Why is the centrifugal force radially outwards?

Question

The tendency of a body placed on a rotating frame (say with a uniform velocity) is to fly off tangential to the velocity of the plane. Why then does the inertia of the body tend to take it radially outwards (which appears to be the centrifugal force for an observer on the plane)? I am talking from an inertial frame, where I can invoke inertia of the body. Shouldn't the inertia be tangential? My question is simply, how do I explain myself that the inertia (or the centrifugal force) has to be outward given that the tendency of the body is to fly off tangentially?

Answer 1

In the rotating reference frame (there is no need to mention observers) bodies do not fly off tangentially, but radially. This is because you need to subtract the rotational motion of the inertial frame.

Therefore, the "fictituous centrifugal forces" in the rotating reference frame are pointing outward.

Note, that in an inertial frame there is no fictituous force whatsoever. The reason bodies tend to fly off tangentially in the inertial frame is just Newtons first axiom, which applies as soon as you remove the centripetal force (e.g., you let loose of the rope).

Answer 2

If you watch a body in a circular motion from a non accelerated frame of view outside the body, the body always has tendency to go tangentially to circle. But we all know that it doesn't go tangentially but it is in circular motion. Hence there must be a force which is changing the direction of body at every instant. This force is the centripetal force. Centripetal force always acts towards the centre, i.e. radially. Now, you must have observed , whenever you may have went past something that it doesn't seem that you have move past it rather it seems that the object is moving towards you. Observer always consider itself at rest. When you are in a train that is accelerating and you are looking outisde the window, you see the trees accelrating in a direction towards you. You will say, no net external force is acting on the trees but they are still moving. Hence to make Newton's Law applicable in this case you have to consider false force on trees in opposite direction of your acceleration. Now, If you observe from the accelerated frame of view of observer inside the body in Circular motion, then to ensure applicability of newton's laws of motion, just like the previous case you consider a false force in direction opposite to your direction of acceleration. Since your acceleration is towards centre due to centripetal force, you consider the false centrifugal force opposite the direction of centre. Thus it acts radially.

Thus for an observer in inertial frame, the body always has tendency to move tangentially and the acceleration of body is towards centre and as you go in the non inertial frame you consider acceleration radially outwards due to above stated reasons. And hence centrifugal force is radially outwards

Answer 3

@Lukas Nullmeir's answer is correct, but not direct enough for my tastes:

1. In an inertial frame there is no centrifugal force, period, full stop, end-of-story. The centrifugal is a so-call fictitious force that appears in rotating frames.

2. In a frame rotating at $\vec \Omega$ (and $m=1$):

$$ F_c = -\vec \Omega \times (\vec \Omega \times r) $$

That expands to:

$$ \vec F_c = -\Big(\vec\Omega(\vec \Omega \cdot \vec r) - \Omega^2 \vec r \Big)$$

Now I can arbitrary define $\vec \Omega = \hat z$ to show

$$ \vec F_c = \vec r - \hat z (\hat z \cdot \vec r) $$

where the right hand side is the vector rejection of $ \vec r$ onto $\vec z$ (https://en.wikipedia.org/wiki/Vector_projection), that is, it's the complement of the vector projection, and that, by definition: points straight outward from the $z$-axis, always.

An example of fictitious forces that I have never seen taught, but helps at least me understand how fictitious they are, is a mass in an inertial frame at rest at

$$\vec r = r \hat x $$ $$ \vec v = 0$$ $$ \vec a = 0 $$

Pretty basic. Nothing is happening and there are no forces.

Now change coordinates to a rotating frame with

$$\vec \Omega = \Omega\hat z,$$ aligned at $t=0$:

So:

$$ \vec r' = r \hat x $$

$$ \vec v' = -\Omega r\hat y $$

$$ \vec a' = -\Omega^2 r\hat x $$

which is uniform circular motion. What forces keep it revolving around the origin?

1. There is an outward centrifugal force:

$$ \vec F_c = \Omega^2 r \hat x $$

which points the wrong way for the orbit. Let's check the Coriolis force:

$$ \vec F_C = -2\vec\Omega \times \vec v' $$ $$ \vec F_C = -2 \Omega(\Omega)r(\hat z \times -\hat y) $$ $$ \vec F_C = -2\Omega^2 r \hat x $$

Which, with the factor of 2 both overcomes the centrifugal force and keeps the mass orbiting in the rotating frames.

So to answer you question: does it always work out? A hard YES.

Physics DOES NOT CARE ABOUT YOUR COORDINATES! And:

$$ \vec F = m\vec a $$

is physics, so you need those fictitious forces to persevere Newton's laws. (Exercise: combine the previous statement with Special Relativity to derive General Relativity).

JK: a better exercise is the following:

Apply this to a mass orbiting a planet (or on a string) with $\vec \omega = n\vec \Omega$ with $n = (-2,-1,0,1,2)$ to convince yourself of that fact.

Regarding it flying off on a tangent. It does...or does it? [Vsauce face].

https://www.youtube.com/watch?v=AL2Chc6p_Kk