# Trying to gain better intuition about fictitious forces

Recently started doing some reading into fictitious forces. It's easy enough to understand when dealing with linearly accelerating frames but I find it more challenging to understand rotating frames like the earth or a merry-go-round. Can someone clarify if I make any errors in the following statements?

If I was sitting still on a rotating merry-go-round some distance r from the centre of rotation, then I would feel a force pointed radially outward (the centrifugal force). This is also easy to understand if you've ever been on a merry-go-round.

However, if I started walking towards the edge of the merry-go-round in a straight line, I would still feel the centrifugal force pointing radially outwards. While it seems that I am walking in a straight line, my path would be a curved line when viewed from an inertial reference point. So there is some force that is causing me to deviate from a straight line (The Coriolis force) Does this force point in the direction of the rotation or the opposite? I would assume it points in the direction of rotation. If im rotating cw, then I have to move left as I move towards the rim to make my path straight so the Coriolis force points right?

Finally, let's say I move in a circle around the merry-go-round. I would still feel a force pushing me outwards but would there be any Coriolis force? The merry-go-round is not rotating under me so i would assume there is no Coriolis force in this instance. However, i imagine it would feel as if there was a force pushing me from behind if i move in the direction of rotation and pushing me from in front if i move opposite the rotation. It would feel the same as if a car was accelerating or braking linearly. Is this true? if so is there a given name for such a force?

• Commented Apr 8, 2023 at 1:51
• Can you make a diagram with a circle?
– user324939
Commented Apr 8, 2023 at 3:09

In an inertial frame, $$F=ma$$. In particular for an observer in an inertial frame, $$a=0 \iff F=0$$.

You want to use the merry go round as your reference frame. This is an accelerated frame. You are at rest in this frame if you occupy a point with fixed coordinates relative to the merry go round.

You want to do physics in an accelerated frame. You want to use $$F=ma$$. You can make it work.

I will assume you have gone through the math, and describe the reasons why it works on a more intuitive level. It will be much the same as your description. But I hope filling in a few more steps will clarify things.

# Object at rest on the merry go round

We start with the simple case of an object at rest in the accelerated frame. An inertial observer sees that a point with fixed coordinates on the merry go round is accelerated. An object at rest in that frame must have a force on it keeping it at rest. You have to hold on to the merry go round to stay at rest. For circular motion, the name of that inertial frame force is the centripetal force.

You can do a coordinate transform from the inertial frame to the merry go round frame. If you work it out, to get your acceleration you have to add an acceleration to the inertial acceleration.

$$a_{merry \space go \space round} = a_{inertial} + a_{transformation}$$

The inertial observer travels tangent to your circle and momentarily matches your speed. As he passes, he sees you at rest and accelerating straight inward. Soon he sees your trajectory curve, but that is a more complex case than at rest. We save it for later.

You are at rest in the merry go round frame. $$a_{merry \space go \space round} = 0$$. You have a centripetal acceleration in the inertial frame. $$a_{inertial} = a_{centripetal}$$. Therefore $$a_{transformation} = -a_{centripetal}$$. This is an outward acceleration.

You can see this by watching the inertial observer. From your point of view, the inertial observer is momentarily at rest and accelerating straight outward. He is unaccelerated in the inertial frame. $$a_{inertial} = 0$$. His merry go round acceleration is $$a_{merry \space go \space round} = 0-a_{centripetal}$$. The name of this outward acceleration is centrifugal acceleration.

$$a_{merry \space go \space round} = a_{inertial} + a_{centrifugal}$$ $$a_{centrifugal} = -a_{centripetal}$$

# Forces on an object at rest

In the merry go round frame, an inertial observer is accelerated outward. The inertial observer must therefore be experiencing an outward force. The name of this force is the centrifugal force.

It is easy to show forces transform like accelerations, but it hides a bit of a cheat. You just multiply the acceleration equation by $$m$$.

$$ma_{merry \space go \space round} = ma_{inertial} + ma_{centrifugal}$$

$$F_{merry \space go \space round} = F_{inertial} + F_{centrifugal}$$

The cheat is that $$F_{inertial}$$ has a physical cause. You are in uniform circular motion in the inertial frame because you hold on to the merry go round and pull yourself toward the center. The inertial observer has no acceleration, and therefore his total force is $$0$$.

In the merry go round frame, the force on the inertial observer is $$F_{centrifugal}$$. Where does this force come from?

It comes from the same place as $$a_{centrifugal}$$ - the coordinate transformation. Forcing $$F=ma$$ to work in an accelerated frame of reference. There is no physical cause. For this reason, this type of force is called fictitious.

Adding this fictitious force makes $$F=ma$$ correctly describe motion in the merry go round frame. For example, you are at rest. Your total force must be $$0$$. You keep yourself at rest by holding on to the merry go round and pulling yourself toward the center. There is also a fictitious centrifugal force acting on you.

$$F_{merry \space go \space round} = F_{centripetal} + F_{centrifugal} = 0$$

One further point.

$$F_{centrifugal} = m \omega^2 r$$

For a given object on our uniformly rotating merry go round, $$F_{centrifugal}$$ depends only on position.

# Forces on a moving object

The transform from inertial frame to merry go round frame gets more complex if the object is moving. You get two fictitious forces: $$F_{centrifugal}$$ and $$F_{coriolis}$$. $$F_{centrifugal}$$ depends on position. $$F_{coriolis}$$ is proportional to velocity.

$$F_{merry \space go \space round} = F_{inertial} + F_{centrifugal} + F_{coriolis}$$

The merry go round is covered with ice. You are wearing ice skates which you keep pointed perpendicular to the radius vector. The skate dig into the ice and keep you at a constant radius. But they slide without friction around a circle of constant radius. You are at radius $$r_0$$, and you have velocity $$-\omega r_0$$ in the merry go round frame so that you are at rest in an inertial frame.

You begin to step outward at a constant velocity. In the inertial frame, you are moving straight outward. But you are moving to successively larger radii. Each ring of the merry go round is moving over the inertial frame faster, but your component of sideways velocity does not change in the inertial frame. You acquire a tangential velocity in the merry go round frame. Because your outward speed is constant, your tangential velocity grows at a constant rate. You have a constant acceleration in the merry go round frame, even though you are unaccelerated in the inertial frame. This acceleration is explained by the fictitious coriolis force.

In this case, the direction of the force is against the direction of the merry go round's rotation. You should be able to see that if you stepped inward, the force is with the direction of rotation.

Hopefully, you can also see that the coriolis force would be the same if you started at rest in the merry go round frame.

Start at rest in the merry go round frame at radius $$r_0$$. In the inertial frame, you are traveling in a circle at angular velocity $$\omega$$, so you are experiencing centripetal force $$F_{inertial} = -m \omega^2 r$$.

In the merry go round frame, you are balancing the centripetal force against the centrifugal force. The total force is $$0$$.

Now skate at constant speed in a circle at this radius in the direction of rotation at angular velocity $$\omega_{skate}$$. In the inertial frame, your speed has increased, so the centripetal acceleration has increased to $$F_{inertial} = -m (\omega+\omega_{skate})^2 r$$.

In the merry go round frame, you are now traveling in a circle of constant radius around the axis at $$\omega_{skate}$$. The total force on you is $$F_{merry \space go \space round} = -m \omega_{skate}^2 r$$. Your radius has not changed, so $$F_{centrifugal} = m \omega^2 r$$.

Lets add up all the forces

$$F_{merry \space go \space round} = F_{inertial} + F_{centrifugal} + F_{coriolis}$$

$$-m \omega_{skate}^2 r = -m (\omega+\omega_{skate})^2 r + m \omega^2 r + F_{coriolis}$$

So

$$F_{coriolis} = 2 \omega \omega_{skate} r$$

$$F_{coriolis}$$ is a fictitious force proportional to your velocity in the outward direction.

• "There is no physical cause. For this reason, this type of force is called fictitious." Perhaps it's better to just say it's fictitious because it doesn't arise from some sort of interaction (i.e. N3L breaks down). One might argue that the physical cause is inertia, but in any case "physical cause" is an ambiguous phrase. Commented Apr 9, 2023 at 5:08

mmesser314's answer is excellent, but I did want to point out one main flaw in your ideas presented.

While it seems that I am walking in a straight line, my path would be a curved line when viewed from an inertial reference point. So there is some force that is causing me to deviate from a straight line (The Coriolis force).

The Coriolis force isn't present in the inertial frame, so it doesn't describe your trajectory in the inertial frame. All motion in the inertial frame can only be described by forces that obey Newton's third law ("real" forces): forces between you and the merry-go-round.

The Coriolis force comes into play in the rotating frame in order to describe your straight line trajectory in that frame. In order to maintain a fixed angular velocity (in the inertial frame), you apply a tangential force in the direction of rotation; the Coriolis force counteracts this so that you observe no tangential acceleration in the rotating frame.

So on the contrary, the Coriolis force keeps you on the straight line trajectory in the rotating frame as opposed to explaining your curved trajectory in the inertial frame.