- #1

phantomvommand

- 249

- 39

- 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

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

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^2The 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