How Does the Coriolis Effect Influence Object Displacement at 31.3°S?

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Homework Help Overview

The discussion revolves around the Coriolis effect and its influence on the displacement of an object dropped from a height of 40 meters at a latitude of 31.3°S. Participants are exploring the dynamics involved in calculating the final displacement due to this effect.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the equations of motion that incorporate the Coriolis acceleration. Questions have been raised about the integration limits and the initial conditions for the velocity vector.

Discussion Status

Some guidance has been provided regarding the formulation of differential equations and the need to incorporate gravitational acceleration into the z-equation. There is an ongoing exploration of the integration process and the implications of successive approximations.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or the methods they can apply. The discussion includes questioning the assumptions made in the initial setup of the problem.

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Homework Statement


An object is dropped from rest at height H = 40m above the ground at latitude 31.3^{o}S. Calculate the final displacement, in magnitude and direction, due to the Coriolis effect.


Homework Equations


<br /> \Omega = \omega \left(<br /> \begin{array}{cc}<br /> 0\\<br /> \cos{\varphi}\\<br /> \sin{\varphi}<br /> \end{array}<br /> \right), <br /> v = \left(<br /> \begin{array}{cc}<br /> v_{east}\\<br /> v_{north}\\<br /> v_{upward}<br /> \end{array}<br /> \right),<br /> a_{C} = -2\Omega \times v =<br /> \left(<br /> \begin{array}{cc}<br /> v_{north}\sin{\varphi}-v_{upward}\cos{\varphi}\\<br /> -v_{east}\sin{\varphi}\\<br /> v_{east}\cos{\varphi}<br /> \end{array}<br /> \right)<br /> <br /> <br />


The Attempt at a Solution



I know that \omega = angular velocity of rotating reference frame (in this case, the earth), and that \varphi = 31.3 degrees, but could someone please give me a few pointers to get started, I'm kind of confused with all this Coriolis business...
 
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You forgot a factor of 2Ω in your expression for the acceleration, but otherwise it is OK. To make things simpler, say x = "East", y = "North" and z = "upward". This question is best answered by successive approximations, here to first order will probably be OK.

First write out three differential equations in the form

\frac{dv_i}{dt}=a_i

where i = x, y, and z.

Do that, and I will guide you to the next step.
 
Last edited:
\frac{dv_{x}}{dt} = 2\omega (v_{y}\sin{\varphi} - v_{z}\cos{\varphi})


\frac{dv_{y}}{dt} = 2\omega (-v_{x}\sin{\varphi})


\frac{dv_{z}}{dt} = 2\omega (v_{x}\cos{\varphi})

Ok, so I suppose this will give me a velocity vector dependent on time t ... is this correct?

Also, when I integrate the above expressions, what will the limits of integration be?
 
What you show is just the Coriolis acceleration, not the acceleration of the free falling mass. You need to add -g to the z-equation and you will have the equations of motion. Let's call these "the original equations of motion". When you integrate, you do so to successively increasing orders of approximation.

Zeroth order
Since the mass is released from rest, to zeroth order, you set vx=0, vy=0 and vz=0. What do the equations become then? Integrate once to get the components of the velocity vector and once again to get the components of the position vector. The limits of integration must be from t = 0 to t = t and the result is something extremely familiar. Use the z-equation to find the time of flight tf.

Do this first and then I will tell you how to move on to the first order correction which is where the effects of the Coriolis acceleration come in.
 

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