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

In summary, the conversation discusses the physics behind a chain falling from a table and reaching the ground. The equations for the acceleration and velocity of the chain are explained, and the textbook's claim that the chain is in free fall once the first joint touches the ground is questioned. It is explained that the normal force from the ground stops the chain and is balanced by the downward gravitational force. The lack of rigidity in the chain allows for the force to be transmitted in one direction only. The force on the ground briefly exceeds gravity and then is balanced by gravity, with the mass of the chain on the ground being effectively dissociated from the falling chain. It is clarified that force does not dissipate, but rather energy does, and that the chain
  • #1
phantomvommand
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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
 
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  • #2
phantomvommand said:
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.
 
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  • #3
PeroK said:
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.
 
  • #4
phantomvommand said:
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.
 
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  • #5
PeroK said:
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.
 
  • #6
phantomvommand said:
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.
 
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  • #7
PeroK said:
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.
 
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  • #8
phantomvommand said:
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.
 
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Related to Analyzing the Fall of a Chain: Problem 103 of 200 Puzzling Physics Problems

1. What is the problem statement for Problem 103 of 200 Puzzling Physics Problems?

The problem statement is: "A chain of length L and mass M is initially held at rest on a table. One end of the chain is then released, causing it to fall off the edge of the table. What is the speed of the chain when the last link leaves the table?"

2. What is the key concept behind this problem?

The key concept is conservation of energy. The chain's potential energy is converted into kinetic energy as it falls, and this energy is conserved throughout the fall.

3. What is the equation used to solve this problem?

The equation used is the conservation of energy equation: PE + KE = Constant. In this case, the initial potential energy (PE) of the chain is equal to its final kinetic energy (KE) when it reaches the ground.

4. How can we determine the speed of the chain when it reaches the ground?

By setting the initial potential energy of the chain (PE) equal to its final kinetic energy (KE), we can solve for the final velocity (v) using the equation KE = 1/2 * M * v^2. This will give us the speed of the chain when it reaches the ground.

5. Are there any assumptions made in solving this problem?

Yes, there are a few assumptions made in solving this problem. We assume that the chain is released from rest, there is no air resistance, and the chain remains straight and doesn't tangle or twist as it falls. We also assume that the chain is a continuous, uniform object with a constant mass and length.

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