Inerital vs Non inertial reference frames: quick conceptual question

[jumbogala]

Homework Statement

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?

Homework Equations

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.

 

[Fredrik]

In the inertial frame, the centripetal force is the friction.

Do you understand what the accelerations are in the two coordinate systems?

 

[jumbogala]

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 non-inertial 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.

 

[tiny-tim]

hi jumbogala!

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.

no, you are confusing yourself by talking about "centripetal force" as if it was a separate force …

"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 = mv²/r radially inward, and

in the co-rotating (non-inertial) frame, there is no acceleration, so F_{total non-inertial} = ma = 0,

which of course only works because F_{total non-inertial} = F_{total inertial} + mv²/r radially outward,

ie we have to add a separate fictitious centrifugal force (mv²/r) to keep good ol' Newton's second law working!

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?

… 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.

you're right …in all frames, all real forces are exactly the same …

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 free-falling (inertial) frame, there is no force and no acceleration, but in a (non-inertial, compared with the rest of the universe!) frame fixed on the Earth's surface, we have to add a fictitious mg force)

 

[jumbogala]

Thank you! I'm still a little confused though.

- Why is there no acceleration in the non-inertial 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?

 

[jumbogala]

Edit: double post, sorry.

 

[tiny-tim]

oops!

Why is there no acceleration in the non-inertial 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?

sorry, i forgot the ant was crawling …

yes, you're right, there is (in the co-rotating frame) a https://www.physicsforums.com/library.php?do=view_item&itemid=86" 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 co-rotating frame is w²/r inward, while its acceleration in the inertial frame is (w + v)²/r inward

the centrifugal force (in the co-rotating frame) is mv²/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 - w²/r} = mv²/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 -\,\Omega\,r, and so experiences:
centrifugal force m\,\Omega^2\,r outward;
and Coriolis force 2m\,\Omega^2\,r inward;
net force: m\,\Omega^2\,r inward, forcing the house to move in a circle round the driver! ​

 

[Fredrik]

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?
...
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.

Maybe you understand it now, after discussing it with tiny-tim, but I'll say what I had in mind. Consider the special case of the ant not moving relative to the rotating disc. (You should make sure you understand that before you start thinking about a more complicated scenario). In the inertial frame, it's doing circular motion with constant velocity. In this case, the acceleration is towards the center, and the only force acting on the ant is friction. So friction is the force that keeps it going in a circle. Another name for that force is "centripetal force". I don't mean that these are two forces with equal values. I really mean that in this case, there's only one force. You can call it friction because of what's causing it, and you can call it centripetal force because of the motion it produces.

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.

 

[jumbogala]

if its (relative) crawling velocity is, for example, w tangentially (so r stays the same), then its acceleration in the co-rotating frame is w2/r inward, while its acceleration in the inertial frame is (w + v)2/r inward

I'm getting w and v mixed up. Is w the angular velocity of the wheel by itself, and v is the velocity of the ant within the rotating system?

Because shouldn't the acceleration of the ant within the co-rotating 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?

 

[jumbogala]

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!

 

[tiny-tim]

hi jumbogala!

(have an omega: ω and try using the X² icon just above the Reply box )

I'm getting w and v mixed up. Is w the angular velocity of the wheel by itself, and v is the velocity of the ant within the rotating system?

Because shouldn't the acceleration of the ant within the co-rotating frame depend on only the ant's velocity within that system, and so should be v^2/r?

my v is the speed of a point on the pot, my w is the speed of the ant relative to the pot, so my total speed of the ant (in the inertial frame) is v+w, my co-rotating acceleration is w²/r, and my inertial acceleration is (v+w)²/r

(if i'd meant ω … i'd have said so! )

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!

difference between w and ω !

(my v = ωr)

 

[jumbogala]

Thank you for clarifying :)

So from your equation here:

m{(w + v)2/r - w2/r} = mv2/r + 2mvw/r

, I plug in μmg for the force of friction as the inertial force, correct? (So basically I'm swapping out m(w + v)²/r for the frictional force). Then I solve for w.

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?

 

[tiny-tim]

hi jumbogala!

So from your equation here: , I plug in μmg for the force of friction as the inertial force, correct? (So basically I'm swapping out m(w + v)²/r for the frictional force).…

sorry, you still haven't got it

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

(co-rotating:) µmg - mv²/r - 2mvw/r = m*acceleration = mw²/r

my (obvious algebra) equation, m{(w + v)²/r - w²/r} = mv²/r + 2mvw/r, simply confirms that these two F = ma equations are the same ​

(and there's no such thing as "inertial force")

 

[jumbogala]

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!)