MOMENTUM AND ENERGY

MOMENTUM AND
ENERGY

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MOMENTUM

Momentum of a free falling body due to constant force of gravity can be collected in a second body either in one increment, or in more increments. The more increments (above two) are used, the more momentum can be transfered from the free falling body to a second body over the same total height of a free fall, as long as the final speed of an increment of a free falling body is higher than the speed of the collecting body at the location of collection.

“Newton’s second law of motion states: The acceleration imparted to a body by a force acting on it varies directly as the force and inversely as a mass of the body, and is in the direction of the applied force. Momentum = mass x velocity > M = mv.”

Ramsay/Eubank “Basic Physics” MacMillan Co Canada 1963, page 51

“Therefore, the total change in momentum depends upon both, force and time. It should be apparent then that the force is proportional to change of momentum.”

Ramsay/Eubank “Basic Physics” MacMillan Co Canada 1963, page 52

We shall take 100 kg mass and drop it 19.5m. We shall first drop the 100kg mass 19.6m in one increment and then in 4 increments of 4.9m each, totaling again 19.6m.

- A -

Lets assume that we have a flywheel, which will collect the momentum of the single increment free fall of a 100kg body as its own momentum of spin. The body falls through the full 19.5m height in 2 seconds. The force of gravity acted on the body for the duration of 2 seconds and a spring apparatus transferred the momentum to the flywheel. The final velocity of the falling object at the bottom of the fall, when its momentum was transferred into the spin motion of the flywheel, was 19.6m/s.

- B -

Lets assume that the same flywheel can collect momentum from the falling body in an four-increment fall. The body falls 4.9m in each free fall increment within one second each. It is stopped at the end of each increment by being jammed between a flywheel spoke and two springs straddling the flywheel. This transfers the momentum of the falling body gained in each increment to the flywheel spin. Instead of once over the total 19.5m fall, we did it four times over the 19.6m fall. Therefore, we allowed the body to accelerate four times during the 19.6m fall, which totals the duration of g acting on the body to 4 seconds.

When we use the standard formula “0.5 x m x v^2” and calculate the theoretical kinetic energy of the object at the bottom of A and B, they match impeccably. Nevertheless, we can bypass the KE calculation and concern ourselves only with the transfer of the momentum of the falling body and the transfer of its momentum to the flywheel, then back from the wheel to the body. Momentum is theoretically preserved in the same manner in which energy is preserved (less friction etc.).

A)

The body in a single increment free fall falls full 19.6m in 2 seconds and attains the final speed of 19.6m/s. Its momentum is therefore M = mv > M =100kg x 19.6m/s = 1960 units of momentum.

B)

The body in a free fall falls in four increments and dumps its momentum into the flywheel at the end of each increment itself being stopped. Each increment yields M = mv > M = 100kg x 9.8m/s = 980 units of momentum per incremeantal fall consecutively translated into the flywheel spin momentum. This multiplied by 4 (for the four-increments) equals 980 units of momentum x 4 = 3920 units of momentum dumped into the flywheel spin during the 19.6m free fall.

The single increment free fall of a 100kg body A dumps 1960 units of momentum into the flywheel spin accelerating it, while the four-increment free fall of a 100 kg body B dumps 3920 units of momentum into the flywheel spin accelerating it. A = B/2

HAMMERWHEEL K2

An actual three stage free fall experiment series, as presented in fig 1 &2, has yielded more energy translated back from the flywheel to the falling body than the energy translated from the falling body to the flywheel. Yet, the momentum has been preserved.

Fig. 1 presents the schematic of the three stage free fall of a body (a bearing ball) within the flywheel.

Fig. 2 presents the schematic of the experimantal rig used. A series of these experiment had been performed by the author sometime in 1998. This series has been further probed during recent months in steel as well as wodden frames.

Fig. 1

Fig. 2

The three-stage drop in the illustrated flywheel brought the bearing ball to an elevation higher than the loading elevation as long as the weight of the flywheel and the weight of the ball were in a rough ratio of 1(ball) : 1.3 (flywheel) so that the flywheel reached circumferential speed needed to "kick" the ball up high enough with the first impact. The wheel did not stop after hitting the ball the first time with one of its spokes, it only slowed down. The ball fell back and got "kicked" again three times total, before the wheel reversed the spin direction and the ball eventually settled at the base.

