Is Centripetal Force Affected by the Earth's Rotation?

[jbriggs444]

I was merely pointing out to Dale, that a $2 r\omega^2$ acceleration on the girl in frame R cannot be all the acceleration.

Of course we all agree that there is another acceleration on the girl in frame R: Centrifugal, as you had suspected.

This agrees with what we already know about Newton's 3rd law

Newton's 3rd law applies only to real forces. It does not apply to inertial forces that arise from a choice of reference frame. Newton's 2nd law is the one that is important here.

[Adopting a positive = inward convention]

$F=ma$
$Coriolis + Centrifugal + Real = ma$
$2mr\omega^2 - mr\omega^2 + 0 = mr\omega^2$

Anyway, that's how I see it, I'll accept correction if I erred. I originally was working in frame S for the girl, so naturally in that frame her Coriolis component would be 0. I generally use frame R for the man on the merry-go-round, and frame S for the girl on the ground.

If one is going to even speak about Coriolis force the implication is that one has adopted a rotating frame that is not tied to the motion of the object of interest. Talking about the Coriolis force on an object using a frame of reference in which it is at rest is not often useful. Of course it will be zero. The object is not moving in that frame. Talking about the Coriolis force on an object using a non-rotating frame is not often useful. Of course it will be zero. There is no Coriolis force in a non-rotating frame.

 

[A.T.]

I generally use frame R for the man on the merry-go-round, and frame S for the girl on the ground.

When you analyse a scenario you should stick with one frame for all analysed objects, otherwise you will just confuse yourself. Of course can you do multiple analyses from different frames, but that means considering all objects in each analysis .

 

[Dale]

Ok, but how is it that the total Coriolis acceleration is $2 r \omega^2$ inwards, when we know a priori that the net acceleration of the girl in frame R has to be $r \omega^2$ inwards? In frame R, the girl rotates with uniform angular speed $\omega$, is that correct? So her net acceleration must be directed towards the center, and have a magnitude of $r \omega^2$ inwards, is that correct? Hence if the Coriolis component is twice that value, $2 r \omega^2$ inwards, then there would have to be another acceleration equal to $r \omega^2$ directed outward, centrifugal? Please explain. Thanks.

Claude

Yes, Coriolis acceleration is $2 r \omega^2$ in and centrifugal acceleration is $r \omega^2$ out for a net acceleration of
$r \omega^2$ in, as expected.