http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9# 

A Rotation with Translation Movement is standalone natural phenomenon

 

For this experiment, two identically thin cylinders which are initially static to the observer are taken. These cylinders are attached with internal mechanical springs that induce a repulsive action between them.

Two experiments are to be conducted.

 

The first experiment.


An action that repels the two cylinders is initiated. This action is induced from their center of mass. Now the cylinders are observed to move in directions opposite to each other. Based on the law of conservation of momentum, the linear velocities of the cylinders are equal. The kinetic energies of the cylinders are deduced using the below derivation: 
 
T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}
T_1=m\frac{v_1^2}{2}+0(\text{no rotation})
T_2=m\frac{v_2^2}{2}+0(\text{no rotation})
 
v_1=v_2\Rightarrow (m\frac{v_1^2}{2}+0) = (m\frac{v_2^2}{2}+0) \text{ , }T_1=T_2

Thus it is inferred that their kinetic energies are identical. Now, lets compare their translation energies and angular momentum.

 
P_1 = mv_1 \text{ } L_1=I\omega_1
 
P_2 = mv_2 \text{ } L_2=I\omega_2
 
v_1=v_2\text{ , }\omega_1=\omega_2=0
 
mv_1=mv_2\Rightarrow P_1=P_2\text{ , }0=0\Rightarrow L_1=L_2

It is inferred that their momentums are identical as well.

This experiment thereby proves that the two cylinders exhibit a symmetric action due to the initial repulsive inducement of their translation movement.

 

The second experiment.


Initiate the same repulsing action as done in experiment 1, but induce the action in different positions. Induce one cylinder from its center mass and the other from its edge. On observing the movement of the cylinders, the following are noticed:

The two cylinders move in opposite directions

The cylinder that is induced from its center mass alone rotates

Assuming their linear velocities are identical according to the law of conservation of momentum, their kinetic energies are determined:

T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}
T_1=m\frac{v_1^2}{2}+0(\text{no rotation})
T_2=m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}(\text{has rotation})
 
v_1=v_2\text{ , }0\ne\omega_2\text{ }\Rightarrow \text{ } (m\frac{v_1^2}{2}+0) \ne (m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}) \text{ , }T_1\ne T_2
 

From the above, equation it is inferred that their kinetic energies are not identical. Now, as in the previous experiment, their translation energies and angular momentum are compared.

P_1 = mv_1 \text{ } L_1=I\omega_1
 
P_2=mv_2\text{ } L_2=I\omega_2
 
v_1=v_2\text{ , }0\ne\omega_2
 
mv_1=mv_2\Rightarrow P_1=P_2 \text{ , }0\ne I\omega_2\Rightarrow L_1 \ne L_2
 

Hence it is deduced that their linear momentums are identical. The values of the angular momentums of the two cylinders are found to deviate and the cylinder initiated from its center mass alone has an angular momentum.

Thus it is assumed that the action is not symmetric, this assumption is made by relating the action to the initial repulsing event. But this presumption fails to follow the law of conservation of angular momentum.

These deviations are however based on faulty assumptions. The linear velocities of these cylinders are not identical.

A single event can not induce two different movements in a single object. One event can trigger only one kind of movement in each object, in this case the cylinders.

The first experiment conforms to this rule, as there is only one translational movement that is induced.

In the second experiment, one cylinder exhibits only one type of movement, whereas, two types of movements are observed for the other cylinder. Though both the cylinders are stimulated using only one event which implies that each cylinder must exhibit only one type of movement, one cylinder exhibits translation movement and rotational movements. These two movements are thus considered as new type of movements. These movements are hence standalone natural phenomenon. So it follows, the movement should have its momentum and follow its own conservation of momentum.

Assuming the movement has a linear and angular momentum, the total momentum of rotation with translation movement is:

 
P_f= \sum P_j +\frac{1}{R_u}[\sum L_k]
 

Where, Pj - linear momentum   Lk - angular momentum   Ru - unit radius

The law of conservation of momentum for the translation movement with rotation is:


\sum P_j +\frac{1}{R_u}[\sum L_k] = Const

This movement has two components: rotation and translation.  How are these two correlated to the momentum? To answer this lets consider the following diagram of a rotating body.


