# Fictitious forces

1. Sep 16, 2013

### stripes

1. The problem statement, all variables and given/known data

Here is the question, copied verbatim from the assignment:

You are sitting on a chair in the rotating co-ordinate system, going round and round. You are sitting at the centre of a turntable in fact. Directly above the turntable, a horizontal distance R from the centre of the turntable, and suspended from the roof of the non-rotating building the turntable is in, is a mass (1) on a string. Also, rotating with the table is a support from which a second mass (2) is suspended by a string.

An inertial observer sitting besides the turntable sees mass (1) stationary, its string hanging vertically. They see mass 2 going in circles and the string at an angle so that a component of the tension provides a net force pointing to the centre of the circle, hence providing the ‘centripetal’ force which allows the mass (2) to go in circles.

The rotating observer sees mass (2) being drawn to the outside by the centrifugal force. This observer sees ALL objects experience a force, the centrifugal force which is proportional to its mass and the distance from the axis of rotation ( rω2 eh?) so this is no surprise. This observer sees the mass (2) stationary with the centrifugal force cancelling the horizontal component of the tension. This rotating observer expects mass (1) to have a centrifugal force too, however this would point OUTWARDS. The rotating observer sees mass (1) to be going in circles which would require a net force to point INWARDS. How do we resolve this dilemma??

2. Relevant equations

3. The attempt at a solution

i honestly don't know. the person going round and round sees a ball going around him. he expects to see a centripetal force that makes the ball go in circles. but he doesn't see the string at an angle. do i need to talk about pseudo-pseudo-forces? i mean really it's cause mass (2) just isn't moving in any frame, inertial or non-inertial, to begin with, so no one sees ANYTHING at all. the way i see it, it gravity and string tension, and nothing more, in no frame of reference anywhere. because it's not moving in ANY frame. but that's not right...how do i explain the fact that the string is not at an angle?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 16, 2013

### TSny

In the rotating frame, there is another fictitious force acting on m1 besides the centrifugal force. Can you identify it?

3. Sep 17, 2013

### stripes

There are only four fictitious forces that we've mentioned (and these are the most common, perhaps the only ones). One due to acceleration of the frame, coriolis, centrifugal, and azimuthal. we have not discussed azimuthal, and it doesn't apply here since the frame is rotating at constant angular velocity (at least we assume it to be so since we won't be talking about azimuthal forces). centrifugal is clear, but what is cancelling it? it's not one due to acceleration. so it's got to be coriolis. but this confuses me further. since the object isn't moving in the rotating frame how does it apply here?

4. Sep 17, 2013

### TSny

[Edited]

Yes, Coriolis.

But, as you said in your first post, m1 is moving around in a circle from the point of view of someone in the rotating frame.

How strong is the Coriolis force and in what direction does it act?

Last edited: Sep 17, 2013
5. Sep 17, 2013

### stripes

I only know the answers because I can figure it out from the information provided. Clearly the Coriolis force acts in the direction opposite of the centrifugal force, and is of equal magnitude. That's because the string makes no angle; it is perpendicular to the ground in all frames.

I suppose I should review the Coriolis force, what it does, and what it is the result of exactly.

Edit: I might not even be correct. This is just how I see it.

6. Sep 17, 2013

### TSny

See if you can calculate the Coriolis force. You should find that it does not have a magnitude equal to the centrifugal force.

7. Sep 17, 2013

### stripes

Well if the centrifugal force points outwards and the Coriolis force points inwards, and the ball is dead in the middle (no angle), can I not conclude they are of equal magnitude?

8. Sep 17, 2013

### TSny

No. If the forces "balance out" on an object, then the object must either remain at rest or move along a straight line with constant speed. But from the point of view of the rotating frame, $m_1$ is moving in a circle. So, in this frame, the forces (including fictitious forces) cannot add to zero.

9. Sep 18, 2013

### stripes

the idea you presented makes sense but in my mind i'm hopping between inertial and non inertial reference frames and it's hard for me to add them up.

to me the reason why the mass is rotating for this person is because the person is rotating. there are absolutely no forces whatsoever acting on that m1. or centrifugal and coriolis cancel. we can't have extra forces lying around, to me it just seems like we gotta have two things cancel or nothing at all.

10. Sep 18, 2013

### stripes

i think of it like this...i'm in an airplane or car, and it makes a sharp turn. i can't see anything in the airplane so it seems like something is pushing me outwards (centrifugal). but if i remain in the same spot even though this turn is being made, then there absolutely must be a force equal to and opposite of the centrifugal force that keeps me still. of course the ball is rotating in circles and not "still" to that person but that's because his whole frame is rotating

11. Sep 19, 2013

### TSny

We have two frames of reference : The “lab” frame which is an inertial frame and the “rot” frame which is rotating relative to the lab frame and is a non-inertial frame. We have an object which is being observed in both frames.

