re: energy producing experiments

[Kator01]

Delburt,

Removed my post, because I had some misunderstanding of your proposal.
Since I am a pragmatist I always try to construct a system which I can build in such a way that I can extract the output.
Also I would avoid strong centrifugal forces so that bearings are not destroyed.

I will re-examine your proposals and check some calculations

Mike

[Delburt Phend]

So let's change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.

After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.

I reposted two sentences; the 160 cm is two 80 cm radiuses, with one kg on each end. The dropped mass (or drive mass) is 40 kg. The 40 kg is accelerating itself and two 1 kg masses, but they are on the end of an 80 cm radius.   

[Tarsier_79]

Since we have a pivot-point around which the "dead-masses" rotate ( regardless of the angle covered ) moment of inertia is involved.
This system can not be compared to a pedulum since a pendulum does not transfer torque.

A pendulum can transfer torque, just as it can have torque transferred to it. I only used a pendulum to show why I=mr². There is link between a weight at a specific radius being accelerated and two weights on an arm being accelerated. Anyone can calculate extra energy using incorrect math.

An atwoods is an easy setup to reproduce. Building a very light one is not as easy. In an atwoods, the mass dropping at radius still acts as if it has rotational inertia from the viewpoint of the atwoods hub assembly, and it acts as if it were linear from the viewpoint of masses accelerating in a gravity field. Both calculations agree with each other (,if you account for the inertia of the hub.).. A simple atwoods does not create energy. I look forward to seeing experiments to investigate the theory though.

[Kator01]

Delburt

This would be 3.27 m/sec/sec acceleration at the surface of the shaft.[/size]

That is what I am not sure abput

Mike

Tarsier: I removed my comments above...it was 3 o clock in the morning so had a misunderstanding
I agree :  a special construction of a pendulum-bearing can transfer torque but this is not what is commonly known as a pendulum.
You have to be more specific about the bearing in use.

Mike

[Delburt Phend]

You were correct the first time; a pendulum bob cannot apply torque through a bearing.

I am going to give you real engineering problem.

I have a 14ft fishing boat. If you kneel on the dock and grab the side of the boat, you can apply torque to the boat so as to bring the nose of the boat around to the proper angle for loading it onto the trailer.

I was thinking about making this application of torque much much easier. I could do this by placing a (horizontal ish) pipe at the center of mass so that an 8-foot wooden pole can be placed in the pipe.

And how would you do it?

Would you place a bearing at the center of mass, of the boat, and wrap a rope around the bearing? Or tie a rope to a vertical pipe at the center of mass?


A light tube with 5 kg on both ends is flipped into the air; it rotates about its center of mass. One end is increased to 20 kg, and it again is flipped in the air. The 20 kg is four times harder to rotate so the center of mass shifts closer to the 20 kg end.

But you say that the 5 kg at 4 times the radius is harder to rotate. Well, if it were then way does the center of mass not shift to that end? The center of mass shifts back to a balance where the two ends are equal: 20 kg * 1 r = 5 kg * 4 r

I make Atwood's machines that produce energy. I made an Atwood's pulley that has three radii (very similar to these numbers): the shaft at about 1 cm r, a 15 cm radius, and a 30 cm radius; all in one pulley. This is three concentric circles all fixed on one pulley: 2 cm diameter, and 30 cm diameter, and 60 cm diameter. 

You can keep the drive mass at 1 cm r.

One kg at 30 cm radius is balanced by 2 kg at the 15 cm radius on the other side.
 
One kg at 30 cm radius is balanced by 30 kg at the 1 cm radius. 

The different drops were timed with a photo gate timer; the photo gate flag was on the pulley head. The times of the gate trips were all the same; sometimes within the same ten thousandth of a second.  Any small addition of mass changed the trip time.

All these had the same trip time.

30 kg at 1 cm balancing 1 kg at 30 cm

30 kg at 1cm balancing 2 kg at 15 cm

30 kg at 1 cm balancing 30 kg at 1 cm

2 kg at 15 cm balancing 1 kg at 30 cm

2kg at 15 cm balancing 2 kg at 15 cm

1 kg at 30cm balancing 1 kg at 30 cm

These are all accelerated by the same mass at 1 cm

All had the same trip time; and 6 additional grams would produce a different trip time. So, there is no way that 1 kg at 30 cm is twice as hard to rotate as 2 kg at 15 cm. Or 1 kg at 30 cm is 30 times harder to rotate than 30 kg at 1 cm.

Now let's look at the energy:

½ 30 kg * 1 m/sec * 1 m/sec = 15 joules

½ 2 kg * 15 m/sec * 15 m/sec = 225 joules

½ * 1 kg * 30 m/sec * 30 m/sec = 450 joules

450 is larger than 15