I used curtain rails of unknown make (bought with my house) for the wheel construction as well as for the guide rails (springs) construction. I used 1/8" dia. farmers tying wire on the outside of the wheel for ballast, and I had built the support structure from 3/4" plywood sheeting fastened together with drywall screews. The wheel was about 12" diameter with 1/4" dia shaft (made from a screwdriver) hung in good high speed roller bearings. The bearing ball was 1.5"dia.

I hope you appreciate the friction inherent to this system. A properly designed flywheel is such, which has as much of its mass on its periphery as possible. A solid flywheel does not work, because not all of its mass gains the speed needed to sufficiently accelerate a bearing ball above the loading position. The shown experiment achieved only three-increment fall.

Compounded cause of altitude loss of an object in a free fall

There are two superimposed components of velocity in a free fall and two reasons for the loss of altitude of a free falling body.

• 1st second > The body falls 4.9m in the 1st second of its free fall attaining the final speed of 9.8m/s. The 4.9m loss of altitude is the only length component during the first second. (As we use our standard units)

• 2nd second > The body falls 9.8m due to its inertial motion attained in the first second + 4.9m for accelerating motion during the 2nd second = 14.7m loss of altitude during the 2nd second. The final velocity of the body at the end of the 2nd second is 2 x 9.8m/s = 19.6m/s, while the total loss of altitude is 19.6m. The total loss of altitude due to acceleration is 9.8m, while the total loss of altitude due to the inertial motion from the 1st second is also 9.8m.

• 3rd second > The body falls 19.6m due to its inertial motion attained in the first two seconds of the free fall + 4.9m for accelerating motion during the 3rd second = 24.5m loss of altitude during the 3rd second. The final velocity attained by the body at the end of the 3th second is 29.4m/s. The total loss of altitude at the end of the 3rd second is 44.1m. The total loss of altitude due to acceleration is 14.7m so far , while the total altitude loss to the inertial motion is 19.6m so far.

• 4th second > The body falls 29.4m due to its inertial motion attained in the first three seconds of the free + 4.9m for accelerating motion during the 4th second = 34.3m lopss of altitude during the 4th second. The final velocity attained by the body at the end of the 4th second is 39.2m/s and the total loss of altitude is 78.4m. The total loss of altitude due to acceleration is 19.6m so far, while the total loss of altitude due to inertial motion is 58.8m.

• ETC.

What is the point of inclusion of inertial motion distance into the formula of energy content calculation, never mind having it squared?

The math of it

Momentum is added to a falling body at each time interval of the free fall by the same amount. M = mgt, where "g" represents 9.8m/s speed increase at the end of each second of free fall. This formula is valid only for friction less accelerating motion. The momentum formula M = mv is valid for linear motion.

Distance of free fall increases with each consecutive unit of time according to s = 1/2g x t^2, where 1/2g stands for the mean speed of a falling body during its first second of free fall.

PE is lost per each unit of time in a free fall as a loss of altitude. The standard PE formula is a differential value calculation, not an absolute value calculation. PE = s - (m x 0.5g x t^2) Where "s" stands for the potential energy difference height. It is valid for single stage acceleration only and only within the gross contemporarry limitations of understanding of what can be obtained as work from gravitational and other acceleration.

ENERGY

"Energy …may be defined simply as ability to do work. ….. The amount of energy a body possesses is related to the amount of work it is capable of doing."

Ramsay/Eubank “Basic Physics” MacMillan Co Canada 1963, page 69.

The same can be stated about the relative momentum of a body.

ETHERGY

I have been advised to introduce a new term, instead of arguing about old terms. The new term is ethergy, as opposed to the term energy. The PE formula coefficient 1/2 does not belong to mass, it does not belong to time, it does not belong to distance, but it belongs to g. The 1/2 in the formula "1/2m x v^2" represents the fact that the numerical constant "9.8" can mean s (2nd second inertial distance of travel), m/s (first sec. final velocity), as well as m/s^2 (2nd second inertial distance of travel squared). The 1/2 represents the numerical value "4.9m/s", which is half g (m/s velocity), which is the mean velocity of a falling body during its first second of a free fall, but which is also the accelerating distance of the first and every consecutive second or travel.

Definition:

Ethergy is energy obtainable from gravitational field by free fall.

