Centrifugal Force: Why is it so real even from non-inertial frame? I have thought about this for quite an amount of time. The questions:
(1) Suppose we tie a ball on a merry go round with a string. Now, we observe the ball from the merry go round and from the ground. As the merry go round, catches speed, the ball moves outwards. From the merry go round, it is easy to explain the motion of the ball, and say it is the centrifugal force that makes the ball move outwards. HOWEVER, from the ground, we also DO see the ball moving outwards. But, we cannot say centrifugal force was responsible, for the outward movement. So, from the ground, some REAL force must be responsible for making the ball move outwards?? Right? Now, what is this real force or component of real force that makes the ball move outwards?
This is another way to think of it.  Suppose A and B standing on a merry go round. C is observing from the ground. Speed of the merry go around increases to such an extent, that A falls off from it in tangential direction. B would say, centrifugal force was too hard for A to bear and so he fell off (which is all right). But, what would C say to that?? C cannot blame centrifugal force for that! C has to blame some REAL force for making A fall off. What is that REAL force?
(2) This question is similar. Suppose we have a bead on a FRICTIONLESS, horizontal rod. One end of the rod is free. The other end is fixed to another vertical rod. Now, we start spinning that vertical rod about it's axis. You get the setup, right? It's somewhat like a "T" rotated 90°. Now, we observe as the rod gains speed, the bead moves outwards. The questions:
(2.1) From the ground, what REAL force is responsible for the bead's outwards movement ? From the rotatory frame, it is easy to say centrifugal force does that.. but how will you explain the outward movement from non inertial frame?
(2.2) From the rotatory point of view, if there is centrifugal force acting on the bead outwards, where is the centripetal force on the bead, acting inwards? The forces should cancel out, and the bead should stay at rest, if you viewed from rotatory frame, right?
(2.3) A time comes when the bead slips off the rod. Now, which direction does it fall off, radial to the rod or tangential to the rod? It should be tangential, as in other cases of swinging a ball and then observing it take a tangential path after releasing it suddenly. Please explain, if you think it falls off in radial direction.
 A: In the absence of any force (external observer) the object will simply remain on its (straight, constant velocity) trajectory. To the co-rotating observer this looks as if the object is "moving away" from him, but seen from the outside this is simply an object with a non-zero velocity relative to the merry-go-round moving happily along in a straight (tangential) line - no forces needed...
A: You are confusing what is real and what is imaginary and as shown in your statement
From the merry go round, it is easy to explain the motion of the ball, and say it is the centrifugal force that makes the ball move outwards.
There are forces acting on bodies which are the same irrespective of the reference  frame, constant velocity, accelerating, rotating etc, from which they are observed.
These are forces which are "real" and a centrifugal force is not one of this type of force.
In your first example the mass moves outwards not because it has any force on it, rather as the mass has inertia and as there is no force acting on it, the mass travels in a straight line as shown in the top gif image which is taken by a "camera" attached to the ground.

When observed in the rotating frame the trajectory of the mass is not a straight line as shown in the bottom gif image where you have to imaging that you are standing on the turntable on the red dot.
That is not due to any extra force acting on the mass it is a result of observing the mass relative to a different frame of reference, in this case one that is rotating.
Newton's laws of motion are good at predicting what happens in inertial reference frames and but do not predict the correct behaviour of the mass in your first example when observations are done in a rotating frame.
In the rotating frame the mass has no force acting on it any yet it accelerates when travelling along a curved trajectory.
So what to do?
Add fictitious/pseudo forces so that Newton's laws of motion can still be used in the rotating frame.
With these ideas hopefully you will note that in your second example all the bead is trying to do is move in a straight line but the horizontal rotating rod is making the bead undergo a curved path by exerting a normal force on it.
The bead tries to reduce the curvature of its trajectory by moving away from the axis of rotation.
Sitting on the axis of rotation one would see the bead accelerate away along the rod but with only a normal force acting on the bead.
To describe the motion using Newton's laws of motion a fictitious force, the centrifugal force, is introduced to account for the outward acceleration of the bead.
A: In this answer I will proceed as follows:
I will discuss inertia in terms of everyday life experience.
After that I will address the specific case mentioned in the question: ball on a merry-go-round, with a string.

