# Gravity problem I need to solve

1. Jul 30, 2014

### Gravity2014

Hi all, I'm not a Physicist, but, I had an idea I wanted to put to bed. If you drop a weight on a rope that turns a generator as it falls, the mgh calculation gives you the force that could be converted to electricity,subject to frictional losses etc. Ignoring getting it back up again, if you connected a pneumatic cylinder to the bottom of the weight so that it struck the ground at the end of the weights journey.My question therefore is, what happens)) as far as I can see, the weight will travel the same distance as before, the motor will turn the same amount, allbeit slowed by the cylinder, yet the cylinder will store force in its air pressure.Please help me get this one correct in my little brain!))

2. Jul 30, 2014

### Staff: Mentor

Hi Gravity2014. Welcome to the famous Physics Forums.

Once the slowly falling weight starts to compress the pneumatic cylinder, the body slows even more, probably stopping short of ground level. So the energy stored in the compressed air subtracts from the energy available to turn the generator.

What you gain on the swings you lose on the roundabout.

3. Jul 30, 2014

### Gravity2014

Thanks for the quick simple reply:-) in the case that the weight drops quickly and the motor doesn't stop turning during compression?)Thanks again:-)

4. Jul 30, 2014

### Staff: Mentor

The weight will fall at a fast rate only if the generator is not powering anything. The more electricity it generates, the slower it must fall.

You won't get something for nothing.

5. Jul 30, 2014

### Gravity2014

Thanks, yes, I didn't want to get into P.M ! I just wanted to understand the principles involved.Thats a big help, thank you. If I understand correctly,the greater the resistance (e.g motor load),the slower the drop.So increasing the weight would only increase the resistance, not the speed.

6. Jul 30, 2014

### sophiecentaur

No; it is the other way round. A high electrical resistance will dissipate less power (it is a lower Electrical Load). The terminology often leads to confusion in the more common case of a source of Volts and a Resistance. P = V2/R. You can always check this as, if you disconnect the load, the resistance becomes infinite and no power is dissipated.

If there is more weight then the generator will tend to speed up (and the weight fall faster). The generator load will receive more Volts from the generator (if it is not a regulated generator). The amount of energy left at the bottom would be mgh - Pt, where the generator produces P Watts and the fall takes time t.

7. Jul 30, 2014

### Staff: Mentor

I don't understand what you've written.

The more light bulbs you try to power with your generator, the more difficult it will be to spin that generator and the falling weight will tend to fall at a slower speed. So in practice you would need to increase that weight to try to restore the proper spin rate of the generator. A slowly turning generator is no use to anyone because it isn't generating the expected voltage and your light bulbs will glow only dimly.

There is no such thing as a free lunch!

8. Jul 31, 2014

### Gravity2014

That makes sense, thank you for your time! So increasing the light bulbs would slow the generator, meaning more weight needs adding.If you didn't add the weight, then more time is taken for the drop,meaning power is lower but energy conversion is the same.

9. Jul 31, 2014

### Staff: Mentor

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Last edited by a moderator: May 6, 2017
10. Jul 31, 2014

### Simon Bridge

The bottom line is that the energy that goes to compress the cylinder is not available to the generator.

The question seems to be about the mechanism ... how?
After all, the generator still turns the same number of times.

The amount of energy generated depends not just on how many times the generator turns but also how fast it turns when it is doing it - the cylinder will slow the falling mass down: you know this already right?

11. Jul 31, 2014

### sophiecentaur

I would add that the energy conversion is not exactly the same in all cases. When the mass arrives at the bottom, it still has some Kinetic Energy. This is 'wasted energy'. It implies that the higher the load (lower the resistance), the slower the mass will fall and the less energy will be wasted. The greatest efficiency will be with the greatest electrical load you connect. Of course, the operating voltage will be less and less as the electrical resistance decreases; no good for lighting but OK for heating.

This wasted energy at the bottom is always a problem with water wheels etc, where the exiting water needs to be moving in order to get it out of the way - taking some KE with it.

12. Aug 1, 2014

### Gravity2014

Thank you, everyone!

13. Aug 1, 2014

### Simon Bridge

No worries.
The complete calculation is that gravitational PE turn into kinetic energy + electrical energy + energy in the cylinder.
Starting from there will avoid a lot of confusion that can result from trying to use force-and-motion reasoning.

Donald Simanek has a collection of examples of how people can get turned around by trying to think only in terms of forces. They make for good exercizes for good physics too: https://www.lhup.edu/~dsimanek/museum/unwork.htm

14. Aug 2, 2014

### Gravity2014

I think we've well covered this one and have an understanding, thank you all who helped me)) I've been to the museum of unworkable devices before, its fascinating)) The question that you've helped me with is a part of my invention, it was wether to include a rebound mechanism (e.g the cylinder) or not. Thanks again.

15. Aug 4, 2014

### Wes Tausend

...

It sounds as though the OP is saisfied with the conclusion of his post.

But other questions are raised. Is there a condition where free lunch seems imminent? Where the rope need never be recovered? Just to add a complication and bit of fun, it might depend on where the weight is falling because of gravity.

Suppose a rapidly orbiting generator lets out a weight which is allowed to fall into a black hole. In some ways we might propose the weight would fall forever, as I have seen proposed as a scenario for an astronaut falling in. The generator could safely remain in a delicately balanced orbit around the black hole. So then it seems we might get our "free lunch" anyway, if the rope is infinitely long, or continuously manufactured. "FREE LUNCH" NOTE BELOW: (my bold)
But perhaps the outer generator end of the rope would slow as the deep far end weight approaches light speed, the rope foreshortening until it moved barely at all, but was now under nearly infinte tension. Then the generator load could be increased to nearly infinite amperage to collect the free power.

More realistic, I believe the rope would likely be torn apart by tidal forces long before one could collect the free lunch. But OTOH the generator might burn up first. Food for thought I guess.

Wes
...

16. Aug 4, 2014

### Simon Bridge

If that question was raised in the thread before now, I missed it.

I'm going to treat this as an orbital mechanics and black-hole question rather than a pmm question, because pmm questions are not allowed.

I take it you want to rig, by thought experiment, a situation where the mass falls forever - generating electricity? Presumably postulating ideal classical components like a massless non-stretchy rope and a frictionless motion generator?

Depends how you are keeping the generator in place.

Off the top of my head:
If it were just an ideal frictionless spindle unwinding the rope, then the spindle and mass would move apart (conservation of momentum) - the spindle climbing to higher orbits. Since the generator is adding some drag to the unwinding - the rope cannot unwind as fast as needed to keep this balance - so the mass ends up dragging the generator after it, or the orbiter ends up dragging the mass along behind it or some combination of the two (it will depend on the initial speed of separation). Either way the whole contraption ends up falling at the same speed after a finite time.

Consider: What started the mass falling in the first place?
You need to input some energy to start the separation... that is the maximum amount of energy you can usefully generate.

Tidal forces would eventually pull such a contraption apart - but it looks like we can add infinite strength and length to the ideal classical rope and still not get energy generated in perpetuity.

17. Aug 4, 2014

### Drakkith

Staff Emeritus
Since the OP's question has been answered I am closing this thread, as it borders on perpetual motion, something which is against PF rules, and is sure to attract crackpots.