It can be collected by any body in a free fall from the gravitational force field and transformed into momentum of a body. The mass of the body and the duration of the collection, i.e. free fall are the two values, which decide how much is collected for further transformations. Ethergy can be equivalent to momentum as noted here, but it is transformable into any other form of energy by other means. It can also be expressed in its basic form as: M = mtg, where g represents the linear increase of speed of a falling body per unit of time (9.8m/sec).

KE calculations have to be revised, because they are currently based "in error" on PE differential formula, a special case, instead of being based on the absolute formula and the concept of momentum. As an example, we have a paradox in the KE and PE energy content calculation of a rocket accelerating (to the moon). The kinetic energy of an accelerating rocket calculated by the standard formula increases nonlinearly with time and faster than potential energy of the fuel dissapears in flames from its tanks. The PE of the rocket also increases nonlineraly to with liner consumption of rocket fuel PE to make the matters even worse.

The momentum of a falling body increases linearly, because the cause of the free fall is an action of a constant (sort of) force with linear time. It makes no sense to take length of free fall path as one of the causes of acceleration. The length of path is a result of a single increment free fall acceleration relative to the input of energy into a falling object. The length of the path (height in our case) can be manipulated in a way (exactly as I did it) by dumping the gained momentum at time intervals, therefore extracting good deal of the inertial speed component out of the falling body's path, dumping its ethergy into another system, a flywheel in my case. This extends the duration of the free fall per height traveled without sacrificing the altitude. The wheel as shown and experimented with is not a closed system. We have to take gravitation as an outside source of ethergy. My wheel is an open system. Any and every subsystem of the universe is an open system. The only exception may be the universe itself but, that one is infinite and therefore the terms closed and open become meaningless.

KINETIC ENERGY

When we go back to the formula for calculation of KE, we can see that this formula had been derived from the differential formula of PE and has nothing to do with KE. KE = M = mv = m x s/t. That is the amount of ethergy behind the capability of a body to do damage or work when "stopped". In the same sense, the amount of damage or work a momentum will cause is equal to its relative capability to perform work.

POTENTIAL ENERGY

PE is a concept within the concept of energy transformations. It does not really exist. When a stone lies on a roof, it exerts constant pressure on the roof, which is constantly counter acted by the structural strength of the roof. The origin of this force g is in the gravitational field, not in the stone. The origin of the PE is in the G field, not in the stone. The stone is constantly being acted upon. This action is called weight. It is claimed that the weight is lost in a free fall. That is so, but that is so due to the fact that there is no counter force opposing g in the free fall. The g acts constantly and constantly attempts to be transformed into other kinds of energy. It does it in the free fall as can be demonstrated, but there are certainly other ways. While a stone lying on the roof does not accumulate ethergy from the gravitational field, a falling stone does accumulate it and transforms it into its own energy of motion. The more it accumulkated, the faster it moves relative to its gravitating counterpart.

We can represent the PE on a man made situation. We can lean our car with its nose against a wall and hold onto gas. The car will push against the wall as long as we press the gass pedal and it could be said that it has a potential energy (equivalent to what?), because it does no work, it does not accelerate or lift the wall. Yet, it is exerting pressure on the wall . But, it needs gas for keeping that force constant and keeping it at all. No gas, no PE. The car will accelerate though, once the wall is removed from its path. Then the car performs what is called work, it will start accelerating and overcomming friction. It will again perform work when it hits another wall. It will perform work twice, once it accelerates and once it decelerates, or better said, its engine will perform work on the car on acceleration and the car will perform work on the wall (and itself) while crashing. As much as this PE of the car leaning against the wall can only be sustained by the burning of the fuel in it, the weight of the stone on the roof can only be sustained by the steady action of energy coming from gravitational field, itself being wasted in some way. The waste comes from redirection of gravitational field component paths of communication. I could also state that the waste goes into the deformation of geometry of the mutual earth stone field. Thermal energy does not enter the causality on this scale of causality.

When a stone falls off the roof, it gains KE by being accelerated by ethergy and yields energy of heat and other when hitting the ground. But, it has no ethergy while sitting on the roof and it has no ethergy while sitting on the ground and it gains no ethergy when you dig a hole under it. It has no capability to perform work. It has capability to transform ethergy into motion and motion into work in a free fall.

please refer to TTF2/FORCE1 for the correct solution.