Faster Than Gravity

The standard model (developed by Cayley many years ago and much reproduced in textbooks) for this problem has the right-hand section of the chain falling freely under gravity, with no tension in that section of the chain. The bottom portion of that right-hand section is successively brought to rest by an inelastic collision (being prevented from falling further through being attached to the stationary left-hand section of the chain). This model is not energy-conserving. Recent experiments have shown that an energy-conserving model is the correct one. In the energy-conserving model, tension is continuous in the chain, and particularly so between the left-hand and the right-hand sections. The tension that exists in the left-hand section of the chain to bring successive portions of the right-hand section of the chain to rest then acts downwards on the right-hand section of the chain, giving the right-hand section an acceleration greater than that of gravity. These clips give a practical demonstration of this effect.

This paper by Wong and Yasui summarises the results. Unfortunately, the original 1989 paper on this topic, by Calkin and March, is only available online on payment of ready money.

At the moment when $B$ is a distance $x$ below the ceiling, the left-hand segment of chain has length $\tfrac12(L+x)$ , and hence has mass $\tfrac{L+x}{2L}m$ , and its centre of mass is a distance $\tfrac14(L+x)$ below the ceiling. On the other hand, the right-hand segment of chain has length $\tfrac12(L-x)$ , and hence has mass $\tfrac{L-x}{2L}m$ , and its centre of mass is a distance $x + \tfrac14(L-x) = \tfrac14(L+3x)$ below the ceiling. Thus the gravitational potential energy of the system is $-\tfrac{L+x}{2L}mg \times\tfrac14(L+x) - \tfrac{L-x}{2L}mg \times \tfrac14(L + 3x) \; = \; -\tfrac{mg}{4L}(L^2 + 2Lx - x^2)$ The left-hand segment of the chain is stationary, while the right-hand segment of chain has speed $\dot{x}$ , and so the kinetic energy of the system is $\tfrac12 \tfrac{L-x}{2L} m \dot{x}^2 \; = \; \tfrac{m}{4L}(L-x)\dot{x}^2$ Conservation of energy tells us that $\tfrac{m}{4L}(L-x)\dot{x}^2 - \tfrac{mg}{4L}(L^2 + 2Lx - x^2) \; = \; -\tfrac14mgL$ (since, initially, $x=\dot{x}=0$ ), and hence $\begin{array}{rcl} \displaystyle (L-x)\dot{x}^2 & = & \displaystyle gx(2L-x) \\ \displaystyle \dot{x} & = & \displaystyle \sqrt{\frac{gx(2L-x)}{L-x}} \end{array}$ Thus the time for the chain to fall is (substituting $x = L\sin^2\theta$ ): $\begin{array}{rcl} T & = & \displaystyle \int_0^L \sqrt{\frac{L-x}{gx(2L-x)}}\,dx \; = \; \sqrt{\frac{L}{g}}\int_0^{\frac12\pi} \frac{\cos\theta \times 2\sin\theta \cos\theta\,d\theta}{\sqrt{2 - \sin^2\theta}\sin\theta} \\ & = & \displaystyle 2\sqrt{\frac{L}{g}} \int_0^{\frac12\pi} \frac{\cos^2\theta}{\sqrt{1 + \cos^2\theta}}\,d\theta \\ & = & \displaystyle 2\sqrt{\frac{L}{g}}\left[ \sqrt{2}E\big(\tfrac{1}{\sqrt{2}}\big) - \tfrac{1}{\sqrt{2}}K\big(\tfrac{1}{\sqrt{2}}\big)\right] \\ & = & \displaystyle \sqrt{\frac{2L}{g}}\left[2E\big(\tfrac{1}{\sqrt{2}}\big) - K\big(\tfrac{1}{\sqrt{2}}\big)\right] \end{array}$ where $K$ and $E$ are the complete elliptic integrals of the first and second kinds respectively: $K(k) \; = \; \int_0^{\frac12\pi} \frac{1}{\sqrt{1-k^2\sin^2\theta}}\,d\theta \qquad \qquad E(k) \; = \; \int_0^{\frac12\pi}\sqrt{1 - k^2\sin^2\theta}\,d\theta$ (see Gradshteyn and Rhyzik for this last evaluation). But since $K\big(\tfrac{1}{\sqrt{2}}\big) \; = \; \frac{8 \pi^{\frac32}}{\Gamma(-{\frac14})^2} \qquad, \qquad E\big(\tfrac{1}{\sqrt{2}}\big) \; = \; \frac{4 \pi^{\frac32}}{\Gamma(-\frac14)^2} + \frac{\Gamma(\frac34)^2}{2 \sqrt\pi}$ we deduce that $T \; = \; \sqrt{\frac{2L}{g}} \frac{\Gamma(\frac34)^2}{\sqrt{\pi}}$ making the answer $3+4+2+2 = \boxed{11}$ .

This was so much fun! Huge respect to Mr. Hennings for posting such an amazing problem!

There is already a really nice solution written by Mr. Hennings, but I had a lot of fun solving this problem, and I want to "catalogue" my solution for future reference (in my own words), if that makes any sense!

