View Full Version : work in cyclic motion

Dave Maurice
09-15-1995, 05:43 AM
Hello friends:

I am sorry to be entering into this so late, and I must confess I
have no knowledge of what started this. However, I have seen several
well-intentioned postings the last few days which have reflected some
of the confusion which commonly arises from considerations of cyclic
motion. It's very easy to get confused with this, but I hope to try
to clarify some of the points that seem to confuse people most
frequently. In so doing I hope to avoid offending anybody; if I
point out a (perceived?) error please don't think I am singling
somebody out.

The most common misconception that arises is that if the final
position is the same as the initial position, no work has been done.
This stems from an (sorry, every word I think of for here sounds
unfairly critical) interpretation of the definition of work, which
most people take to be W = F delta X. However, a more illustrative
definition in my opinion is that W = the integral of FdX. This
seemingly simplistic (even semantic) distinction makes a profound

Rather than insult your intelligence with a review of this (because I
can hear the lightbulbs clicking on all over the world I know it
isn't necessary) I'll instead risk the wrath of a few courageous
folks who unfortunately overlooked this in their attempts to explain.
Friends, this isn't to draw attention to your errors, but is done
because those errors are the ones we ALL are inclined to make, and
because the examples will clarify the necessity of integration over
the path rather than mere attention to the endpoints.

If a hand presses against a spring and compresses it some distance,
then the hand performs work on the spring. If the hand relaxes and
the spring presses the hand back to the original position, then the
spring perofms work on the hand. Now that they are returned to the
original position, the temptation is to say that the net wok done on
the spring is zero. But it isn't; in fact the return to the original
position in no way reduces the work done on the spring. If you wish
to see this mathematically you can graph it; for a constant rate
spring you will get a triangle whose area is a measure of the work
done. An "easier" way is to make the situation initially more
complex. Imagine that we had a thingamajig (all inclusive term for
mechanical translation device - e.g. a gear) to transform that spring
compression (the motion thereof) into the movement of yet another
piece. When the spring is allowed to return to its initial position,
that other movement would still have taken place, right? And in
fact, this principle is, I am sure you will agree, rather closely
related to the principle behind the Carnot cycle (and the internal
combustion engine).

Back to our example. Upon return to the starting point, the net work
done on the spring is the work done in compressing it. The net work
done on the hand by the spring is the work done in moving the hand
back to the starting position. These do not "sum" to zero.

The swimmer will be the second example. If we cling to the idea that
work is defined by F delta X, then one rotation of an arm would seem
to yield zero. We can't do that. In fact, we can't directly apply
the definition I gave, as it is "phrased" for direct application to
rectilinear motion. For rotational motion it becomes W = integral of
tau d theta, where tau is the torque and theta the angular
displacement. And this will give the type of results that are
desired; if we rotate the arms twice with a constant stroke we get
twice the work, not a return to zero. Now admittedly the tethered
swimmer problem is fairly complex (how to determine tau?) but this is
the approach that will give the mechanical work done by the swimmer
on the water.

Thank you for your attention. I hope I haven't bored or offended to
many of you; if any clarification is desired I will do what I can.