re: energy producing experiments

[Delburt Phend]

From experiments like the hammer or tennis racket toss; the center of mass is the point on the object that travels at the same speed even while the object is put in rotation. The tube is put in rotation by catching its end.

The center of mass of the tube, with a mass of 5 kg at 30 cm and a 2 kg mass at 100 cm, is at 50 cm.    We assume the tube has no mass; for easier math.

So, with the center of mass at a 50 cm radius and moving 3 m/sec: then the 5 kg mass with a 30 cm radius would be moving 1.8 m/sec after the tube is caught on the end.

The 2 kg mass has a radius of 100 cm and therefore would be moving 100 cm / 50 cm * 3 m/sec = 6 m/sec.

After the tube is caught on the end: the 5 kg mass would be rotating faster that the 2 kg so it would have to transfer its motion to the 2kg so that they could rotate at the same rate. You could say that the 5 kg is placing torque on the 2 kg. But for you the 2 kg is inertia and would be (2kg * 10 dm * 10 dm) / (5 kg * 3 dm) 200 / 15 harder to move. Then why does the 5 kg lose the same amount of momentum as the 2 kg gains.  5 kg * 3 m/sec = 15    5 kg * 1.8 m/sec = 9   15- 9 = 6   and 2 kg * 3 m/sec to 2 kg * 6 m/sec is also 6. The same thing is the same thing. It is not about torque and r² inertia it is about momentum.

The 5 kg's momentum changes from 15 to 9 kg m/sec; and the 2 kg's momentum changes from 6 to 12 kg m/sec.

The total initial momentum was 7 kg * 3 m/sec = 21 kg m/sec; and the final linear Newtonian momentum is   5 kg * 1.8 m + 2 kg * 6 m/sec = 21 kg m/sec.

Why does the 2 kg gain the same amount of momentum that the 5 kg loses when the 2 kg is allegedly 200 / 15 harder to move?

When you swing a baseball bat you place torque on it to accelerate it; but if you miss the ball then you place torque on the bat to decelerate it. So the application of torque does not have to be acceleration. Why would it not be legitimate to say that the 2 kg torques the 5 kg. Then the 5 kg would be the inertia and it would be 45 / 20 harder to move than the 2 kg.

Newton's view is that the 5 kg pushes the 2 kg and the 2 kg pushes the 5 kg; and it all works out beautifully.  But your 200 / 15 and 45 / 20 has no math for a real-world event.

[Kator01]

Delburt,


[quoteyou have a one meter low mass tube that has 2 kg attached to one end and 5 kg attached at 30 cm from the other end.
The tube is moving sideways at 3 m/sec so that both the 2 kg and 5 kg masses are moving at the same speed (3 m/sec).


The tube is caught on the end near the 5 kg. The tube's end is then on a bearing and the tube is forced to rotate.
The 5 kg now has a 30 cm radius, and the 2 kg has a 100 cm radius.

You need to be consistent with your arguments. You changed systematics.


We talk about intertia torque which  means angular acceleration of masses at a distance from its center of rotation.
We accelerate by applying torque on the axis. Here we experience moment of inertia ( m*r² ) of the masses to be accelerated


Your last example describes a bar with two weights at different position from a center bearing which moves translational then
hits an obstacle and then at almost the same instant is suddenly connected to a bearing around which it rotates.


1) you changed systematics by describing the conversion of a translational moving system to a rotational one.
2) the transition between both systems is irrational in regard to a practical solution


Since there is no practical solution to this transition you need to change the system as follows:


The bar is mounted on the bearing which is fixed to the ground of the earth and another mass with the same mass as the earth hits the system a one end at 3m/s.


Got the point ?


The same irrationality


If we design a system like this which accellerates a rotational system at rest by applying a force ( translational[/font][/size] moving mass M hits the system) at a point on the circumference where the mass is,
then we have inertia of the mass involved and not moment of inertia. The masses then rotate freely at the final speed by translational[/font][/size] acceleration multiplied[/font][/size] [/size]by[/font][/size] the time of physical impact leaving again the system without torque.


Wasn't the topic "pure rotational systems set in motion by angular acceleration" ?


Suddenly we are a lost in a discussion caused by the compulsion to find another counter-argument to defend a claim of creating OU.


Mike

[Delburt Phend]

It is not irrational and in fact it is deadly simple; and the solution is given.

You could place objects on a cart and drive the cart up against a wall. Fix one end of an object to the edge of the cart. You could test and see if the linear speed of the center of mass of your chosen shapes proceeds at the same rotational speed. You would need a means of reducing friction.

And this event of the tube mentioned does not conserve energy; so, I guess it is a third way of making energy

[Kator01]

No,


there are to cases to be considered:


1) elastic impact


2) no elastic impact


3) premise:
    configuration must be symmetric, i.e. two identical masses attached to the edge- better to 2 bearings on a shaft on two levels ( one above the
    other) , shaft mounted in the mass-center of the car.


ad 1)
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car will be repelled -> situation develops into chaotic behaviour , behaviour can not be predicted because of timing difference between angular accelleration of the the masses on the edge and the time lapsed during repelling of the car.




ad 2)


momentum of the car will be used up by deformation of the wall and the car ( molecular destruction and heat) . Only initial momentum of the objects attached to the edge of the car is left. No gain possible




Mike

[Delburt Phend]

Suppose you were to swing a rod from a 20 meter string as in a physical pendulum. The end of a 1 meter rod is attached to the end of the 20 meter string. Thirty cm down from the attachment of the rod is a 5 kg mass fixed to the rod. And 100 cm down from the string attachment is a 2 kg mass fixed to the rod.

This 20.5 meter pendulum is swung into a pin at the down swing position. The pin is at the point of the attachment of the rod to the 20 meter string.

If the speed of the center of mass of the rod is 1 m/sec before contact with the pin at the downswing position: then what it the speed of the center of mass after contact with the pin?