The human psychology has the following quirk. When something is everywhere, you are likely to end up unaware of it. It becomes so internalized that it is never in your conscious thought.
In the following I take a lot of time to discuss inertia. Not because the subject is difficult, but because the human psyche tends to overlook inertia.

A potter's wheel is made to be quite massive; the wheel tends to rotate at a constant velocity because of that bulk. If the wheel would have just the minimum to be rigid enough it would have a relatively small mass. The bulk really helps the potter's wheel.
There is almost no penalty to making a potter's wheel have a lot of bulk; If the bearing of the axle is good quality there is little additional friction. The amount of effort it takes to sustain the desired rotation rate is hardly larger when the wheel has more bulk.
Of course, a potter's wheel with more bulk takes more effort to get going: more bulk means more total inertia.
Inertia is the absolute core of theory of motion. For sure: Inertia is forceful, but at the same time we cannot think of inertia in terms of force.
Inertia offers opposition to change of velocity, but at the same time inertia does not in any way offer resistance to velocity.

If inertia would be an opposite force then any force would be countered by an equal and opposite counter-force, and then motion would be impossible. This shows that trying to think of inertia as an opposite force is not a viable option.
If you could have some object without any inertia then the slightest push would be sufficient to accelerate it to infinite velocity.
With inertia you get a response in between the above two extremes: the response to application of force is that the acceleration follows a law of proportion. The stronger the force (pushing a given mass), the larger the resulting change of velocity. The larger the bulk of an object (given a particular force that is applied), the smaller the resulting change of velocity.

Example from everyday life:
The circumstance:
You try to open a door, but you feel that one corner is seriously stuck. You feel the door flex a little, so you know the door isn't latched, but one corner isn't moving. To get that door to open you take a step back and then you move forward briskly, slamming your shoulder against that door. You know from experience that that will do it.
By moving forward briskly you gave yourself a velocity towards that door. As you impacted that door it was up to the door to decelerate you. That deceleration takes a force, and the larger the deceleration the stronger the required force. As you slam yourself against that door: the peak force is easily larger than any static force that you are capable of producing.
Moving a heavy object
Another example is moving a heavy piece of furniture, say a heavy lounge chair. If that chair will hardly budge you can move it by bumping into it. You take a step back, and then you give yourself a bit of velocity towards that chair. So the common center of mass of you and the chair has a velocity. As you bump up against that chair you and the chair proceed as effectively a single mass, moving at the existing velocity of the common center of mass. The friction will bring the chair to a standstill almost immediately, but the chair did move.
Hammering in a nail
One of the most violent decelerations that we know from everyday life is using a hammer to hammer in a nail. Let's say that with every blow the nail is driven two or three nail widths into the wood.
You swing the hammer to give it a high velocity. At the instant that the hammer strikes: it is up to the nail to provide the force to decelerate the hammerhead to zero velocity. A massive amount of friction force has to be overpowered; the nail decelerates the hammerhead so hard that the peak force is sufficient to overpower the friction, and the nail is driven into the wood.

So I took a lot of time, discussing multiple examples.
My point is: in daily life you are dealing with inertia all the time, often using it to your advantage. You are never not dealing with inertia. The effect of inertia is very, very consistent. Inertia is among the most predictable things. Precisely because of that your conscious thought tends to be not aware of it; it's all muscle memory.


We have that inertia is the property that change of velocity requires a force.
As an object is pulled along, resulting in circumnavigating motion:
At every point in time the velocity component parallel to the circumnavigating motion is a constant velocity. In circumnavigating motion it is the motion component that is perpendicular to the circumnavigating motion that is subject to acceleration. A force is required for that acceleration.

So: I'm submitting that it is about recognizing the central role that Inertia plays. With that recognition everything falls into place.

