# Conservation of Energy?

1. Apr 26, 2012

### Hypo

Hallo everyone.

I had a equation about energy in a form of "rotational kinetic energy".

Imagine this situation with me so I can explain the question I'm trying to get at:

Let say I had a rotating wheel that is connected to other wheels directly... I supplied 1000J(simple number for me to understand clear) to the wheel for it to spin!(Try to ignore energy loss).

So 1000J is converted from kinetic energy to rotational kinetic energy... Lets say I had two wheels connected to the main wheel thats supplying 1000J of energy will that energy be divided into by 2? 500 and 500?

And obviously if 1 wheel's connected with no energy loss it will convert the 1000J directly right? If so please give out a detailed reason why both of those wheel can't get 1000J? I feel confused because "WORK" is mind bothering!

*NOTE: Ignore energy loss for a bit please lets keep this all under the realm of"theory".

2. Apr 26, 2012

### tiny-tim

Welcome to PF!

Hallo Hypo! Welcome to PF!
I'm not sure what you mean by "supplying energy".

Suppose you supply 1000 J to the first wheel, then you disconnect the power supply.

Then you connect the first wheel to two other wheels in such a way that the first wheel stops, and all its energy goes into the other two.

Isn't it obvious that they get 500 J each?

(If all three wheels are connected from the start, I don't see how you can say you're supplying energy to the first wheel first, it's having to be shared between all three wheels)

3. Apr 26, 2012

### Filip Larsen

If you look at a system of rotating wheels, then the total kinetic energy (which is equal to the total rotational kinetic energy and total linear kinetic energy) of the system is the sum of the kinetic energy of each wheel. If the system starts with zero kinetic energy and you provide 1 kJ, then the kinetic energy of the system will be 1 kJ assuming that there are no frictional loss and that there are no increase in potential energy.

In order to find out how those 1 kJ is distributed on each wheel you must know more about their shape and mass. If you have two identical wheels directly connected such that they both rotate with same angular speed afterwards, then each wheel will have 500 J of rotational kinetic energy. In general, if the two directly connected wheels have moment of inertia around the axis of rotation of I1 and I2, then the rotational kinetic energy of each wheel is K1 = 1/2 I1ω2 and K2 = 1/2 I2ω2, where ω is the angular speed. Since total rotational kinetic energy is K = 1/2 (I1+I22 = 1 kJ the angular speed ω can be found from K and then inserted into the rotational kinetic energy for each wheel, giving K1 = K I1/(I1+I2) and K2 = K I2/(I1+I2). In other words, the ration of rotational kinetic energy of the wheels are K1/K2 = I1/I2.

You can find a list of moments of inertia for different shapes at [1].

[1] http://en.wikipedia.org/wiki/List_of_moments_of_inertia

4. Apr 26, 2012

### Rap

You also have to conserve angular momentum as well as energy. You cannot figure out how the energy will be shared unless you take into account that the angular momentum must be shared also. Then you can figure it out.

Suppose K is the rotational kinetic energy of the first wheel and M is the angular momentum of the first wheel. Then by conservation of energy, $$K=\tfrac{1}{2}I_1\omega_1^2+\tfrac{1}{2}I_2\omega_2^2$$ and by conservation of angular momentum, $$M=I_1\omega_1+I_2\omega_2$$ where subscripts 1 and 2 refer to the other two wheels. Using these two equations, you can find the two unknowns, $\omega_1$ and $\omega_2$

5. Apr 26, 2012

### Hypo

Thank you everyone.

Basic conclusion from all the formula's and laws is that energy is shared equally in the system if ALL the physical states are the same: Size, Mass, etc...

Sadly this is kinda of a burden because "Energy" is a complex process I think that limitless us in every way we go.

I believe their's more to study about "Energy". I thought for some reason in that system the 1000J will be given to all three wheels but I said to my self thats too good to be true.

6. Apr 26, 2012

### Hypo

Re: Welcome to PF!

Thanks Tim.

7. Apr 26, 2012

### Filip Larsen

In this particular situation, where the OP increases the rotational energy of a system from zero to some value K presumably by applying a torque to a shaft, there is not conservation of angular momentum.

8. Apr 26, 2012

### Hypo

The thing is... Can I notice a difference? I mean looking at the main wheel that supplies 1KJ spinning so so fast and looking at the identical wheels that are connected directly to it spinning fast as well thought they would get the same amount of speed + energy.

Now their is NO WAY they would get the same amount of energy right? I mean the 1KJ they will never get that amount unless I supplied more energy to the main wheel.

The thing is I wonder the most about is for example when the wheel's are connected how is that energy exactly shared?

I'd like to imagine this clearly.