Anyways, let's jump right in. We have to construct an expression that allows us to find the gravitational potential energy when the end of the string has fallen some given distance. When the right side of the string has shortened by some amount $\ell$ , this length is "transferred to the left side of the string system. Thus, the length of the right side at some time $t$ is given by $\frac{L}{2} \ - \ \ell(t)$ and the left side is $\frac{L}{2} \ + \ \ell$ . We can then say that to find the gravitational potential at some instant in time, we can integrate over each point lying on the longer side the rope (the left side), multiply that number by $2$ , to get the energy for both sides, then subtract the energy from $0$ to $2\ell$ , in order to take out the missing component of the rope. Thus, we have:

$\text{d}m \ = \ \frac{m}{L} \ \text{d}L$

$U(\ell) \ = \ 2\displaystyle\int_{0}^{\frac{L}{2} \ + \ \ell} \ -\frac{xmg}{L} \ \text{d}x \ - \ \displaystyle\int_{0}^{2\ell} \ -\frac{xmg}{L} \ \text{d}x \ = \ \frac{2mg\ell^2}{L} \ - \ \frac{mg\Big( \frac{L}{2} \ + \ \ell \Big)^2}{L} \ = \ \frac{mg}{L} \Big( \ell^2 \ - \ L\ell \ - \ \frac{L^2}{4} \Big)$

The kinetic energy is a whole other story. We must consider the fact that only the right side of the chain is moving

$K(\ell) \ = \ \frac{1}{2} \frac{\frac{L}{2} \ - \ \ell}{L} v^2$

Where the velocity is given with respect to how the distance between the top of the chain and the ceiling is changing in time, so $2\ell$ (I was a bit unclear on this , but it was the only way my solution wouldn't be off by a multiplicative factor of $2$ ). Anyways, putting this together:

$U_0 \ = \ U(\ell) \ + \ K(\ell) \ \Rightarrow \ -\frac{mgL}{4} \ = \ \frac{mg}{L} \Big( \ell^2 \ - \ L\ell \ - \ \frac{L^2}{4} \Big) \ + \ \frac{1}{2} \frac{\frac{L}{2} \ - \ \ell}{L} v^2 \ \Rightarrow \ \frac{g\ell^2}{L} \ - \ g\ell \ + \ \frac{\frac{L}{2} \ - \ \ell}{2L} v^2 \ = \ 0$

$v \ = \ \sqrt{\frac{g\ell \ - \ \frac{g\ell^2}{L}}{\frac{\frac{L}{2} \ - \ \ell}{2L}}} \ = \ \sqrt{2L\frac{g\ell \ - \ \frac{g\ell^2}{L}}{\frac{L}{2} \ - \ \ell}}$

$\frac{1}{v} \ = \ \sqrt{\frac{1}{2L}} \sqrt{ \frac{ \frac{L}{2} \ - \ \ell} { g\ell \ - \ \frac{g\ell^2}{L} }}$

Now we integrate over $\frac{1}{v}$ to find the total time. Since $v$ is a function of $a \ = \ 2\ell$ , we have to do a change of variables:

$T \ = \ \displaystyle\int_{0}^{L} \ \frac{1}{v(a)} \text{da} \Rightarrow \ \ell \ = \ \frac{1}{2}a \ \Rightarrow \ \text{d}\ell \ = \ \frac{1}{2} \text{d}a$

And so we get:

$T \ = \ \displaystyle\int_{0}^{\frac{L}{2}} \ \frac{2}{v(\ell)} \text{d}\ell \ = \ 2 \displaystyle\int_{0}^{\frac{L}{2}} \ \sqrt{\frac{1}{2L}} \sqrt{ \frac{ \frac{L}{2} \ - \ \ell} { g\ell \ - \ \frac{g\ell^2}{L} }} \text{d}\ell$

Now, we will do another change of variables. Let $\omega \ = \ \frac{2}{L} \ell$ . We now have:

$T \ = \ \sqrt{\frac{L}{2g}} \displaystyle\int_{0}^{1} \ \sqrt{ \frac{ \frac{L}{2} \ - \ \frac{L}{2} \omega} { \frac{L}{2} \omega \ - \ \frac{L}{4} \omega^2 }} \text{d}\omega \ =\ \sqrt{\frac{L}{2g}} \displaystyle\int_{0}^{1} \ \sqrt{ \frac{ 1 \ - \ \omega} { \omega \ - \ \frac{1}{2} \omega^2 }} \text{d}\omega \ = \ \sqrt{\frac{L}{g}} \displaystyle\int_{0}^{1} \ \sqrt{ \frac{ 1 \ - \ \omega} { 2\omega \ - \ \omega^2 }} \text{d}\omega$

$\Rightarrow \ \sqrt{\frac{L}{g}} \displaystyle\int_{0}^{1} \ \omega^{-1/2} (1 \ - \ \omega)^{1/2} (2 \ - \ \omega )^{-1/2} \text{d}\omega$

This integral looked VERY similar to the Beta function, but I couldn't exactly figure out how to express it in terms of the Beta (or Beta's) so I used Wolfram Alpha to simplify. It gave me:

$\Rightarrow \ \sqrt{\frac{L}{g}} \displaystyle\int_{0}^{1} \ \omega^{-1/2} (1 \ - \ \omega)^{1/2} (2 \ - \ \omega )^{-1/2} \text{d}\omega \ = \ \sqrt{\frac{L}{g}} \ \frac{2 \sqrt{\pi} \Gamma\Big( \frac{3}{4} \Big)}{\Gamma \Big( \frac{1}{4} \Big)}$

By the Legendere Duplication Formula:

$\Gamma \Big( \frac{1}{4} \Big) \ = \ \frac{\pi \sqrt{2}}{\Gamma \Big( \frac{3}{4} \Big)} \ \Rightarrow \ T \ = \ \sqrt{\frac{2L}{g}} \ \frac{\Gamma\Big( \frac{3}{4} \Big)^2}{\sqrt{\pi}}$

And then we have:

$A \ + \ B \ + \ C \ + \ D \ = \ 3 \ + \ 4 \ + \ 2 \ + \ 2 \ = \ 11$