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What exactly is the reactive centrifugal force (split) 
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#1
Jan2913, 09:43 AM

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Mod note: This thread on the reactive centrifugal force was split from this thread on the (fictitious) centrifugal force.



#2
Jan2913, 12:54 PM

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So one could say that the reaction force to the centripetal force on one astronaut is that astronaut's (centripetal) force on the other astronaut arising from the structural tensions within the space station structure. In that sense, all the forces are centripetal. AM 


#3
Jan3013, 04:03 AM

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#4
Jan3013, 08:45 AM

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What exactly is the reactive centrifugal force (split)
AM 


#5
Jan3013, 08:48 AM

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Thanks
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You guys are going to have to agree to differ. The 'other guy' is not actually wrong, in this case. Can't you both see that?



#6
Jan3013, 09:16 AM

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#7
Jan3013, 08:48 PM

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Newton's third law does not talk about tensions. Newton talks about "actions" and "reactions which mean "changes in motion". The third law says "To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts." In other words, changes in motion cannot occur in just one body. The change in motion (momentum) of one body requires an opposing change in motion of another body. The "reaction" to any "action" is an equal and opposing change in momentum of another. AM 


#8
Jan3013, 09:27 PM

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Let [itex]\stackrel{\rightarrow}{F_I}[/itex] be the force on object [itex]I[/itex]. Then, if the only forces are twobody forces (which is the case for many problems), we can write: [itex]\stackrel{\rightarrow}{F_I} = \sum_J \stackrel{\rightarrow}{F_{IJ}}[/itex] where [itex]\stackrel{\rightarrow}{F_{IJ}}[/itex] is the force on object [itex]I[/itex] due to object [itex]J[/itex]. The assumption of equal and opposite forces is just the assumption that [itex]\stackrel{\rightarrow}{F_{IJ}} =  \stackrel{\rightarrow}{F_{JI}}[/itex] Ultimately, it's certainly true that changes of momenta have to cancel, because the total momentum can't change due to internal forces. However, it seems to me that accounting for the changes in momenta is more systematic by using pairwise forces, and using the principle of equal and opposite forces. 


#9
Jan3013, 09:58 PM

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Since, in nonrelativistic interactions, the duration of interaction is the same for both interacting bodies, the changes in momentum of the interacting bodies are equal and opposite: Body 1: Force exerted by body 1 on body 2 = F1 = dp2/dt ⇔∫F1dt = ∫dp2 = Δp2; For body 2: F2 =  F1 ⇔ ∫F2dt = Δp1 = ∫F1dt = Δp2 AM 


#10
Jan3013, 09:59 PM

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Consider a person standing on the ground. There is a gravitational force down and a normal force up, the net force is zero and therefore there is no change in momentum. So according to you there is no normal reaction force on the ground and also no gravitational reaction force since there is no change in momentum. This is NOT how Newtons third law works. It is the individual forces which are in action reaction pairs, not the net force. 


#11
Jan3013, 10:02 PM

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Your statement is a misrepresentation of Newtons third law. 


#12
Jan3013, 10:18 PM

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In this setup, it's not the case that the change in momentum of [itex]O_1[/itex] is equal and opposite to the change in momentum of [itex]O_2[/itex], because [itex]O_1[/itex] is not the only object acting on [itex]O_2[/itex]. It's certainly true that the total change in momentum is zero, if you add up the changes for all three objects. But they don't come in action/reaction pairs. To me, what comes in pairs is "the force on [itex]O_1[/itex] due to [itex]O_2[/itex]" and "the force on [itex]O_2[/itex] due to [itex]O_1[/itex]" 


#13
Jan3113, 03:12 AM

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I also don't understand what you mean by the gap between understanding and truth here: 


#14
Jan3113, 03:27 AM

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#15
Jan3113, 05:36 AM

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If you still believe in the possibility that terminology can be unfortunate and can give people the wrong idea, then why do you object so strenuously when I suggest that YOUR terminology can be unfortunate and can give people the wrong idea? 


#16
Jan3113, 05:52 AM

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As I pointed out, one of your first posts made complaints about what you considered a bad way to think about models, when you said that the phrase "fictitious forces" is unfortunate, and can give people the wrong idea. 


#17
Jan3113, 06:23 AM

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#18
Jan3113, 06:45 AM

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