Small unbalance paradox...

[iacob alex]

.....or more well-defined,small mechanical/gravity unbalance paradox,is no more than a simple logical thought,based on a single supposition,belief without proof ("make it greater and greater").

      This one is intended to help some developments ,regarding a possible gravity powered device ,gravity motor or as you like to name it.

      For the beginning,we count upon a two arms lever in a particular/vertical position,with:

         -two equal masses ("m").

         -two unequal arms ("R"/short arm,"R+h" long arm).

     If the lever is in the vertical position (long arm is up,short arm is down ),we have an unbalanced circumstance...an inverted pendulum equivalent,ready to fall down .

     Now,we suppose:

          -"h"/arm length difference is constant.

          -"m" is constant.

          -"R" is ever increasing...this is the "point"!

       The masses "m",at the end of the two arms,after a full 180 degrees rotation (due to gravity unbalance,one mass/on "R+h" long arm/ plays driver,the other mass/on "R" short-arm/will play the follower)will gain greater and greater velocities...so kinetic energy.

       The paradox,seems to be this simple fact...if we make it cyclical:

             -the input("h"/"m*g*h") is constant.

             -the output (kinetic energy of the two masses),can grow more and more.


        This "apparent" point of view ,if tested, can be really a paradox...


             All the best! / Alex

[exnihiloest]

No paradox. "R" ever increasing doesn't imply that the angular speed w increases or is constant.
m*g*h=1/2*m*v² still applies, v=R*w thus the angular speed reduces because R increases.

[iacob alex]

          Hi !
     Sorry:greater fall,greater speed,greater conversion of potential energy into kinetic energy.

     The fall is (h+R+R).

     The "remake" of the small unbalance is reffering to "h"("m*g*h" ,a small jump up) only .

      If h=constant ,and R is ever increasing ,have we a "single answer"(same input,same output?!)...

        All the best! / Alex

[exnihiloest]

          Hi !
     Sorry:greater fall,greater speed,greater conversion of potential energy into kinetic energy.

     The fall is (h+R+R).

     The "remake" of the small unbalance is reffering to "h"("m*g*h" ,a small jump up) only .

      If h=constant ,and R is ever increasing ,have we a "single answer"(same input,same output?!)...

        All the best! / Alex

m*g*h is the only input (a mass fall from 2*R+h and the other climbs to 2*R).
If R is kept constant, after one turn and if we suppose there are no losses, you are again at your start point: m*g*h input is still available, but you didn't provide useful work during the turn. If you had provided work during the turn, the arm would not have been able to come back at its start point.
And we see that this principle doesn't depend on R. To increase R plays no role in the principle.
With an increased R, it is just the angular speed which will be reduced, in order to maintain the same mean angular momentum that the m*g*h input allows.

[iacob alex]

     Hi EX- !

I said that ,if:
    -h=constant    ...~input
    -R is ever increasing   ...~output
    -h<<<R
    -we play two equal masses,two unequal arms(R+h  vs. R)

... we have a similitude with an "Atwood machine with a heavy pulley" (see the topic,and the recommended site,especially),so a "free fall of a torque difference" ...resembling to the free fall of a mass on the vertical line (longer and longer distance ,due to an increasing R...).

     To "remake" the starting point ,we need a jump up~ "h".

     This "remake"  is "assisted" by a greater energy storage ~ "h+R+R" fall down.

     A simple test can be a short way,so to dispense with any contest...

            All the best! / Alex