Challenge Questions

Using:

a = acceleration of pendant+chain

m= mass of each link

L = distance of pendant movement from initial position in metres

G = acceleration due to gravity = 9.8 ms²

µ = coefficient of friction of table = 0.2

Total mass hanging off the table when has moved distance L = (21+100L)m

Total mass still on table when has moved distance L = (1-L).100m

So total force on pendant + chain = force due to g acting on mass hanging over the table – force due to friction acting on remaining chain on table = total force acting to accelerate total mass of pendant + chain.

(21+100L).m.g – (1-L).100.m.µ.g = 121.m.a

Using µ = 0.2 and simplifying gives

a = g/121 +(g.120.L)/121

Now the end of the chain will come off the table when L=1. To calculate the time at which this will occur we can integrate the above equation between L=0 and L=1 in which case we get

Integral of g/121 +(g.120.L)/121 = (g.L)/121 + (g.60.L²)/121

Calculating for L=1 and noting that result is 0 when L=0 gives 4.9 seconds

This could also be calculated iteratively if we assume that the acceleration is constant until another link drops over the side in which case this gives a figure of around 5.2 seconds, however, this is not compatible with assumption of infinite flexibility at the table edge.

Where is time in your equations? Get off the mushrooms.

That is one of the funniest responses I have read on this forum - I will qualify this by saying that this is not a significant achievement. I would give you a GA but I could not be bothered to login.

Off mushrooms now. Equation for acceleration OK but rest is crap. I re did the culculations using approximations on spreadsheet and got 1.79 secs.

Assuming this answer is right for the 100 links to fall off the table in a line then the right answer is HALF the time as a Chain holding a pendant will be HALF its maximum length because it is closed into a LOOP.

So using the 5.2 second time for constant acceleration the time is actually 2.6 seconds.

Problem did not state that the chain was closed into a loop. Not all pendants have to be hung around the neck.

Making the usual assumptions, I make it 1.76 secs.

Codey

Good Challenge Question. I have a solution to a similar problem. My problem is the same except that at the corner of the table, my chain makes a 90 degree turn sort of like it was making an elastic impact with a plane at 45 degrees to the table top surface. All the horizontal kinetic energy is transferred to vertical kinetic energy. I think for the real chain, there will be a tendency for the chain to retain some of this horizontal kinetic energy and, therefore, the solution I have will be somewhat in error. Maybe if I can get some more insight and spend a little more time, I can get a better solution.

There are three energy terms to consider. First is the potential energy of the chain and pendant. Second is the energy lost to the friction from the table's surface. Third is the kinetic energy which is just the difference between the potential energy and the lost energy.

Let W₀ equal the weight of the chain, W1 equal the weight of the pendant, β is the ratio of W₁/W₀, X₀ is the length of the chain (100 cm), X is the distance the trailing end of the chain has moved, μ is the coefficient of friction (0.2), and g is the gravitational constant (980 cm/sec²).

The potential energy is just the weight of the pendant times X plus the weight of the chain that overhangs the table times one half X. PE = W₀(βX + X²/(2X₀)).

The drag force from friction is the weight of the chain still on the table times the coefficient of friction = μ W₀X/X₀. The energy lost is the area under the drag force versus distance curve. Lost energy = μW₀(1+X/X₀)X/2.

The kinetic energy is ½ M V² = ½ W₀(1 + β)/g (dX/dt)² = W₀(βX + X²/(2X₀)) - μW₀(1+X/X₀)X/2. W₀ can be divided into each side so the weight per unit length is not relevant.

This gives us an equation for dX/dt versus X. Using an Excel spread sheet to numerically integrating this (closed form solution was too hard), I get the time to be 1.10 seconds with a final velocity of 287.3 cm/sec.

Thanks,

Jim

Jim

Absolutely correct in all respects GA from me). For the situation as described the chain velocity at the corner will only change to vertical while the tension at the corner exceeds ρ.(dv/dt)² (where ρ is the linear density of the chain, and v is the horizontal velocity).

It follows that the assumed trajectory cannot be maintained until the end leaves the table unless the chain passes through a smooth tube with a right-angle bend (or some equivalent arrangement).

Calculation of the time and length at which the usual assumption fails should be straightforward. But the motion of the chain after that time is complex - I doubt the existence of an analytic solution. [What is certain is that the chain will take longer to fall than the existing equations suggest]

BTW, both the literature and the web are littered with false solutions to problems involving dynamic chains.
x y z fail to apply to this very problem.
However, the most academically mishandled problem of this kind is likely to be the so-called "coiled chain" problem. a; for example completely misses the complexities. b is unusual in recognising the some of the issues, however, even here the authors fail to recognise that existence and effect of the lateral constraint set by the pulley in their experiment2 means that this does not provide support for the application of their simplified theory to a coiled chain.

Nicely over-viewed Physicist

I used the question mark earlier in reference to the shape of a chain at such a change of direction. "All you have to do" is work out the two r's (inc.rate of change) and subtract the energy lost in angular acceleration. Similarly a chain will 'stick' to a pulley at speed, giving the similar ? shape/losses. A pulley model would at least give one r that is an r, as opposed to some sort of inertia modified catenary.

Q: "How long will it be before the end link falls off the table?"

A: It won't ! The jeweler is much two/to/too quick to allow it. As quickly as he "releases it", he safely re-grabs it before link #40 is off the edge.

The End.

Why , oh why must you deem her to be stupid...? (the OP did not state her hair color)

I somehow think there may be something unrealistic or inconsistent about the assumptions stated at the end of the challenge. Like jim35848 (post# 6 & 16), I'm feeling uncomfortable about the horizontal momentum acquired by the chain as it is sliding on the table. What happens to it when sliding off the edge in accelerating mode? Maybe one should assume a frictionless pulley and then let the radius tend to zero. I'm afraid I'm poor at that kind of mathematics, but it's an interesting problem all right, at least for thinking about the physical aspects. =Teesquare=

Now let's solve the problem in Jupiter environment and see if the answer changes.