Re: energy problem

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Wed, 23 Jan 2002, Carl E. Mungan wrote:

> You may have already seen this in The Physics Teacher (Nov. 2001),
> but in case you missed it, here's a problem which might amuse you:
>
> A long uniform chain is hung over a frictionless pulley. The left
> part of the chain rest on a table, and the right part rests on the
> floor. Soon after the chain is released, it reaches a constant speed.
> Find that speed, given that the table is height h above the floor.
>
> The solution using momentum is given in the Feb. 2002 issue. Your
> assignment is to find a solution using energy. I'd be interested in
> hearing from anyone who wants to take a crack. Carl

First of all, note that since this is a problem in Newtonian
mechanics and since the problem statement only "gives" h, the
answer (if there is one) can *only* be sqrt(gh) times some
dimensionless constant.

I don't approve of the solution given in TPT because it is
incomplete.  One might easily wonder, "What about the fact that
momentum is a vector?  What about the force applied by the pulley
itself (which *does* depend on the speed of the chain)?"

Here are a few possible ways to approach this problem that should
be far easier to follow and that make it clear that the primary
point of analysis has to do with what is going on at the bottom of
the short side.

Let,

    lambda = mass per unit length of the chain
         h = height of the table above the ground
         d = height of the pulley axis above the table
         v = speed of the chain
         R = radius of pulley

Then,

  T1 = Tension at the bottom of the long leg
     = 0

(because nothing hangs from the bottom of the long leg)

  T2 = Tension at both endpoints of contact with the pulley
     = lambda(h+d)g

(because T2 - T1 must equal the weight of the long leg and because
the pulley is not accelerating)

Now, to find T3 we *could* apply the impulse-momentum relationship
to a small length of chain at the bottom of the short leg.

  T3 dt = lambda(vdt)v   => T3 = lambda v^2

Then, we simply require that the short leg be in equilibrium

  T2 - T3 = lambda d g   =>   v = sqrt(gh)

I would call this the "momentum" solution because of the method I
used to obtain T3.

I can easily change it to an "energy" solution by applying the
pseudowork-kinetic energy relation to a short length of chain dx
at the bottom of the short leg.  Since the CM of the short length
of chain moves up a distance dx/2 while the force T3 is applied,
we find

   T3(dx/2) = (1/2)(lambda dx)(v^2)   => T3 = lambda v^2

The rest proceeds as before.

One final important note:  If we calculate (what I call) the
external work done on the short length of chain (i.e., T3 dx), we
find that it is *twice* the pseudowork and, therefore, twice the
bulk kinetic energy gained by the short length.  Therefore, we
must be "dissipating" (that is, causing the internal energy of the
short length of chain to be increasing by) an amount of energy
equal to the kinetic energy gained.  Moreover, we dissipate
another identical amount of energy at the lower end of the long
chain as it comes to rest.  This dissipated energy is accounted
for by the reduced gravitational potential energy of the chain.
Thus we might write:

  Amount of energy dissipated at the bottoms = Amount of GPE lost
                              2 (1/2 dm v^2) = dm g h

 From which we find, once again, v = sqrt(gh).

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm