Is energy really conserved?

[Ben tesoriero]

What is "steady state", "no motion", and "resulting torques"? Since you have posted a labeled diagram, why don't you use those symbols in your text?

Can you write down the formulas for energy input, energy output and stored energy using the symbols you have defined? Where is the maths that you already have had checked by someone?

Also, what does actually happen in your experiment? And how did you ensure that the two wheels have exactly equal but opposite angular momentum?

Steady state in this case im meaning constant angular rotation of ω, ie the resulting torques have caused a fixed deflection in the shafts.

 

[weirdoguy]

You ignored other questions @A.T. asked. Where is the maths?

 

[A.T.]

Steady state in this case im meaning constant angular rotation of ω,

When $\omega \neq 0$, the total kinetic energy stored on the system is greater than when $\omega = 0$. That is where the energy partially goes, that you provide by applying $\tau$ over some angle.

Where is your math showing energy disappearing?

 

[Dale]

Steady state in this case im meaning constant angular rotation of ω, ie the resulting torques have caused a fixed deflection in the shafts.

OK, so you don't seem to understand what it would take to show that energy is not conserved.

To show energy non-conservation you have to measure the total energy of the system $E=KE+PE+U$ where $KE$ is the system kinetic energy and may be further subdivided into linear and rotational parts, $PE$ is the potential energy of the system in any external fields and may be further subdivided into parts for different external fields, and $U$ is the system internal energy including any thermal energy, internal fields, or other internal locations of potential energy.

You have to measure the $E$ at two different points in time, $E_1$ and $E_2$, as well as any work $W$ done on the system and any heat $Q$ that flows into the system.

Then to show non-conservation you would have to show that $E_1+W+Q \ne E_2$. The difference would also need to be large enough that it could not be attributed to measurement errors. You haven't shown ANY of that, not even a measurement of $E_1$.

 

[Ben tesoriero]

Ive tried to keep it simple and given a simple way to try it for yourself using two angle grinders.

Maybe explaining it like this will help. Take two scenarios, in the first a base case, start with both wheels turning at 8000 rpm, ramp the rotation shown as ω in the diagram up from 0 to 120rpm over 10 seconds then ramp down to rest over 10 seconds. The wheels remain turning at 8000 rpm at the end.

Repeat for the second scenario but this time hold the rotation at 120rpm for say 10000 seconds, then ramp down to rest as before. The wheels remain turning as per the base scenario.

You have 10000 seconds of energy going into the system difference between the two scenarios, but no difference between the ramp up, ramp down or final states of the scenarios.

 

[PeroK]

During those 10000 seconds, if energy is going into the system but the system maintains a constant mechanical energy, then that energy must be dissipated through friction, air resistance and noise.

It's no different from having to keep your foot on the accelerator to maintain constant speed of a motor vehicle.

 

[Dale]

but no difference between the ramp up, ramp down or final states of the scenarios.

The heat is different. If the experiment is done in an insulated chamber then the 10000 second chamber will have a higher temperature than the 10 second chamber. $E_1\ne E_2$. So we are back to my previous comment

If you ignore where energy goes then of course you will think that energy is not conserved.

 

[sophiecentaur]

try it for yourself using two angle grinders.

How can you even suggest such an experiment? Where would you get two identical grinders with no (or at least equal) friction at every point in the drive). It's like the equivalent experiments, in olden times, that lead people to fail to grasp Newton's first law of motion. The errors from the equipment must be well in excess of the error in energy that you're trying to show.

You haven't presented your experiment and you don't acknowledge the maths in the theory. In no way is this 'Physics".

 

[Herman Trivilino]

You have 10000 seconds of energy going into the system difference between the two scenarios

You mean energy is being transferred for a period of time equal to 10 000 seconds. That doesn't describe the amount of energy transferred. Just because the time periods are the same doesn't mean the energy amounts are necessarily the same.

 

[A.T.]

You have 10000 seconds of energy

You seem to be confusing energy with time.

Does this not meet the definition of energy being destroyed?

No, more like a waste of time.