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Leigh notes that the sudden change of direction at the edge of the table
presents some problems, which can be partly solved with a smooth radius at
the edge. To avoid the hanging portion of the rope (when it gets moving)
from departing from a straight vertical, a suitable contstraint (a bent
frictionless tube) can be placed at the edge of the table. Once that is
done, the solution is a piece of cake.
Oh, yes, the acceleration of the rope at the edge of the table is, of
course, infinite, but if we take the limit as Leigh's table edge radius
goes to zero we have an infnintessimal portion of the rope undergoing
infinite acceleration there, and we can deal with that by a bit of hand
waving. (Or simply ignore it, for few students will notice.)
The problems ask for v(t), y(t) a(t) a(v) v(y) of the rope. From these you
can easily find how long it takes to leave the table, what its speed was
as the last portion leaves the table, etc. etc.
I've got extensive notes on this, which I don't care to convert to a form
for e-mail right now, but you guys can check my answers.
Using conservation of energy (remembering that the potential energy of the
portion over the edge is mgh where m is its mass considered to be at its
midpoint), you get v = (g/L)(y^2 - yo^2)^1/2, where yo is the initial
little piece hanging over. If nothing hangs over the edge initially, the
rope doesn't go anywhere.
The total mass of rope is M, the portion hanging over the edge at any time
is y.
This result suggests that if we *could* ignore yo, we could integrate v =
(g/L)y to get y = exp(gt/L), from which v = (g/L)exp(gt/L) and a =
(g/L)^2exp(gt/L).
Using free-body diagrams and F = dp/dt, we find that a = (g/L)y. This
agrees with the "quick and dirty" method in which you simply observe that
the net force accelerating the total mass of the rope is the total mass
times the acceleration, so (rho)gy = (rho)La where rho is the linear
density M/L of the rope, and therefore we conclude that a = (y/L)g. The
initial length yo doesn't appear because the acceleration depends only on
the applied forces, hence only on y.
Integrate the equation for a from yo to y to get v = (g/L)(y^2 -yo^2)^1/2.
Writing a as the second derivative of y and equating it to [(g/L)^2]y we
have a differential equation. Guess a solution of form y = Aexp (gt/L) +
B exp(-gt/L), we get y = (yo/2)[exp(gt/L) + exp(-gt/L)] This agrees with
the solution given in Kleppner and Kolenkow, problem 3.15.
These solutions can better be expressed:
y = yo cosh(gt/L)
v = (g/L)yo sinh(gt/L)
The time it takes to leave the table is
t = (L/g)^1/2 ln(y/yo) + [(y/yo)^2 - 1]^1/2
Since I have to teach this course next term, I'd appreciate being informed
of any errors in the above. :-)
-- Donald
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Dr. Donald E. Simanek Office: 717-893-2079
Prof. of Physics Internet: dsimanek@eagle.lhup.edu
Lock Haven University, Lock Haven, PA. 17745 CIS: 73147,2166
Home page: http://www.lhup.edu/~dsimanek FAX: 717-893-2047
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