Dropped object in a rotating frame

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etotheipi
Homework Statement
A rotating space station (at ##\omega##) has a radius R. Alice climbs up a tower of height H and drops an object from rest, in the rotating frame. Calculate the velocity and horizontal distance to the tower when it hits the floor. It is given that ##R \gg H##
Relevant Equations
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I solved this in an inertial frame, but now I want to do it in the rotating frame. As far as I can tell the equation of motion is $$\vec{F}_{cent} + \vec{F}_{cor} = mr\omega^2 + 2m\vec{v} \times \vec{\omega} = m\frac{d^2\vec{r}}{dt^2}$$The solutions take a different approach. They state that the Coriolis force is $$F_{cor} (t) = 2m \omega^2 R t \omega = 2 m \omega^3 R t$$and they simply integrate this w.r.t. time. There are two things I don't understand. Why have they ignored the centrifugal force (or used it in a weird way, that I can't see), and also where did their expression for the Coriolis force come from? Thanks :smile:
 
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Seems, to me, that the problem is suggesting some things:
The rotating of the object create some artificial gravity, that will accelerate the body by -w²*(R+h) ~ -w²*R
so a = -w²*R
v = -w²*R*t
And the object was dropped in some point where the angular rotating vector is perpendicular to the velocity that will rise up, so |Fcor| = 2mwv = 2mw*w²*(R)*t = 2mw³Rt

I am just trying to construct any space sphere station with this properties, but is a little hard.
 
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LCSphysicist said:
Seems, to me, that the problem is suggesting some things:
The rotating of the object create some artificial gravity, that will accelerate the body by -w²*(R+h) ~ -w²*R
so a = -w²*R
v = -w²*R*t
And the object was dropped in some point where the angular rotating vector is perpendicular to the velocity that will rise up, so |Fcor| = 2mwv = 2mw*w²*(R)*t = 2mw³Rt

I am just trying to construct any space sphere station with this properties, but is a little hard.

Thanks for the reply, I think you might be right.

It looks like they're approximating the centrifugal force to be constant at ##mR\omega^2##, even though it will actually vary between ##mR\omega^2## and ##m(R-H)\omega^2## which is maybe a fair assumption if ##H \ll R##. More worryingly they also seem to be neglecting the contribution of the horizontal component of velocity to the Coriolis force, which again is perhaps a fair(ish) assumption if the horizontal displacement ##d_x## is small.

If we make these two assumptions, then we do indeed find that the Coriolis force is approximated as a constant ##2m \omega^3 Rt##.

Funnily enough this approximate method does give the same results as I got for the analysis in the inertial frame, so maybe it's not so bad after all. I might try and finish off the vector analysis too and see if it gives a similar thing :smile:
 
LCSphysicist said:
I am just trying to construct any space sphere station with this properties, but is a little hard.

And also just in case I didn't specify, the platform of height ##H## is inside the space station, i.e. the object falls from the centre out towards the edge :wink: