jbriggs444 said:
To a good approximation, yes. But "g" denotes the local apparent acceleration of gravity. It already contains a deduction for centrifugal force.
Sorry, I mean that “g" is simply the intensity of attractive gravitational force divided by mass, that is :
## g = GM/R^{2}##
not the apparent local acceleration, already corrected for the centrifugal acceleration, which has a maximum value at the equator . When considered in the rotating r.f. , on which we live, it is obvious that you have to sum
vectorially the a.m. ##\vec g## with the apparent centrifugal acceleration. The resulting acceleration gives you the local vertical.
jbriggs444 said:
Yes. Let us work this out for the case of someone rotating at −ω as measured against a rotating frame which is itself rotating at +ω.
The object's velocity measured against the rotating frame is −ωR. The Coriolis force mv×R will then be given by −2ω^2R.
Why do you say : “the Coriolis force mvxR ...“? This fictitious force, which is to be considered in the rotating frame only, is given by a cross product , and a factor -2m :
## \vec F_{cor} = -2m\vec\omega\times\vec v ##
in particular, factor “-2” comes out by accurately deriving velocity vectors against the rotating r.f. , please refer to the book by Morin, already cited.
## \vec F_{cor}## is perpendicular to the plane which contains ##\vec\omega## and ##\vec v## , of course. Its modulus , is given in our case by : ## F_{cor} = 2m\omega^{2}R## , right.
jbriggs444 said:
At first glance, one might look at this and think "that's twice as much force as is needed to balance centrifugal force. So half that fast and he's canceled centrifugal".
Whoah there. Do the accounting more carefully. You are either jumping frames or missing a term.
No, that’s not my reasoning here, it would be wrong. May be I have not been clear here, apologises for that.
I am just comparing the intensity of centrifugal acceleration (let’s take off mass) ##\omega^{2}R ## with the intensity of Coriolis acceleration ## A_{cor} = 2\omega^{2}R## , which I can write ##2\omega v ## , isn’t it ?
So factor 2 is correct in this contest.
But now, let’s see an example regarding what you said. Let’s take a rotating carousel which, as seen from the above by a inertial observer, rotates
counterclockwise, and put a bead on it , at a certain distance R from the center. Suppose friction is completely
absent here, so that the bead keeps its “fixed” position relative to the external inertial observer; but for an observer sitting at the center of platform and rotating with it, the bead seems to describe a circumference
clockwise , with tangential speed v =ωR .
Taking into account the directions of vector ##\vec\omega## (oriented from platform toward the inertial observer above it, because platform rotates counterclockwise) and of vector ##\vec v## , oriented clockwise, and also the “-“ sign, vector ##\vec F_{c}## points toward the center, while centrifugal force, also considered by the NON inertial observer, points outward. Because the first ( put m=1 for simplicity) has module ## F_{cor} = 2\omega^{2}R## and the second is ## F_{centrifugal} = \omega^{2}R## , the resultant is a force directed toward the center , that is centripetal, of module ## F_{centripetal} = \omega^{2}R## .
So, the non inertial observer thinks he has found the centripetal force that, in his r.f. , causes the bead to move in circular uniform motion of radius R , and tangential speed v = ωR .
But this is actually one more demonstration that fictitious forces are...
fictitious! They don’t exist, neither in inertial frames nor in non inertial frames. No real force acts on the bead of the example , exception made for vertical equilibrated forces. So all effects that we attribute to inertial fictitious forces are, in reality, due
due to the “bad” reference frame used to describe motion.
Anyway, we live on a Non-inertial r.f. , the Earth, so we are obliged, in a certain sense, to include fictitious forces in our description of motion on Earth, if we can’t ignore, at least locally and for a short time, the non-inertiality of Earth.
Sorry for my bad English, and thank you for attention.