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Old 01-19-2009   #19 (permalink)
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Lightbulb Using basic physics to understand why perpetual motion machines can't work

Many previous posts in this and other threads have given good explanations of the impossibility of perpetual motion machines. However, I think that to understand these explanations, one needs to know some physics basics definitions and principles

The fundamental quantities in physics are:
  • \mbox{mass}
  • \mbox{distance}
  • \mbox{time}
From this, we define:
  • \mbox{velocity} = \Delta \mbox{distance} / \mbox{time}
  • \mbox{acceleration} = \Delta \mbox{velocity} / \Delta \mbox{time}
  • \mbox{force} = \mbox{mass} \times \mbox{acceleration}
  • \mbox{energy} = \mbox{work} = \mbox{force} \times \Delta \mbox{distance}
\Delta, pronounced “delta”, means change in a quantity. For example, “January 19, 2009 6:50:00 PM GMT” and “January 19, 2009 6:53:15 PM GMT” are times, their delta is 195 seconds.

Other principles don’t lend themselves to definition with simple formulae. An important one, possible the most important one is
  • “simplification” – when making calculations, one can chose simple, but entirely accurate one, over more complicated ones
If one has a good working grasp of these definitions as principles, the impossibility of perpetual motion machines is intuitive and obvious. I’ll now use them to address a few of ozi-rock’s questions.
Quote:
Originally Posted by ozi-rock View Post
am I right in saying that if there was no other forces on it other than those of the weight it would keep spinning?
Yes. Any physical system – bodies traveling in space, wheels, pendulums, etc. – will continue doing what it’s doing if no forces act on it, because \mbox{force} = \mbox{mass} \times \mbox{acceleration} = 0 means \mbox{acceleration} = \Delta \mbox{velocity} / \Delta \mbox{time} = 0, which means \Delta \mbox{velocity} = 0, which can be rephrased “nothing changes”.

This also holds if the system is subject to forces, but the forces are exactly equal and opposite. Because objects resting on the surfaces on planets, such as the “unbalanced wheel” and axle in question, are subject to the force of gravity, this applies to them.
Quote:
Originally Posted by ozi-rock View Post
I mean it would keep getting energy from the falling weights?

Could it not be possible to overcome these forces with greater ones?
Here the simplification idea is important.

In a physical system like weights on a wheel, an elevator, etc., how the weight get from one height to another doesn’t matter. Because the force of gravity is always downward, only distance in the up-down direction is multiplied by it. The acceleration of gravity near the Earth’s surface is about 9.8 m/s/s, so its force on a 1 kg weight is about \mbox{force} = \mbox{mass} \times \mbox{acceleration} = 1 \,\mbox{kg} \times 1 \,\mbox{m/s/s} = 9.8 \,\mbox{kg m/s/s}. To reduce writing, the unit kg m/s/s is called the Newton, abbreviated N. The wheel lowers and raises the weight the same distance with each revolution. Let’s say this distance is 1 meter. For its downward trip, then \mbox{energy} = \mbox{work} = \mbox{force} \times \Delta distance = 9.8 \,\mbox{N}  \times -1 \,\mbox{m} = -9.8 \,\mbox{N m}. Again, the unit N m, has a short name named after a famous scientist, the Joule, abbreviated J. For the weight’s upward trip, \mbox{energy} = \mbox{work} = \mbox{force} \times \Delta distance = 9.8 \,\mbox{N}  \times +1 \,\mbox{m} = +9.8 \,\mbox{J}. So, for the entire revolution, \mbox{energy} = \mbox{work} = -9.8 \,\mbox{J} +9.8 \,\mbox{J}= 0 \,\mbox{J}. No energy is used or created, no work done.

Here the “simplification” principle is important. With the unbalanced wheel, with its arms, hinges, and stops, the upward and downward path of each weight, and thus its up-down and left-right distance and work/energy done/used, is complicated. We don’t need to calculate all this, though, only calculate work/energy for each revolution.

Nearly all perpetual motion machine claims invite the reader to attempt the more complicated calculations, or worse, not calculate them, but try to imagine them. Because they’re complicated, nearly everyone attempting them will make a mistake, finding that the machine does negative work/gains energy with each cycle of the machine. The reader’s mistake, however, is not physically real, and the machine really behaves as described by simple calculations, or by complicated ones performed without mistakes.


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