Okay, let's do the maths. First, a reminder of
exactly what ImaginalDisc is asking.
Does each chain slip over the edge at the same rate...
I thought so too, but I considered that when chain B's first link hits the groud...
Perhaps I don't understand the math well enough, but I thought that the last link of each chain would fall over the edge at the same rate of speed.
I asked at what rate of speed the last link would slip off the edge of the precipice at.
Hokay?
Now then, let R be the total length of the chain, let t be time, let r be the distance the chain has has fallen as a function of t (and so r', r'' are the velocity and acceleration as functions of time). Let T be the time that ImaginalDisc is asking about: the time at which the last bit of chain slips over the edge. And let g be Galileo's constant.
Now, unless you're Art Vandelay, and if you can still remember how to draw a force diagram, the force acting on the chain is rgD, where D is a constant in kilogram / meters (the "density" of the chain).
The mass to be shifted by this force is RD.
The acceleration of the chain is therefore given by:
r'' = rgD / RD = rg / R
Hence (for
some c and d) we have
r = ce
kt + de
-kt
where k is the square root of g / R.
Likewise, we have r' = cke
kt - dke
-kt
c and d are determined by our boundary conditions. For starters, we want the velocity at time 0 to be 0. Inserting t = 0 into our formula for velocity, we get
r'(0) = (c - d) (ke
0 + ke
0)
But of course we require r'(0) [i.e.the velocity at time 0] to be
zero, so this means
c = d
This means that our formula for distance can be rewritten as
r = c(e
kt + e
-kt)
So, how do we determine c?
We can't without extending the question slightly.
The thing is that we'd like our boundary condition for distance to be that at time 0 there is
no chain hanging over the edge. But if we do that, the system will never move.
In order for questions about the dynamics of the system to have any meaning at all, if we start with r'(0)=0 we must assign a small but non-zero value to r(0). And whatever it is, c must be twice that, as we can see by substituting t = 0 into the equation for r.
Now, this tells us something interesting about the dynamics of the situation: if we halve c, the distance travelled in n seconds (for any n < T) is also halved. We can make ImaginalDisc's time T as large as we choose by making c as small as we choose. As c tends to zero, T tends to infinity.
Fortunately, ImaginalDisc doesn't want us to calculate, T, but r'(T). Now here we have some hope. We know that the dynamics of the situation fix r(T), since r(T) = R. And it fixes r''(T), since r''(t) = g. Hence we may at least hope that as c tends to zero, r'(T) tends towards some constant.
In fact, we can get at the answer through the very fact that as c tends to 0, T tends to infinity. For if this is true, then given our formula for distance at time T:
r(T) = c(e
kT + e
-kT)
we can write
r(T) tends towards ce
kT as c tends towards 0.
since
ce
-kT tends toward zero as c tends towards zero
and T tends towards infinity.
In the same way, in the limit (c = 0) we have
r'(T) = kce
kT
and
r''(T) = k²ce
kT
But we know that r(T) = g --- that when the last bit of the chain is unsupported, the thing is in freefall. So we can write
g = k²ce
kT
Divide the equation
r'(T) = kce
kT
by the equation
g = k²ce
kT
and we get r'(T) / g = 1 / k
Notice how c just vanished from the equation! Hooray!
Hence
r'(T) = g / k
Now recall that we defined k to be the square root of g/r.
Hence r'(T) is the square root of g times the square root of r.
Therefore, Art Vandelay is a complete twat without one single redeeming merit.
