Re: [Phys-l] nifty question (solution)

[richard lindgren <ral5q@virginia.edu>]

Suppose you move the two support points closer and closer together, so  
that you have almost two parallel hanging cables or in the limit  you  
might imagine one hanging cable. Since the length of the cable does  
not change, the center of mass of the hanging would would not change  
when the mass was added. I guess I would conclude that as the support  
points move closer and closer together, the amount that the CM moves  
upward is less and less. At first i though that maybe the CM of the  
cable would not change. How does the thermodynamic argument predict  
that the CM does not change for a single hanging cable? My answer  
would be that the string does not stretch so in the this limiting case  
there is no work done on the hanging cable so the potential energy  
does not increase. Basically the answer is trivial for a single  
hanging cable. Just thinking out loud.
Richard


On Sep 24, 2010, at 8:00 AM, Carl Mungan wrote:

> Okay, looks like most people got it, and several people even
> anticipated the second question below. Reference for these questions
> is given in the last line below.
>
> > Consider an ideal (ie. cannot stretch, and can bend freely) cable
> > suspended from two points of equal height (that are closer together
> > than the cable's length). The cable adopts the shape of a catenary
> > (hyperbolic cosine).
> >
> > Now suppose you hang a weight from the center of the cable. Does the
> > cable's center of mass move up, down, or remain at the same height?
>
> The center of mass goes up. Here's a simple explanation:
>
> Suppose instead of suddenly attaching the hanging mass m to the
> cable, I instead apply a downward force F to the center of the cable
> that slowly increases from 0 up to mg. (This guarantees the process
> is quasistatic, just like what we do in thermo when we compress a
> piston in a gas-filled cylinder.) Clearly I do positive work on the
> cable. Thus the energy of the cable increases. Since the process was
> quasistatic, there is never any KE of the cable. Further, the cable
> is ideal and thus doesn't stretch, so it cannot gain elastic PE. All
> that's left is for the cable to gain gravitational PE, and thus its
> COM must rise.
>
> From the same readings, here's another (much easier) nifty question.
> Suppose instead of pulling downward on the center of the cable, I
> pull upward, until the center of the cable is at the same height as
> the two ends of the cable. What upward force F am I applying at that
> instant?
>
> Reference: AJP 58:1110 (1990). -Carl
> --
> Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
> Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
> mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/
> _______________________________________________
> Forum for Physics Educators
> Phys-l@carnot.physics.buffalo.edu
> https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

Dr. Richard A. Lindgren
Research Professor of Physics
Director of Master of Arts in Physics Education program.
Department of Physics
University of Virginia
Charlottesville, VA 22901 USA
ral5q@virginia.edu
Office 434-982-2691
Fax    434-924-4576