Is Centripetal Force Affected by the Earth's Rotation?

[Dale]

Likewise in frame R, the girl standing on the ground observing the merry go round has a 0 Coriolis acceleration as she is at rest wrt to her frame S.

She is at rest with respect to S, she is moving with respect to R.

The Coriolis acceleration is $2 v \times \omega$ where v is measured in R, not S. In R, she is moving perpendicular to $\omega$, and her speed is $|r \omega|$ so the total Coriolis acceleration is $2 r \omega^2$ inwards.

 

[cabraham]

She is at rest with respect to S, she is moving with respect to R.

The Coriolis acceleration is $2 v \times \omega$ where v is measured in R, not S. In R, she is moving perpendicular to $\omega$, and her speed is $|r \omega|$ so the total Coriolis acceleration is $2 r \omega^2$ inwards.

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

 

[D H]

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.

You've explained it yourself. The net apparent force in the rotating frame is $m r\omega^2$ directed inward. The centrifugal force is $m r\omega^2$, directed outward. That's the right magnitude, but exactly the wrong direction. To yield that apparent net force, there must be some other force (or forces) with twice this magnitude but directed inward. This is the Coriolis force.

 

[Khashishi]

You can find a frame of reference with no absolute rotation. If you want to have the simplest physical model for multiple rotating objects, you better use a non-rotating frame. You can only cancel out the Earth's rotation with a merry-go-round if your merry-go-round has the same axis as the Earth, so this is only possible at the poles. Away from the poles, you can find a parallel axis, but a parallel axis isn't good enough to cancel out the rotation.

 

[cabraham]

You've explained it yourself. The net apparent force in the rotating frame is $m r\omega^2$ directed inward. The centrifugal force is $m r\omega^2$, directed outward. That's the right magnitude, but exactly the wrong direction. To yield that apparent net force, there must be some other force (or forces) with twice this magnitude but directed inward. This is the Coriolis force.

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. There has to be another component acting such that the NET acceleration on the girl in frame R is $r\omega^2$, oriented inward, i.e. centripetal acceleration. Since Coriolis in frame R, wrt the girl, computes to be ##2 r\omega^2## inward, then there has to be an acceleration of ##1 r\omega^2## outward. The net result would be ##1 r\omega^2## inward. This agrees with what we already know about Newton's 3rd law. Since the girl's net acceleration is centripetal, we expect ##1 r\omega^2## inward as the net final result. We also know that in the R frame, a centrifugal force is added to keep Newton's laws intact. This force has the same magnitude as the centripetal, i.e. ##1r\omega^2## inward, but oriented outward. The Corioils computes to ##2 r\omega^2## inward, so that subtracting $1r\omega^2$ outward, the centrifugal component, we get what I believe is the correct final result, ##1 r\omega^2## inward, which is expected as it equals the centripetal component.

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. But there is no reason why we cannot view the girl as being in frame R. We must add the centrifugal component, which we have. Best regards.

Claude

 

[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.