About circular motion:
Let me return to the potter's wheel. Put some circular dish with a high rim on that potter's wheel, pour some water in the dish, and spin the potter's wheel at a constant velocity.
With that constant angular velocity the surface of the water becomes concave. The final state, with constant angular velocity and a corresponding constant shape of the surface of the water, that state is called 'solid body rotation'. Motion with a constant angular velocity has a uniformity to it that is remarkably close to the uniformity of non-accelerating motion. That degree of uniformity is quite unique, and it plays a role in why people feel compelled to suppose the existence of some centrifugal force.
However, the prime organizing principle for theory of motion is inertia.
Inertia: in order to cause change of velocity a force is required.
A: (1): If the cable that links the rotating object to the center is broken, the object moves in a straight line. In reality, if the cable is broken before the movement starts, there is no rotation at all, only straight movements.
There is some tangential force to initiate the movement from the rest. And a centripetal force along the cables to avoid the scattering of the objects linked to the center. The rotation that we see is only possible due to that force.
(2) If the rod is frictionless, the only possible force acting on the bead is tangential to the movement. So there is a tangential force what causes an initial tangential velocity of the bead from the rest.
If the rod stops spinning just after that, the bead keeps its initial straight line movement. But if it continues rotating, the point of contact with the bead is progressively shifted outward, where the radius and consequentely the tangential velocity is greater.
In this case, there is no centripetal force, (due to absence of friction), and the bead is free to move outward all the way, with increasing tangential velocity, until lose contact with the rod.
A: In the case of the bead on the rod the force that causes it to move outwards is the reaction force of the rod acting on the bead that causes it to move in the way it does (I mean that is the only net force component here, so it has to be this). We can see this by looking at Newton's second law in polar coordinates:
$$\mathbf F=m\mathbf a=m\left(\ddot r-r\dot\theta^2\right)\,\hat r+m\left(r\ddot\theta+2\dot r\dot\theta\right)\,\hat\theta$$
In the case of the bead on the rod, in the plane of the rod's rotation we have a force $\mathbf f_\text{rod}=f\hat\theta$, so we end up with (note $f$ is not constant)
$$f=m\left(r\ddot\theta+2\dot r\dot\theta\right)$$
If the rod rotates at a constant angular velocity so that $\ddot\theta=0$, then at any instant in time we can determine the instantaneous radial velocity of the bead:
$$\dot r=\frac{f}{2m\dot\theta}$$
And of course we also have from the radial component being $0$:
$$\ddot r=r\dot\theta^2$$
A: There are (only) four forces in nature.

*

*Gravitational force

*Electromagnetic force

*Weak force

*Strong nuclear force

If you cannot explain the origin of a force from this list then it is not a real force. This is that simple.
A: The problem you are running into is the word "outward". Outward is not a fixed direction in space, it is the radial basis vector in polar coordinates which changes from point to point. So if you write your physics equations in terms of "outward" then you are using polar coordinates rather than standard Cartesian coordinates.
The Lagrangian of a free particle in polar coordinates is $$L = \frac{1}{2}m\left( \dot r^2 + r^2 \dot \theta^2 \right)$$ and the Euler-Lagrange equations give us $$\ddot r = r \dot \theta^2$$ $$\ddot \theta = -\frac{2 \dot r \dot \theta}{r}$$
This first equation is essentially a fictitious force, although many authors would not classify it as one. It arises mathematically the same way that fictitious forces do, and it is proportional to the mass and undetectable by accelerometers as is the case with fictitious forces. However, it is not due to acceleration of the coordinate system, but simply the fact that the coordinate system is non-Cartesian.
So, regardless of whether you choose to classify it as a fictitious force or not, the fact remains that effect exists in polar coordinates. There is an outward directed acceleration of $r \dot \theta^2$ even in the non-rotating frame. This is the "force" that causes objects to move "outward" even from the non-rotating perspective.
A: This is just food for thought important enough to share as an answer, rather than a comment: World renown astrophysicist Kip Thorne discuses how a rotating universe would be the equivalent of a rotating Earth. During the Gravity Probe-B experiment.
https://www.youtube.com/watch?v=Wvnf6BUsyyo
