I have a question that I can't seem to get my head around. I'm really rusty on my physics, but I'm sitting here trying to think this out, but I'm missing a really big part of the equation.

So lets say I have a turbine in space. It's purpose is to convert the energy given to it, in joules to watts. So I realize we will have friction, and we need to overturn the static friction before turning can be achieved.

I want the device to turn at a relatively high speed, therefore F > Ff. However, this will create a situation where the device will constantly accelerate, because there is F and not being apposed by friction, Fn = ma.

So here is my thought process for the situation.

Force needed to start turning...
F = Fn - Ff
Energy needed to start turning...
W = Fd
W = (KEf - KEi) where KEi = 0
a = F/m

Since there will always be force, and constant supply of energy, the device will accelerate to infinity, supplying infinite energy in watts? I understand this to be fundamentally impossible... So where am I going wrong? I feel like this will be a facepalm moment, but I just can't seem to remember what I'm missing here.

Best Regards.

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  • $\begingroup$ It will not constantly accelerate. Once force is applied, the device will quickly reach an equilibrium where the force of friction--it may be more appropriate to say the resistance--equals the applied force. Thus the acceleration will only be momentary. $\endgroup$ – Klik Jul 14 '14 at 5:06
  • $\begingroup$ How does the resistance "catch up" with the applied force? In other words, how does resistance increase with acceleration? $\endgroup$ – user3471141 Jul 14 '14 at 5:53
  • $\begingroup$ The resistance increases proportionally to the speed of the rotation, not to the acceleration. Take a bicycle, put it in the highest gear and flip it upside down so you can rotate the pedal by hand. It takes a lot more effort to spin the pedal bicycle fast than it does to rotate it slowly, right? A bicycle has friction just as a turbine has friction (and resistance). $\endgroup$ – Klik Jul 14 '14 at 7:07
  • $\begingroup$ What resistance increases with speed in space? $\endgroup$ – user3471141 Jul 14 '14 at 7:51
  • $\begingroup$ I will make an answer below and I will edit it in the morning. $\endgroup$ – Klik Jul 14 '14 at 8:38

With Regards to the Energy Conservation

To turn a turbine requires energy. A turbine will convert the energy that is required to turn it (mechanical energy), into another form of energy (electricity). This is why there is resistance in turning a turbine; if there was no resistance and it produced energy, it would defy the law of conservation of energy.

Let's say that for every rotation of the turbine we generate 1 joule of energy. If we rotate the turbine two times in one second then we have an output of 2 joules per second--equal to 2 watts.

So, as we increase the speed of the rotation of the turbine we are generating more energy per second. Since we know that the turbine is converting mechanical energy into electrical energy, we can deduce that in order to produce more energy per second, we need to put more energy per second into the rotation of the turbine. Therefore we can assume that in order to rotate the device faster we will incur more resistance from the turbine.

With Regards to the Acceleration

The difficulty you are having in understanding is that you are caught up in the fact that there is an acceleration term in the equation. I.E. Force = mass x acceleration; joules = force x distance = mass x acceleration x distance.

The thing is, the device will only accelerate until the two opposing forces are equal: force applied = force of resistance.

Let's say that the turbine is 10 kg. Initially you put enough force to accelerate the object by 1 m/s^2--you are giving a force of 10 newtons (1 N = kg x m/s^2).

The resistance of the turbine when its speed is 0.5 m/s is 10N. But when you increase the speed of rotation, the resistance the turbine provides will be greater. Let's say that when the speed of rotation is 0.6 m/s the resistance is 12N.

So if you provide a force of 12N to rotate the turbine, the turbine will accelerate to 0.6 m/s where the applied force will equal the force of resistance and the turbine will cease to accelerate as long as the applied force and resistance is maintained.

