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

difference.

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.

Dave

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

difference.

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.

Dave