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My book says that

when a frame rotates uniformly, pseudo/fictitious force must be taken to count. This is centrifugal force.

  1. Also, while Googling, I found another pseudo force Coriolis force. So, what's the difference between them? And what's their intuitions?

  2. Also, the book mentions uniform; what if it were non-uniform?

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The Coriolis force $\vec F_{\text{coriolis}} = -2m \, \vec \omega \times \vec v$ only depends on velocity. The centrifugal force $\vec F_{\text{centrifugal}} = -m \, \vec \omega \times (\vec \omega \times \vec r)$ only depends on position.

Finally, if the object is not rotating uniformly ($\dot {\vec \omega} \ne 0$), then yet another fictitious force comes into play, the Euler force $\vec F = -m \, \dot{\vec \omega} \times \vec r$.

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Look in Wikipedia http://en.wikipedia.org/wiki/Coriolis_effect.

For understanding intuitively the Coriolis force effect, assume an object moving according to a static (inertial) frame of reference, in the plane perpendicular to the rotation axis, and along the radius,

In the rotating frame, see the animation in Wikipedia, the Coriolis force imposes an acceleration perpendicular to the radius, s.t. the object is seen as getting a tangential velocity component,

To the difference, for understanding the centrifugal force, imagine an object in movement perpendicularly to the radius from the point of view of the static observer.

For the rotating observer, the object is deflected in the direction of the radius.

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Suppose you're standing at the edge of an enormous merry-go-round. You'll feel a "force" that seems to be pushing you outwards. This is the centrifugal force.

The centrifugal force is the force you feel because you're at a certain distance away from the axis of rotation. It is written as $\vec{F}_{\rm centrifugal} = m~\vec{\omega} \times (\vec{r} \times \vec{\omega})$

Now imagine you're hovering above the merry-go-round, stationary with respect to the Earth, and I've taken your place on the merry-go-round. I now start walking towards the center of rotation at a constant speed. When I'm at the edge, I'm moving with some tangential speed, say $v_{\rm tan}$. When I'm closer towards the center, there is nothing to stop moving me with that same speed, so even though I think I'm travelling in a straight line, you see I'm veering off-course, and I travel along a curved path. This apparent force is called the Coriolis force. $F_{\rm Coriolis} = 2m (\vec{v} \times \vec{\omega})$. As you can see, this doesn't depend on your position on the merry-go-round, just on your velocity and the carousel's angular velocity.

The Coriolis force is the force you feel because you're moving in the rotating reference frame in the radial direction.

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  • $\begingroup$ You don't "feel" the fictitious Coriolis force, the fictitious centrifugal force, or any other fictitious force for that matter. They're fictitious forces. $\endgroup$ – David Hammen Dec 8 '14 at 23:10
  • $\begingroup$ @DavidHammen: ??? if you sit on a swing ride, your butt will be pressed into the seat; it feels as if there's a force pushing you into it; sure, if you cut the chain, you'll be in free fall and conclude the force was fictious, but you felt it nonetheless $\endgroup$ – Christoph Dec 8 '14 at 23:32
  • $\begingroup$ @Christoph - That's a real force. It's called the normal force. There's nothing fictitious about it. All reference frames agree on that force, which is the key factors that distinguish a real force from a fictitious force. $\endgroup$ – David Hammen Dec 9 '14 at 0:12
  • $\begingroup$ @DavidHammen: So, should we also correct people that claim to 'feel the force of gravity'? After all, the real force is the normal force of the surface of the earth that prevents us from falling freely... $\endgroup$ – Christoph Dec 9 '14 at 0:40
  • $\begingroup$ @DavidHammen, when you're in a rotating frame, the fictitious force is real to you for all intents and purposes. It's only when you step out of that frame that you realize it was a fictitious force. $\endgroup$ – Pranav Hosangadi Dec 19 '14 at 5:00

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