9. Apr 27, 2012

### Rap

Well, there is always conservation of angular momentum, I guess the question is whether it all goes to the other two wheels or not. I am having trouble picturing the mechanism, so I can't really say.

10. Apr 27, 2012

### Filip Larsen

Correct. The rotational energy in the wheels have to come from somewhere, so if you provide 1 kJ then the wheels will increase their rotational energy with 1 kJ. If any of the energy you provide gets lost to friction or converted into potential energy then the increase in rotational energy of the wheels would be less than 1 kJ.

I have provided you with some equations in post #3 that show how rotational energy are distributed between two wheels in a simple system. Are you able to understand what I explain in that post? I assume in that post that you know about rotational energy, but if you don't you may want to read up on that concept. Wikipedia has a short introduction to the concept [2]. If that doesn't help you may want to pick up a textbook on introductory physics.

[2] http://en.wikipedia.org/wiki/Rotational_energy

11. Apr 27, 2012

### Filip Larsen

In an isolated system (i.e. a system that does not exchange force (F = 0) or torque (N = 0) with its environment) you do indeed have conservation of momentum and angular momentum which follows from Newtons law [3] (dp/dt = F = 0) and Eulers law [4] (dL/dt = N = 0).

However, in this case you have a system of rotating bodies to which energy is provided from outside the system. If, the energy is provided to the system by a person applying torque N over time to to a shaft in the system in order to induce rotation in the system, then the change in angular momentum is equal to the torque (dL/dt = N) which means that angular momentum is not constant.

[3] http://en.wikipedia.org/wiki/Newton's_laws_of_motion
[4] http://en.wikipedia.org/wiki/Euler's_laws_of_motion

12. Apr 27, 2012

### Rap

Filip - ok, right. I was talking about a closed system, you are not. Again, there seems to be differing ideas of the OP.

1. Wheel 1 has energy 1000J and somehow transfers all of this energy to wheels 2a and 2b, with no dissipation or changes in potential energy. (closed system)

2. Wheel 1 is continually supplied energy, some of which is transferred to wheels 2a and 2b, until the total energy of wheels 2a and 2b is 1000J, again with no dissipation or changes in potential energy. (open system)

Any I missed? I was assuming case 1, you are assuming case 2. Can we first clarify what the OP is?

13. Apr 28, 2012

### Filip Larsen

If you want to model the final energy state of the wheels you call 2a and 2b then it surely is most interesting to have a system boundary that includes those two wheels. In your case 1, wheel 1 is not rotating in the final state so it might as well be outside the system. By the way, in the model I used, there is no wheel 1; the energy and torque simply comes from an unspecified source outside the system of the two wheels 2a and 2b.

Also note, that in your case 1, even if you can say there is conservation of momentum it does not provide any extra information or constraint over conservation of energy since wheel 1 already has been specified to have 1 kJ of rotational energy before and zero afterwards, i.e. conservation of energy under the given circumstances alone specifies that the rotational energy of wheel 2a and 2b afterwards must equal to 1 kJ and, thus, their angular speed can be determined from conservation of energy alone. You could of course translate the rotational energy of wheel 1 into the equivalent angular momentum, use conservation of angular momentum and translate angular momentum back to rotational energy of wheel 2a and 2b, but since the OP (as I read it) is asking about the energy distribution between 2a and 2b you may as well stay with energy.

14. Apr 28, 2012

### Rap

Only if you assume that the angular velocity of 2a and 2b are equal, as you did, but which I do not find in the OP. Let's call the two wheels a and b, rather than 2a and 2b. Conservation of energy states that: $$\tfrac{1}{2}I_a\omega_a^2+\tfrac{1}{2}I_b\omega_b^2=1000J$$ and you cannot solve that single equation for the two angular velocities, nor the individual energies.

15. Apr 29, 2012

### Filip Larsen

I agree that is not easy to extract the exact situation from that post and that you need another constraint to find exactly how the rotational energy is distributed between the two wheels, at which point I used equal angular velocity as a constraint to model a mechanical linkage between the two wheels and you used conservation of angular momentum as a constraint in the situation where three wheels are used.

16. Apr 29, 2012

### jartsa

Let us consider a train that is 100 coaches long.

The locomotive pulls with force 100 Newtons the first coach.
The first coach pulls with force 99 Newtons the second coach.
The second coach pulls with force 98 Newtons the third coach.

We observe that each coach causes a 1 Newtons decrease in the pulling force.

When a coach moves a distance of 1 meters, an energy of 1 Newtons * 1 meters turns into heat, if we have a non-accelerating train, or that much energy turns into kinetic energy of the coach, if the train is an accelerating train.