# Coriolis effect problem

1. Jun 6, 2010

### QuanticEnigma

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$\Omega = \omega \left( \begin{array}{cc} 0\\ \cos{\varphi}\\ \sin{\varphi} \end{array} \right), v = \left( \begin{array}{cc} v_{east}\\ v_{north}\\ v_{upward} \end{array} \right), a_{C} = -2\Omega \times v = \left( \begin{array}{cc} v_{north}\sin{\varphi}-v_{upward}\cos{\varphi}\\ -v_{east}\sin{\varphi}\\ v_{east}\cos{\varphi} \end{array} \right)$$

3. 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...

2. Jun 6, 2010

### kuruman

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: Jun 6, 2010
3. Jun 6, 2010

### QuanticEnigma

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

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

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

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?

4. Jun 7, 2010

### kuruman

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.