The mass of the rotating body is concentrated on its radius (Ru). An initial force Pf strikes the body which is at a distance h from the center of mass of the body. It is known that the moment of inertia is:

 
I=m(R_u)^2
 

Which implies that:  

 
R_u=\sqrt{\frac{I}{m}}
 

Assuming this momentum is applied to the angular movement:

 
P_f\frac{h}{R_u+h}
 

So, the angular momentum is:

 
L = P_f\frac{hR_u}{R_u+h}
 

Translation momentum is then equal to:

 
P = P_f-P_f\frac{h}{R_u+h} = P_f\frac{R_u}{R_u+h}
 
Lets sum these parts and check full momentum:
 
P_f=P_f\frac{R_u}{R_u+h}+\frac{1}{R_u}\times P_f\frac{hR_u}{R_u+h}=P_f(\frac{R_u}{R_u+h}+\frac{h}{R_u+h})=P_f
 
 
 

Adding both these equations, the total momentum is determined.

 Following the new law of momentum conservation for translation with rotation movement, the translation and angular velocities of the cylinders have a different value on experiment 2.

Translation and angular velocities for cylinder 1 equal to:

 v_1=P_f\frac{1}{m}
 
\omega_1=0
 

Translation and angular velocities for cylinder 2 equal to:

v_2=P_f\frac{R_u}{m(R_u+h)}
 
\omega_2 = P_f\frac{hR_u}{I(R_u+h)}
 
R_u=\sqrt{\frac{I}{m}}
 

Hence, based on the previous statement translation with rotation movement corresponds to the consolidated translation and rotation movements. It is thus inferred that one of the movement is the result of the other primary complex movement.

 

The above experiments are simulated. 

 

The experiment 1.

The simulation elicits results that relate to the theory. 

Initial results: Before start

 

Values during experiment:

 
 

The experiment 2.

Initial results:

Before start

Values during experiment:

Let's simulate same result from experiment using additional torque.


Initial results:
Before start

Values during experiment:

It is inferred that the experiment 1 and 2 with additional torque produced the same results.

 

The red rod in experiment 2 follows these equations:

 
P = m_r\times v_r=F_i\times t_i
 
L=I\times \omega=F_i\times t_i \times R
 

Where,  P -  rod's translation momentum, mr - rod's mass, vr -rod's translation velocity, Fi - initial pulse force, ti - initial pulse force time L - rod's angular momentum, I - rod's moment of inertia, w - rod's angular velocity, R - rod's unit radius

The red rod in experiment follows these equations:

P = m_r\times v_r=F_i\times t_i_1
 
L=I\times \omega=\tau \times t_i_2
 

Where  P -  rod's translation momentum, mr - rod's mass, vr -rod's translation velocity, Fi - initial pulse force, ti1  - initial pulse force, ti2 - initial torque time L - rod's angular momentum, I - rod's moment of inertia, w - rod's angular velocity, tau - torque.

These experiments use different parameters for the equations of angular momentum. For this equation the simulator for experiment 2 uses same initial pulse force. However, for experiment with additional torque simulator uses another parameter (torque).  The torque should have a pair on real world. However, the simulator can trigger unpaired torque for one body.

experiment 1 + additional torque = experiment 2

The two rods have same mass and moment of inertia.
However, in experiment 2, the rotating rod exhibits extra torque, which is simulated easily.

Observe experiment 3

The simulator applies extra translation force for rotation rod, which is not the case in real world. A fraction of applied force is spent for rotation movement (torque) then this fraction of force should be detected from the force applied for tranalation. The sum of these rotation and translation fractions of applied force should be equal to applied force for non-rotating rod. (3-rd Newton’s law)


The translation velocities of these rods should vary. But they are observes, to be equal on simulator. Because simulator is unaware about rotation with translation movement and hence, the independent rotation and translation movements make it difficult to make inferences from experiment 2. The classical mechanics should include new standalone translation with rotation movement to describe natural phenomenon correctly.

 

The following are the animations recorded for experiment 2:

 
This animation follows classical mechnics laws: 
 
 

This animation follows theory of standalone rotation with translation movement.

 

 

The Natural Experiment 2.

 

3 successful experiments were conducted with 2 pencils.
In these experiments pencils with rotation movement have lower velocity than pencils without rotation.

 

The theory is CORRECT.

The simulator is WRONG.

 

Equipment: 2 pencils thread and thin rubber band 3''

The rubber band is repulsing 2 objects (2 pencils).
The mass of the rubber band is much less that the mass of the pencil

 


The followng are the snapshots of the experiment dynamics.

 



 
Links to experiments movies (avi files)
Experiment 2_1
Experiment 2_2
Experiment 2_3

Links to experiments pictures (zip files)
Experiment 2_1
Experiment 2_2
Experiment 2_3



Conclusion

The experiments using the pencils prove the theory that rotation with translation movement is standalone natural phenomenon. This new movement is shown to have its own law of momentum conservation. This theory conforms to the existing classical mechanics laws. This new theory that is framed through this experiment proves existing classical mechanics laws and gives additional natural phenomenon explanations.


 

 email: abelov0927@gmail.com