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  • $\begingroup$ What stops the device from Accelerating past the force of resistance? Could you express this mathematically? I understand you included some math behind it, but I'd like a good example from energy put in to energy put out with some varying speeds. $\endgroup$ – user3471141 Jul 16 '14 at 6:37
  • $\begingroup$ also one more thing. You say that at 0.5 m/s the resistance is 10N. at 0.6 m/s, 12N. So I deduce that at 1.0 m/s, 20N of resistance. But you state at t=1, 1 m/s, the force given is 10N. If the force of friction is greater than the force of forward movement... wouldn't the turbine not be spinning? $\endgroup$ – user3471141 Jul 16 '14 at 7:00
  • $\begingroup$ It is 1 AM where I am right now, I will provide more information tomorrow. In the meantime, could you please read up on free body diagrams? It would make it easier for me to explain this to you. This is turning into a lot more work than I anticipated, but I'm willing to give you some more time provided you do some research on your own. $\endgroup$ – Klik Jul 16 '14 at 7:13
  • $\begingroup$ Right, I'm familiar with free body diagrams. Believe it or not, I did study lots of physics. But it's been a few years and for some reason I can't get my head around this. It's really embarrassing, especially cause I study statics. $\endgroup$ – user3471141 Jul 16 '14 at 7:55
  • $\begingroup$ What I said was that the acceleration of the object was 1m/s^2. It is not the actual acceleration, the actual acceleration will be determined by the Fnet force. Acceleration is part of force and cannot be separated. Let's say that 12N is applied to the turbine. At the current speed it will provide 10N of resistance, so the Fnet is 2N (note that these are vector units, but for simplicity, in this explanation the resistance is always directly opposing the applied force). Since Fnet is 2N, the device will accelerate by 2N ÷ 10 kg (its mass), and so it will accelerate by 0.2m/s^2 at that instant $\endgroup$ – Klik Jul 16 '14 at 22:49

A turbine that converts kinetic energy to electrical energy would always have some kind of operational friction - the magnetic field that must be turned through to generate electricity.

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According 2 me a turbine in the space will keep on rotating unless someone or something disturbs it so we can get a non-renewable source from this project

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You want to put in joules and take out watts? Watts = joules/second. Done.

As for friction, if it is turning in space, why is there any?

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  • $\begingroup$ It's a turbine... Turbines have friction from the gears and such working to change mechanical energy into usable energy no? Lets say this device is bolted down to an even bigger object (not that it should matter too much). Right, watts are in joules/second. But I'm not trying to figure out why, I'm trying to mathematically approach a turbine in space. Where mechanical energy is converted to usable energy. $\endgroup$ – user3471141 Jul 14 '14 at 6:04
  • $\begingroup$ Lets say that I have 30 Joules of energy that I put into the system. Lets say the distance is 1 meter. The work done at 360 degrees is 30(J) = F(1m). F = 30N. Lets say Ff = 10N so Fnet = 20N. a = (30N)/(1kg) the mass of the turbine is 1kg. So a = 30m/s^2... This would mean that, with no other apposing force, it will accelerate forever. $\endgroup$ – user3471141 Jul 14 '14 at 6:09
  • $\begingroup$ This produces another issue... KEout = 1/2 mv^2. This is the energy produced by the velocity of the turbine. So KEout = 1/2 (1kg)(20m/s^2*1.0s)^2 = 200J VS 30J. There is definitely something wrong with what I'm thinking here. I wish for correction! $\endgroup$ – user3471141 Jul 14 '14 at 6:23
  • $\begingroup$ YOUR turbine has gears and stuff. Mine was floating in space and spun up with magnetic fields and energy extracted that way. Anyway, the flaw is in the thinking about force times distance. With your first 30 joules you have it spinning at 30 RPM. Now in the next second say you add 30j. It is already going 30 RPM. What happens to your distance traveled while the 30 joules are added? Your force is applied over a greater distance and in order for the work to be the same, F is smaller. Not long and F=the friction force. This is why you can't pedal a bicycle to Lunar escape velocity (?). $\endgroup$ – C. Towne Springer Jul 14 '14 at 6:31
  • $\begingroup$ Put it another way; at what speed does friction force times distance = work put in per second. $\endgroup$ – C. Towne Springer Jul 14 '14 at 6:36

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