I Forward momentum of mass due to inertia

AI Thread Summary
The discussion explores the dynamics of a rotating axis with attached rods and balls, focusing on how inertia affects momentum transfer. As the axis rotates, balls move outward due to inertia, impacting the ends of the rods and creating upward momentum in the assembly. This upward movement is momentarily countered by the balls' weight when they reach the rod ends, leading to a complex interaction with external forces, such as an electromagnet. The conversation clarifies that while there may be a brief upward jump of the axis, it does not constitute a reactionless drive because the system's weight and forces ultimately balance out. The conclusion emphasizes that the added weight from rotational inertia complicates the motion, requiring significant force to counteract it.
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Help me understand a concept I came across by accident. So there is an axis (red) that is rotating with two rods attached to it (45 degrees from axis and 90 degrees with respect to one another) now if the balls at first are located closest to the red axis , as the axis begins to rotate the balls are dragged by inertia towards the ends of their rods, as they hit the ends of the rods they will transfer their momentum to the rods themselves, so there will be a push which will cancel out because there is an exact opposite action at the other side at the other rod, but since the rods are at 90 degrees , will there not be a force midway between them at 45 degrees which is the center of the red axis, so what I am asking is will the axis jump upwards as the balls hit the ends of the rods?
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Conservation of momentum in the vertical direction demands an answer of Yes, provided there are no components of externally applied forces in that direction to invalidate application of the conservation rule.

Note there must be some externally applied forces, otherwise we cannot have that 'the axis begins to rotate'. However we can assume that those forces have no components in the vertical direction.

Given that, the assembly has a positive upwards momentum, being that of the balls. When they hit the ends and either stick there or rebound, they lose upward momentum. To conserve momentum, the vertical rod and arms must begin to move upwards.
 
@andrewkirk ok exactly that is what I thought too, so if this then is so , how come this is then not a "reactionless drive" ? because as the balls hit the rod endings the axis jumps, but if the balls are then dragged back down say by a electromagnet attached to the vertical red axis, then again the axis moves upward slightly , and as the balls hit the rod ends once more the axis again should move upwards ?
 
As the balls move upwards the axis moves downward. If the balls are moving downwards the axis moves upwards. When the balls stop moving up or down, so does the axis.

That's why it's not a reactionless drive. When you turn on your electromagnet you reverse the downward progress your axis made.
 
I think I now understand @Ibix I think at first if you start with the balls close to red axis as you then start to spin the axis the balls hit the rod ends and at first the axis indeed jumps so for a brief moment it works like a reactionless drive but then now the balls are at the rod ends and since they are now at their furthest position from the axis due to rotational inertia they become much heavier than when they were close to axis, so now in order to pull them down with electromagnet attached to axis , the magnet would have to pull the weight equal not just to the balls but the ball weight + added weight due to rotational inertia, this added weight would make the pulling of the balls to exert a larger reaction on the magnet side attached to the axis so the axis would then be dragged back by the amount it jumped up in the first place
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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