Inerital vs Non inertial reference frames: quick conceptual questionby jumbogala Tags: coriolis force, fictitious forces 

#1
Nov1310, 01:30 AM

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1. The problem statement, all variables and given/known data
I'm doing a problem in which an ant crawls in a circle on a spinning pottery wheel. Say I'm looking at the friction which holds the ant in place. It keeps the ant from slipping. Looking at it in the inertial frame of reference, I know that the centripetal force points towards the center so friction must pull the ant out, away from the center. But in the non inertial frame, is the frictional force the same one as before? Like in magnitude and direction, or does it change? Also, how do you know which direction it acts? 2. Relevant equations 3. The attempt at a solution I'm not sure. I don't think the frictional force is the same, because it's now a force within the rotating reference frame. But I'm not sure why, because the "real" forces (the ones that are due to interactions) should not change whether we are looking at it from an outsider's POV or the ant's. Please tell me if this should go in advanced physics instead. 



#2
Nov1310, 01:40 AM

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In the inertial frame, the centripetal force is the friction.
Do you understand what the accelerations are in the two coordinate systems? 



#3
Nov1310, 01:48 AM

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Wait... the question asks how fast the ant can walk without slipping. So I thought if I was doing it in a non inertial frame, I would need to calculate the centripetal force mv^2/r and equate that to the frictional force, then solve for v. What do you mean they're the same? (Sorry, it's been a long time since I did 1st year physics). Basically I want to use my answer here to check my noninertial answer.
And I don't think I do get what the accelerations are in the 2 coordinate systems =\ I couldn't tell you what they are. 



#4
Nov1310, 02:22 AM

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Inerital vs Non inertial reference frames: quick conceptual question
hi jumbogala!
"centripetal force" is never a separate force in an inertial frame, it is just a confusing (but accurate! ) name for an existing force (in this case, friction) for another example, consider a mass on the end of a string rotating in a horizontal circle under the top of the string: the only two forces are the tension (at an angle to vertical), and the weight (vertical): some people call the horizontal component of the tension the "centripetal force" … and they're right, it is! … but i fail to see how that helps in the inertial frame (in all these cases), F_{total inertial} = ma = mv^{2}/r radially inward, and in the corotating (noninertial) frame, there is no acceleration, so F_{total noninertial} = ma = 0, which of course only works because F_{total noninertial} = F_{total inertial} + mv^{2}/r radially outward, ie we have to add a separate fictitious centrifugal force (mv^{2}/r) to keep good ol' Newton's second law working! the only exception is the addition of separate fictitious forces (in uniformly rotating frames, that means the centrifugal force, the Coriolis force and the Euler force, the latter two of which are zero in most exam questions ) (btw, gravity can be considered a fictitious force … in a freefalling (inertial) frame, there is no force and no acceleration, but in a (noninertial, compared with the rest of the universe!) frame fixed on the Earth's surface, we have to add a fictitious mg force) 



#5
Nov1310, 02:31 AM

P: 400

Thank you! I'm still a little confused though.
 Why is there no acceleration in the noninertial frame? (Is it because the ant is walking at a constant speed?)  I understand where the centrifugal force comes from. But isn't there a coriolis force here too? Since the coriolis force is 2mωv', where ω is the angular velocity of the pottery wheel, and v' is the ant's velocity in the frame. The ant's velocity is not zero so shouldn't this force be present? 



#6
Nov1310, 02:34 AM

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Edit: double post, sorry.




#7
Nov1310, 02:58 AM

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yes, you're right, there is (in the corotating frame) a Coriolis force as well as a centrifugal force if its (relative) crawling velocity is, for example, w tangentially (so r stays the same), then its acceleration in the corotating frame is w^{2}/r inward, while its acceleration in the inertial frame is (w + v)^{2}/r inward the centrifugal force (in the corotating frame) is mv^{2}/r outward, the Coriolis force is 2mvw/r outward, and these two together are a fictitious force which makes up for the difference in accelerations in the two frames … m{(w + v)^{2}/r  w^{2}/r} = mv^{2}/r + 2mvw/r here's an amusing example from the PF Library about a house being observed in the frame of the driver of a car moving uniformly in a circle … The house has tangential velocity [itex]\,\Omega\,r[/itex], and so experiences: 



#8
Nov1310, 03:14 AM

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If it's not obvious to you that the acceleration is toward the center, think about F=ma and what it feels like to spin a weight attached to a string around in a circle. The force you apply is toward the center, and by F=ma, F and a must be in the same direction, so the acceleration must be toward the center. A more rigorous way to show it is to write down the mathematical expression describing the motion, and find its derivative. In the rotating coordinate system, the acceleration of an ant that's not moving relative to the disc is zero by definition of the rotating coordinate system. That's why in this coordinate system, there must be a force that's equal and opposite to the friction. This force is called the centrifugal force. 



#9
Nov1310, 03:15 AM

P: 400

Because shouldn't the acceleration of the ant within the corotating frame depend on only the ant's velocity within that system, and so should be v^2/r? And shouldn't its acceleration in the inertial frame depend on the angular velocity of the wheel, plus the ant's angular velocity relative to the wheel? I didn't think you could add w and v, without converting some things first? 



#10
Nov1310, 03:26 AM

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My book defines the coriolis force as 2mwv
and the centrifugal force as m(w^2)r So I am having trouble figuring out what exactly is meant by w and v and why your eqns look different than theirs! 



#11
Nov1310, 03:33 AM

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hi jumbogala!
(have an omega: ω and try using the X^{2} icon just above the Reply box ) (if i'd meant ω … i'd have said so! ) (my v = ωr) 



#12
Nov1310, 03:51 AM

P: 400

Thank you for clarifying :)
So from your equation here: The maximum speed of the ant before slipping should be the same in both reference frames, right? So if I wanted to do this in an inertial reference frame to check my answer, I could use m(w+v)^{2}/r = μmg and solve for w as well? 



#13
Nov1310, 04:11 AM

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hi jumbogala!
the µmg is a real force, and is completely separate from these (something)^{2}/r things the full F = ma equations are: (inertial:) µmg = m*acceleration = m(w + v)^{2}/r (corotating:) µmg  mv^{2}/r  2mvw/r = m*acceleration = mw^{2}/r my (obvious algebra) equation, m{(w + v)^{2}/r  w^{2}/r} = mv^{2}/r + 2mvw/r, simply confirms that these two F = ma equations are the same(and there's no such thing as "inertial force") 



#14
Nov1310, 04:24 AM

P: 400

Ohh, I see. I think I get it now! µmg is there because it is a real force. If there were other real forces acting on the ant we'd have to add them to µmg.
Thank you so so much for all your help, this is much clearer to me now. And thanks as well to Fredrik  your explanation above makes a lot of sense :) (And sorry that this quick question turned into a rather long one!) 


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