|Apr4-12, 10:01 AM||#18|
Weight change with rotation ?
With a ring, you would have to integrate over the angle, and the result is 0.
The same can be seen just from symmetry of the ring.
>> B/ Now I move up one mass of 1 m and move down one mass of 1 m, this give energy [...] +19.64491438591486051612 J
This is way too large (and roughly equivalent to the energy you get from just moving one mass by one meter).
You move your masses down by 100m two times?
In general, you should check your formulas. One example: (R-99m) just appears in the last expression. But how did the mass come to this position?
"1/(R+1-99)-1/(R+1)" <- this is a movement by 99 meters, not 100
Anyway: You just go from one potential energy to another. By construction, you have to arrive at 0 if you do that in a circle. (a-b)+(b-c)+(c-a)=0.
|Apr4-12, 10:40 AM||#19|
G is an imitation of g and his formula and value are very doubtful...
|Apr4-12, 11:48 AM||#20|
I have corrected
2*m*G*M*H/100*(1/(R-100)-1/(R)) = 1964.52227397479936678121
m*G*M/1*(-1/(R+1)+1/R-1/(R-1)+1/R) = +3.08348993657429924912e-6
m*G*M*H/100*(1/(R-99)-1/(R+1)+1/(R-101)-1/(R-1)) = 1964.52227397494456785185
m*G*M/1*((1/(R-99)-1/(R-100))+(1/(R-101)-1/(R-100))) = -0.08363513764493721675e-6
The sum is :
1964.52227397479936678121-3.08348993657429924912e-6-1964.52227397494456785185+3.08363513764493721675e-6 = 1e-27 it's ok, the sum is 0.
|Apr6-12, 05:34 AM||#21|
I have thought about something I don't understand. When a screw is in water with bottom thread end with vertical surface. The torque on the bottom surface is canceled all along the thread with a differencial of pressure (up/down surface of the thread). With the pressure of water = f(h) it's ok, all torque are canceled, but now with pressure = f(h,Δh²) why the bottom torque is not greater than the differencial torque ? The difference is not big but must be to 0 I think ?
NB. We are for example on Earth with gravity = 9.81 m/s². The screw can only turn.
|Apr6-12, 09:34 AM||#22|
All is fine I understand the problem.
But for this thread I will calcultate if the torque is 0 :
A circular thread (can only turn) in sandwich between 2 circular fixed thread. Up and down small film of water. See the drawing. The thread has only film of water up and down.Normally, this thread don't turn because the height of water is the same. But really, one side is more far from Earth (the up side). This give a torque ?
|Apr6-12, 11:42 AM||#23|
I drew the developpant of the helicoid (thread). This helicoid is in a sandwich between 2 fixed helicoids. Only film of water Up and Down surfaces.
The down torque = T1 = Radius_of_helicoid * Down_surface * density_of_water * height * Gt * Mt * ( g + (1/(R+250) - 1/R ) )
The up torque = T2 = Radius_of_helicoid * Up_surface * density_of_water * height * Gt * Mt * ( g + (1/(R+350) - 1/(R+100) ) )
g = 9.81
Mt = Mass of Earth
Gt = Gravitational constant
Up_surface = Down _surface
k = Radius_of_helicoid * Down_surface * density_of_water * height * Gt * Mt
T1 = k * ( g + (1/(R+250) - 1/R ) ) = 9.8099999999938410347 * k
T2 = k * ( g + (1/(R+350) - 1/(R+100) ) = 9.80999999999384122804 * k
There is a little difference ?
|Apr6-12, 05:08 PM||#24|
I find the problem, the force is perpendiculary to the surface and movement.
Another question is with a fixed column of water with a fixed mass under it. Like the mass is more at left, the pressure at left in the column is higher than at right. The water move inside the column ? Or another force balance this ? (No other gravity, only mass M and water). Mass of M is 100 or higher than mass of water for example, because there are forces in water alone, the up of water attrack the down, idem for the sides, maybe already a movement in water alone.
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