Power Consumption Calculation Verification

[lxman]

I believe you gentlemen are corrupting me. This is starting to make sense.

I am reworking my equations (again) and I will post a graph here momentarily that I believe wraps everything into a nice package.

 

[lxman]

OK,

From scratch, here are my equations. I will not go into the derivation details, if anyone sees a difficulty with any of my equations, I will address that in a subsequent post.

m = 1kg
a(t) = -cos(t) red
v(t) = -sin(t) green
s(t) = cos(t) - 1 blue
KE(t) = 1/2 * m * v(t) ^ 2 brown
W_tot(t) = 1/2 - 1/2 * cos(t) ^ 2 black
P(t) = sin(t) * cos(t) purple

http://www.freeimagehosting.net/uploads/th.290bc3ee4a.png

What I see happening here is the following:

v, s, KE, W_tot and P all begin at zero, as they should. a begins at -1, which I believe is stretching things a bit, but to make everything else work, I left it that way. Now, KE represents the instantaneous energy of the object, and W_tot represents the net work done over time, which in my system KE and W_tot happen to be equivalent. W_tot oscillates regularly between 0 and a positive value. What illustrates the negative force here (and the positive one as well, is P. P is the derivative of W_tot with respect to time, so P is the instantaneous power at time t. Therefore P illustrates where the negative energy is coming from.

How does this sound?

 

[milesyoung]

Looks good but:

KE represents the instantaneous energy of the object

I don't know what you mean by this. KE is simply the kinetic energy of the point mass system.

W_tot represents the net work done over time

W_tot is the net work done on the point mass system at some instant in time.

which in my system KE and W_tot happen to be equivalent

They're equivalent in general. That's the Work-Energy Thorem.

Edit:
Note that we're talking about a change in kinetic energy for this to hold.

In this case, the change in kinetic energy is equal to KE since the kinetic energy of the point mass is zero initially.

 

[lxman]

Points taken, milesyoung. Thank you.

So, what this shows is that in order for my point mass to continue its oscillations I have to affect it with alternating positive and negative power. The sum of this power over time will be zero, of course. But nevertheless it does have to be there.

Therefore, back to my earlier postulation (a 1kg mass in deep space with a variable thrust rocket engine on each end), in order to calculate the total fuel that I would require to operate the system for time t, I would have to integrate the absolute value of P over time t. That mass of fuel would then be substituting for the spring.

Is this a reasonable conclusion?

 

[milesyoung]

Note that doing positive work on the point mass system means transferring energy to it, and doing negative work means transferring energy from it.

If you make the assumption that you're unable to recover the energy you receive from the point mass system as you do negative work on it, in the sense that you cannot use this energy for doing positive work on the point mass system, then the net positive work done on the point mass system should be what you're after.

This is just off the top of my head and I'll leave it to you to prove me right or wrong. I really need some sleep :)

Hope you find your answers. Cheers.

Edit:
If you decide to look up something relating to system descriptions and work, beware that some might define work the other way around, i.e. positive work done on a system means a transfer of energy out of the system. This should be clear from their definition of work though.

 

[lxman]

Yeah, that works.

Thanks much for taking the time.