Proof of Action Without Equal and Opposite Reaction

 c. April 2004, Revised August 2006
by Montalk 

Abstract

Newton's Third Law states that for every action there must be an equal and opposite reaction. This is commonly interpreted to mean that it is impossible for an isolated system to propel itself one way without pushing upon its environment in an equal and opposite way. It also is taken to imply that linear momentum is always conserved, and angular momentum is always conserved. This gives rise to the Law of Conservation of Linear Momentum, and the Law of Conservation of Angular Momentum. Here it will be analyzed whether these laws are still valid in systems whose reactions are part angular and part linear. If linear can be converted to angular, then there will be a net linear component allowing a device to move without pushing against its environment.

Calculations

Below is a calculation by Luis Alberto Pérez analyzing a system where a ball collides with a bar, sending the second into translational and rotational motion:

Discussion

A violation would come about if one assumes the ball has solely linear momentum when in motion, which then must necessary become angular when the bar rotates upon collision. However this is incorrect because relative to the rotational axis of the bar, the ball does have some angular momentum of value m*v*l -- part of which transfers to the bar's rotation upon collision and part of which stays with the ball as it moves onward. Likewise, the ball has linear momentum to begin with, and transfers some of this to the bar's own translational velocity after collision. So the ball starts with both linear and angular momentum, and each separately and conservatively is transferred in part to the bar. If the ball is aimed right at the axis, there is no initial angular momentum, and as expected afterwards the bar does not rotate, merely translate. If the ball is aimed at the tip of the bar, then it has less linear momentum and more angular momentum relative ot the bar, and as expected it would rotate much and translate little.

This is most evident in a simplified situation where in its initial state, the ball moves but the bar is stationary, where v1p and v2p and w are the final velocities of the ball, the bar, and the rotational frequency of the bar. Here are the mathematica-formatted equations:

In the simplest case where m1 = m2 and L = R, Conservation of Linear Momentum is most apparent. Of course, this was built into the assumption used to construct the equations, but as explained in the preceding paragraph, that assumption makes sense.

This shows that while other systems might lead to a violation, at least in this particular example there is none. I do find it counter-intuitive, however, that the ball's translational velocity alone gives rise to both a linear momentum and angular momentum, and that these values or the ratio between them, rather than being absolute in themselves, depend on the position of the bar, whether the bar's axis is in line or off center from the ball's line of motion.

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