1. The motion of the object relative to the lab frame is generally different than the motion relative to the rot frame.

(For example in your problem, $m_1$ remains at rest in the lab frame, but it moves in a circle at constant speed relative to the rot frame.)

2. Thus, the acceleration $\vec{a}_{lab}$ of the object relative to the lab frame is generally different than the acceleration $\vec{a}_{rot}$ of the object relative to the lab frame.

(For example in your problem, $m_1$ has zero acceleration in the lab frame, but it has a centripetal acceleration relative to the rot frame.)

3. For the inertial frame, Newton’s second law holds: $\sum{\vec{F}} = m \vec{a}_{lab}$ where the left side is the sum over all of the “real” forces acting on the object.

(For example in your problem, $m_1$ has two “real” forces acting on it: the tension in the string and the force of gravity. These forces add to zero which agrees with the fact that the acceleration of the object is zero in the lab frame.)

4. For the rot frame, Newton’s second law $\sum{\vec{F}} = m \vec{a}_{rot}$ does not hold if the left hand side is the sum of the “real” forces acting on the object. This is because the left hand side is the same as for the lab frame, but the right hand side has changed because $\vec{a}_{rot} \neq \vec{a}_{lab}$.

(For example in your problem, the sum of the real forces (tension and gravity) on $m_1$ equals zero, but $\vec{a}_{rot} \neq 0$.)

5. However, if you allow the rotating frame to include “fictitious forces” $\mathscr{F}$ as well as “real” forces F, then the rotating frame can still apply Newton’s second law in the form $\sum{\vec{F}} + \sum{\vec{ \mathscr{F}}}= m \vec{a}_{rot}$

(For example in your problem, there are two real forces and two fictitious forces acting on $m_1$ in the rot frame. The real forces are tension and gravity and the fictitious forces are Coriolis and centrifugal.)

---------------------------------------------

So, you should try to verify that $\sum{\vec{F}} + \sum{\vec{ \mathscr{F}}}= m \vec{a}_{rot}$ holds true for $m_1$ in the rot frame.

12. Sep 19, 2013

### stripes

I'm thinking quite simply, the coriolis force acts inward (i.e., opposite of centrifugal), but is greater in magnitude than the centrifugal force which allows it to rotate in rot frame. I'm going to put that down as the assignment is now due. the magnitude as a proportion of anything else i'm given is a little harder to wrap my head around.

13. Sep 19, 2013

### arildno

"the coriolis force acts inward"
Dead wrong. The radial component of the Coriolis forces will be outward or inward depending on travelling east or west.
We call this the Eötvös effect:
https://en.wikipedia.org/wiki/Eötvös_effect

14. Sep 19, 2013

### stripes

then honest to goodness i cannot to save my life see what is making the mass (1) not make an angle. why on earth is the string not angled? the assignment question even exclaims that there absolutely, positively must be a net force pointing inwards.

unless there are more forces i have not accounted for, it seems the coriolis force is the last force to add here. thus, by the instructor's wording, it must be THE net force inwards.

15. Sep 19, 2013

### TSny

That's correct. For mass $m_1$, the Coriolis force will be inward and greater than the centrifugal force. That's true no matter which direction the rotating frame is rotating (clockwise or counterclockwise as viewed from above in the lab frame).

16. Sep 19, 2013

### TSny

Yes, in the rotating frame there must be a net force inward. The "real" forces cancel to zero, but the sum of the centrifugal force outward and the Coriolis force inward yields a net inward force (centripetal).

There are not any more forces. The net inward force is the magnitude of the Coriolis force minus the magnitude of the centrifugal force.

17. Sep 19, 2013

### TSny

For the mass $m_1$ described in the problem, the Coriolis force will be inward. If the rotating frame rotates so that its angular velocity vector is upward in the lab frame, then as $m_1$ passes in front of the observer in the rotating frame, it will pass from his left to his right. If the rotating frame spins the other way so that its angular velocity vector is downward in the lab frame, then $m_1$ will pass from the rotating observer's right to his left.

Either way, the Coriolis force on $m_1$ will be inward toward the axis of rotation of the rotating frame.

Last edited: Sep 19, 2013
18. Sep 19, 2013

### stripes

makes a lot of sense, but arildno's response is a little troubling. according to one hw helper i am dead wrong, but according to you, i was correct...

19. Sep 19, 2013

### TSny

What to do? :uhh:

You could look up the formula for Coriolis force and apply it to your problem. That way, you can see for yourself whether or not the Coriolis force for $m_1$ is inward or outward. If you have any trouble applying the formula, show your work and we can help you out.