prime 2}}{2 l_{0}}+\frac{l_{0}}{2}\right) \frac{M d l^{\prime}}{L_{0}}=\frac{1}{\left(L_{0}-l\right)} \int_{l}^{L_{0}}\left(\frac{l^{\prime 2}}{2 l_{0}}+\frac{l_{0}}{2}\right) d l^{\prime} \\
& =\frac{1}{\left(L_{0}-l\right)}\left[\frac{l^{\prime 3}}{6 l_{0}}+\frac{l_{0} l^{\prime}}{2}\right]_{l}^{L_{-} 0}=\frac{L_{0}^{2}+L_{0} l+l^{2}}{6 l_{0}}+\frac{l_{0}}{2}
\end{aligned}
$$

The position of the upper part of $\mathrm{CM}$ when it contracts to a length $L_{0}-l$ is $H_{c m-u p p e r-f}=$ $\frac{l^{2}}{2 l_{0}}+\frac{l_{0}}{2}+\frac{1}{2}\left(L_{0}-l\right)$. The change in the CM during the contraction process is: $\Delta H_{c m-u p p e r}=$ $H_{\text {cm-upper }-i}-H_{\text {cm-upper }-f}=\frac{L_{0}^{2}+L_{0} l-2 l^{2}}{6 l_{0}}-\frac{1}{2}\left(L_{0}-l\right)=\frac{\left(L_{0}-l\right)\left(L_{0}+2 l\right)}{6 l_{0}}-\frac{1}{2}\left(L_{0}-l\right)$.

The acceleration of the $\mathrm{CM}$ of the upper part is $a_{C M}=\frac{F L_{0}}{M\left(L_{0}-l\right)}=\frac{g L_{0}}{L_{0}-l}$. From the work energy theorem we get the equation $v_{\text {upper-f }}^{2}=2 a_{C M} \Delta H_{\text {cm-upper }}$, hence

$$
\begin{gathered}
v_{\text {upper }-f}^{2}=2 \frac{g L_{0}}{L_{0}-l}\left[\frac{\left(L_{0}-l\right)\left(L_{0}+2 l\right)}{6 \alpha L_{0}}-\frac{1}{2}\left(L_{0}-l\right)\right]=2 g\left[\frac{L_{0}+2 l}{6 \alpha}-\frac{1}{2} L_{0}\right] \\
=\frac{2 g}{3 \alpha} l+\left(\frac{1}{3 \alpha}-1\right) g L_{0}
\end{gathered}
$$

Therefore, $A=\frac{2 g}{3 \alpha}$ and $B=\left(\frac{1}{3 \alpha}-1\right) g L_{0}$.

Note that for $l=L_{0}$, we have $v_{\text {upper }-f}^{2}=L_{0} g \frac{1-\alpha}{\alpha}$ and for $l=l_{0}=\alpha L_{0}$, we get $v_{u p p e r-f}^{2}=$ $L_{0} g \frac{1-\alpha}{3 \alpha}$, hence, the moment we release the spring its velocity is finite (not zero, the meaning is that it accumulate this velocity in time that is much shorter than the contracting time $t_{c}$ ) and it decreases to $\frac{1}{\sqrt{3}}$ of the initial value when $l=l_{0}$.

Final answer: $A=\frac{2 g}{3 \alpha}$ , $B=(\frac{1}{3 \alpha}-1) g L_{0}$