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Wednesday, 1 March 2017
quote [ Simply put, it is a novel arrangement that has a linear action of a mass escaping a rotating wheel while the centrifugal force acts upon the remaining liquid mass to maintain balance, thereby converting the expected linear reaction of imbalance into an angular (rotational) reaction of decreased wheel speed. ]
Finally built and tested. It works.
I admit it's odd to me.
Centrifugal balancers(LeBlanc Balance) have been around for almost a hundred years using the "outside acceleration" of the centrifugal force. We've known since Newton that the first two laws of motion and Conservation of Momentum don't behave as expected/apply in non-inertial systems(where there is an outside source of acceleration). And we've known just as long that an inertial release from a spinning object has a reaction caused by the sudden change in mass distribution relative to the axis of rotation. And yet ... combining these facts in one demonstration unit causes cognitive dissonance. It's a truly odd situation
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endopol said @ 8:21pm GMT on 2nd March
Ignore external forces for a second. Before detaching, the Earth (i.e., everything except the ball) and ball were orbiting the Earth-ball center of gravity, a point slightly removed from the Earth center of gravity:
r_E * m_E = r_B * m_B
Here m_B and m_E are the respective masses of the ball and the Earth, and r_B and r_E are their respective distances from the Earth-Ball CG. r_B is approximately the radius of the Earth, or 6.4*10^6m, so r_E is approximately 4.6*10^-19 meters (assuming an Earth mass of 1.4*10^25 pounds)
At detachment, the Earth and ball fly off tangent to their respective orbits. Since they were both orbiting at ω ~ 1 revolution/day, or 7.3*10^-5 radians/second, the ball flies off at
v_B = ω * r_B ~ 467 m/s
the Earth flies off at
v_E = ω * r_E ~ 3.4*10^-23 m/s.
In the opposite direction. So there is a non-zero impulse on the Earth. In fact, observe that
v_E = (v_B / r_B) * r_E = v_B * m_B / m_E
So conservation of momentum, in the usual sense, applies
endopol said @ 1:11am GMT on 3rd March
Ignore external forces for a second. Before detaching, the ball and the Earth (i.e., everything except the ball) were orbiting the Earth-ball center of gravity, a point slightly removed from the Earth center of gravity. They obey a mass-balance equation,
(1) r_E * m_E = r_B * m_B,
where m_B and m_E are the respective masses of the ball and the Earth, and r_B and r_E are their respective distances from the Earth-Ball CoG. r_B is approximately the radius of the Earth, or 6.4*10^6m, so r_E is approximately 4.6*10^-19 meters (assuming an Earth mass of 1.4*10^25 pounds).
At detachment, the Earth and ball fly off tangent to their respective orbits. Since they were both orbiting at ω ~ 1 revolution/day, or 7.3*10^-5 radians/second, the ball flies off at
(2) v_B = ω * r_B ~ 467 m/s,
and the Earth flies off at
(3) v_E = ω * r_E ~ 3.4*10^-23 m/s
in the opposite direction. So there is a non-zero impulse on the Earth. In fact, observe that
v_E = ω * r_E = (v_B / r_B) * r_E = v_B * m_B / m_E.
So conservation of momentum, in the usual sense, applies.
<-- Entry / Current Comment
endopol said @ 8:21pm GMT on 2nd March
Ignore external forces for a second. Before detaching, the ball and the Earth (i.e., everything except the ball) were orbiting the Earth-ball center of gravity, a point slightly removed from the Earth center of gravity. They obey a mass-balance equation,
(1) r_E * m_E = r_B * m_B,
where m_B and m_E are the respective masses of the ball and the Earth, and r_B and r_E are their respective distances from the Earth-Ball CoG. r_B is approximately the radius of the Earth, or 6.4*10^6m, so r_E is approximately 4.6*10^-19 meters (assuming an Earth mass of 1.4*10^25 pounds).
At detachment, the Earth and ball fly off tangent to their respective orbits. Since they were both orbiting at ω ~ 1 revolution/day, or 7.3*10^-5 radians/second, the ball flies off at
(2) v_B = ω * r_B ~ 467 m/s,
and the Earth flies off at
(3) v_E = ω * r_E ~ 3.4*10^-23 m/s
in the opposite direction. So there is a non-zero impulse on the Earth. In fact, observe that
v_E = ω * r_E = (v_B / r_B) * r_E = v_B * m_B / m_E.
So conservation of momentum, in the usual sense, applies.
