Analyzing the Fall of a Chain: Problem 103 of 200 Puzzling Physics Problems

[phantomvommand]

Homework Statement

A chain of length 1m is loosely coiled close to a hole in a table of height 1m. One end of the chain is pulled a little way through the hole and then released. After what times will the 2 ends of the chain reach the floor?

Relevant Equations

F = dp/dt
Kinematics equations

My attempt:

At first, only a small part of the chain has fallen through. Let that part have mass m, speed v, and length x. Suppose the chain has a mass per unit length of u.

To accelerate a small length of chain on the table to speed v, Force needed = v dm/dt = v (dm/dx) * (dx/dt) = uv^2.
Force acting on on the chain due to gravity = uxg
Net force = uxa

uxg - uv^2 = uxa
a = g - v^2/x.

Noting v^2 = 0 + 2ax,
a = g/3.

Very first part of the chain takes time t = sqrt(2L/a) = sqrt(6L/g) = 0.78s. (L = 1m = height of table)
When the first part touches the ground, the velocity of the chain is v = at = sqrt(2Lg/3) = 2.56m/s

My Equation for last joint of chain to reach the floor: L = vt + 1/2 (g/3)t^2
Textbook's Equation for last joint of chain to reach the floor: L = vt + 1/2 (g)t^2

The textbook claims that once the first joint touches the floor, the chain is in free fall. Why is that so? Isn't there an upwards force from the ground on the chain, which acts to decelerate each joint of the chain hitting the ground to 0 velocity? Thus, shouldn't there still remain the upward force due to changing mass?

An extension: If height of table > chain, is the following analysis correct:
- Acceleration for when the last joint of the chain leaves the table is the same as above. (a = g/3)
- As the chain falls through space (as L > length of chain), it's acceleration is g (free fall).
- Upon the first joint touching the floor, its acceleration = g/3.

FYI: This is problem 103 of 200 Puzzling Physics Problems

 

[PeroK]

The textbook claims that once the first joint touches the floor, the chain is in free fall. Why is that so? Isn't there an upwards force from the ground on the chain, which acts to decelerate each joint of the chain hitting the ground to 0 velocity? Thus, shouldn't there still remain the upward force due to changing mass?

The normal force stops the chain as it touches the ground, after that it's balanced by the downward gravitational force of the chain lying on the ground. The lack of rigidity of the chain is important here.

 

[phantomvommand]

The normal force stops the chain as it touches the ground, after that it's balanced by the downward gravitational force of the chain lying on the ground. The lack of rigidity of the chain is important here.

Thanks for the help. I am not sure if I understood you correctly, but this is what I gathered:
The net force acting to slow down the chain is negligible, because the chain is allowed to "crumble" due to its lack of rigidity.

 

[PeroK]

Thanks for the help. I am not sure if I understood you correctly, but this is what I gathered:
The net force acting to slow down the chain is negligible, because the chain is allowed to "crumble" due to its lack of rigidity.

The net force is not negligible: it slows the chain from its maximum speed to rest on the surface.

 

[phantomvommand]

The net force is not negligible: it slows the chain from its maximum speed to rest on the surface.

Then why would the rest of the chain that is still falling through air be at "free fall", if there is a force acting on the bottom of the chain to slow it down? Did you mean the force acting on the chain on the floor is not transmitted to the falling chain, due to the chain's lack of rigidity? If so, how does the non-rigid chain dissipate the force? Thank you very much for your help.

 

[PeroK]

Then why would the rest of the chain that is still falling through air be at "free fall", if there is a force acting on the bottom of the chain to slow it down? Did you mean the force acting on the chain on the floor is not transmitted to the falling chain, due to the chain's lack of rigidity? If so, how does the non-rigid chain dissipate the force?

A force doesn't dissipate. Energy dissipates. A force is transmitted. The chain may transmit a force in one direction only. Think of two masses joined by a light string. The first mass may pull the second in one direction only, but cannot push it in the opposite direction via the string.

The force on the ground briefly exceeds gravity (enough to bring a falling link to rest) and then is balanced by gravity. The mass of chain on the ground is effectively dissociated from the mass of falling chain. If the chain were rigid that would not be so.

 

[phantomvommand]

A force doesn't dissipate. Energy dissipates. A force is transmitted. The chain may transmit a force in one direction only. Think of two masses joined by a light string. The first mass may pull the second in one direction only, but cannot push it in the opposite direction via the string.

The force on the ground briefly exceeds gravity (enough to bring a falling link to rest) and then is balanced by gravity. The mass of chain on the ground is effectively dissociated from the mass of falling chain. If the chain were rigid that would not be so.

Thank you for the detailed explanation. This has cleared everything up.

 

[haruspex]

Noting v^2 = 0 + 2ax,

You should not assume acceleration is constant.
The general solution of the ODE is $2gx^3=3x^2v^2+c$.
In the present case, when t=0, x=0 and v=0, leading to the equation you obtained.