What exactly is the reactive centrifugal force (split)by A.T. Tags: centrifugal, force, reactive, split 

#109
Feb813, 07:58 AM

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In any case, that doesn't change the analysis for the astronaut and floor during the shear wave propagation. The floor is exerting an unbalanced centripetal force on the astronaut which is accelerating centripetally. The reaction to that force is the centrifugal force on the floor which is also unbalanced and therefore the floor is accelerating centrifugally. 



#110
Feb813, 09:58 AM

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Why does the normal force is always a bit less than the gravitational force? 



#111
Feb813, 10:14 AM

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If you add the inward gravitational force and the outward normal force together, the resultant is the small centripetal force required to keep you on this circular path. If normal force were equal to gravitational force then you would not be accelerating. You would find yourself following a straight line path tangent to the rotating earth and would quickly find yourself in outer space. 



#112
Feb813, 07:32 PM

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You could increase the radial tension artificially. For example in a bicycle wheel, you could tighten the spokes. This would have no bearing on the centripetal acceleration experienced by the rotating bicycle wheel. You would then have a centripetal force due to rotation plus a static radial tension. AM 



#113
Feb813, 09:26 PM

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#114
Feb913, 06:19 AM

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After the bolts are cut, the floor continues to rotate about its centre of mass, not the centre of mass of the space station. When it is rotating with the space station, its centre of mass is accelerating but when it is freed (bolts cut) its centre of mass stops accelerating. [itex]\vec{L} = \vec{r}\times\vec{p}[/itex] [itex]\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt}\times\vec{p} + \vec{r}\times\frac{d\vec{p}}{dt} [/itex] [itex]\frac{d\vec{L}}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times\frac{d\vec{p}}{dt} [/itex] [itex]\frac{d\vec{L}}{dt} = 0 + \vec{r}\times\frac{d\vec{p}}{dt} = \vec{r}\times\vec{F}[/itex] If [itex]\frac{d\vec{L}}{dt}=0[/itex] then [itex]\vec{F}[/itex] is entirely radial: [itex]\vec{r}\times\vec{F}=0[/itex] and this is called the centripetal force. AM 



#115
Feb913, 06:58 AM

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If you are not so familiar with stress and strain then perhaps it will help to think of the astronaut as standing on a spring. While the spring is decompressing the "top" continues to accelerate centripetally while the "bottom" and the COM accelerate centrifugally. It isn't quite the same since the shear wave is progressing laterally while the spring is expanding radially, but both are governed by Hooke's law, so they are closely related. Again, the term "reactive centrifugal force" is common and you should be familiar with it. You have valid reasons for disliking it, but those reasons don't make the term go away. I sympathize with your position since I have similar objections to other terms that just won't go away, no matter how I wish they would (e.g. "relativistic mass"). 



#116
Feb913, 02:31 PM

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It seems to me that the source of the "force" is the relaxation of the tension, so there is no net force on the centre of mass of either the astronaut or the floor. The relaxation of the tension creates a compression wave that propagates through the floor material much the same way that a stretched spring would oscillate if you were swinging it around and then let it go. The duration of those oscillations would depend on the damping forces within the spring. Are you suggesting that the centre of mass of the spring would still be undergoing centripetal acceleration while the oscillations continued? AM 



#117
Feb913, 04:55 PM

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That is the key difference between the floor example and the spring example. They both follow Hooke's law, but for the spring example, the wave would be a compression wave propagating radially rather than a shear wave propagating laterally. In both cases, the forces are radial. If it is easier for you to grasp the spring example then we can do that. 



#118
Feb913, 08:56 PM

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It is very easy to create an acceleration that is outward from the centre of rotation at a given moment. You would just have to have an extended object rotating about a central point at speed ω and also rotating about its centre of mass at speed ω+Δω. This is what happens in some amusement rides (eg. Tilt A Whirl). You can set up compressed springs and then release them, sending mass outward. Of course, these are not third law pairs to the centripetal force. AM 



#119
Feb913, 09:59 PM

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So if you have trouble understanding shear waves we can do compression waves. But I don't know why you would say that the physics cannot depend on the nature of the wave. Do you have access to a statics textbook? I am glad to help you with shear, but if you prefer we can switch to the compression example. Just let me know your preference. 



#120
Feb913, 11:06 PM

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AM 



#121
Feb1013, 03:41 AM

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If you want a scenario where the centrifugal reaction causes centrifugal acceleration in the rotating frame, then take a turntable where a mass on rollers pushes a sliding mass outwards . If you want a scenario where the centrifugal reaction causes centrifugal acceleration in the inertial frame, then take the rocket on a circular path. Of course, none of this is really important, because Newton's 3rd applies regardless of acceleration. 



#122
Feb1013, 07:46 AM

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When the bolts are cut the shear stress in the floor next to the bolts is relieved, but not in the rest of the floor. A shear wave ripples through the material at the speed of sound in the material relieving the shear stress as it goes. Until the shear wave reaches it, the material is in the same state of stress as it was before the bolts were cut. I can go into more detail if you like, but A.T. is right, none of it is important. The centrifugal reaction force depends on the point of application of Newton's 3rd law, not any acceleration. You can argue with me and A.T. if you like, but even if you do convince us that it is bad terminology, that wouldn't make the bad terminology go away. 



#123
Feb1013, 08:37 AM

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Centripetal force on a body defined as is dp/dt of the body where dL/dt of the body = 0. But if dL/dt for the rocket + fuel = 0 the rocket would have to constantly be increasing its rotational speed because it is constantly losing mass. AM 



#124
Feb1013, 09:53 AM

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#125
Feb1013, 03:41 PM

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#126
Feb1013, 03:50 PM

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