Physics+ Newton Laws, Problem 2: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 2

Problem

A chain of infinitesimally small links with overall length L and mass M is stretched on a table with its one end hanging off the edge of a table.
Half of a chain is lying on the table, while another half is hanging down.
The chain is stretched on a table along a straight line perpendicular to the edge of a table.

Initially, the chain is held at rest. Then at time t=0 it is let go, so the hanging part of a chain pulls the rest of it off the table.

The experiment ends at time t=T when the chain's endpoint B will be at the table's edge, where endpoint A was in the beginning of movement.

Assume that table legs are long enough, so chain, when it slides all the way down, does not reach the floor until it is fully off the edge of a table.
There is no friction between a chain and a table.
The acceleration of free falling is g.

What will be the speed of a chain at time t=T?
In other words, what will be the speed of the endpoint B, when this endpoint will be at the table's edge?

Solution

Let's try to solve this problem using only the Newton's Laws.

Assume that at a moment in time t the chain slides off the table by the length s(t).
Then the mass of the part of a chain that is not supported by a table will be
(½L+s(t))·M/L
It will pull down the part of a chain still on the table with the force
F = (½L+s(t))·(M/L)·g
This force pulls the chain of mass M with acceleration
a(t) = d²s(t)/dt² = s"(t).

Knowing the force, the mass and the acceleration, we can use Newton's Second Law getting a differential equation for s(t):
(½L+s(t))·(M/L)·g = M·s"(t)
or
s"(t) − s(t)·g/L − ½·g = 0

Initial conditions are
s(0)=0; s'(0)=0

To solve the above differential equation is not easy, which makes the whole approach less practical.

Let's try to solve it differently.
Assume, the length s(t) is an independent variable.
We know the force of gravity F(s(t)) in terms of s(t):
F(s(t)) = (½L+s(t))·(M/L)·g

Multiplied by infinitesimal increment of the length of the chain ds(t) that slides from the table during the infinitesimal increment of time dt, it will give an infinitesimal increment of work performed by this force
dW = F·ds

Integrating this from s(0)=0 to s(T)=½L, we will get the total work performed by the force of gravity
W_[0,T] = ∫_[0,½L]F(s)·ds =
= ∫_[0,½L](½L+s)·(M/L)·g·ds =
= ∫_[0,½L]½L·(M/L)·g·ds +
+ ∫_[0,½L]s·(M/L)·g·ds =
= ½M·g·s|_[0,½L] +
+ (M/L)·g·(s²/2)|_[0,½L] =
= ¼M·L·g + ⅛M·L·g =
= ⅜M·L·g

All this work was done by a gravitational field. The potential energy of a chain in a gravitational field has decreased by this amount, and this work was converted into kinetic energy of the chain.

Since the chain, when it completely slides from the table at time T has kinetic energy E=½M·v², where v its speed, ⅜M·L·g = ½M·v²
Therefore,
v² = ¾L·g
from which we derive the value of the chain's speed at the moment it completely slides off the table
v = √¾L·g
Notice that the speed of a chain at time t=T does not depend on the mass of a chain.

Yet another, even simpler approach to this problem is to use only a loss of potential energy as a reason for an increase in kinetic energy.

Let's take the surface of a table as the level zero of potential energy of a chain.

In the beginning, at time t=0 half a chain is on this surface, the potential energy P₁ of this half is zero:
P₁ = 0

Another half of a chain is below the level zero and, therefore, its potential energy is some negative value P₂.

The chain is at rest and, therefore, its total energy is
E = P₁ + P₂ = P₂

At the end of motion at time t=T the first half of a chain that had zero potential energy P₁ takes the place of the second one in the beginning of motion and its potential energy will be
P'₁ = P₂
The second half of a chain moves down by half the length of chain ½L, decreasing its potential energy to level P'₂.

Now the chain is moving and has some kinetic energy K.
Therefore, the total energy of a chain is
E = P'₁ + P'₂ + T =
= P₂ + P'₂ + T

From the Law of Energy Conservation
E = P₁ + P₂ = P'₁ + P'₂ + T

Knowing that
P₁ = 0 and P'₁ = P₂
we conclude
E = P₂ = P₂ + P'₂ + T

We see now that
P₂ = P₂ + P'₂ + T
Hence, canceling P₂,
0 = P'₂ + T
which allows to find kinetic energy T
T = −P'₂

Therefore, the loss of potential energy is the negative potential energy P'₂ of the second half of a chain at the end of motion.

It can be calculated by intuitively obvious method that considers all the mass concentrated in the middle of this half of a chain hanging below the surface of a table by ¾L, which leads to its potential energy
P = −½M·g·¾L = −⅜M·g·L.

Alternatively, we can calculate the potential energy of this half of a chain directly integrating potential energy of each infinitesimal part of it.
Let x be a distance of the point of the chain's second half from the table (level zero).
Then
P'₂ = −∫_[½L,L](M/L)·g·x·dx =
= −(M/L)·g·(x²/2)|_[½L,L] =
= −(½−⅛)M·g·L =
= −⅜M·g·L

In all cases the loss of potential energy is ⅜M·g·L.
From the Law of Conservation of Energy follows that kinetic energy of a chain must increase by the same amount, which leads to same value of a speed of a chain at time t=T
v = √¾L·g