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

[QuanticEnigma]

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

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

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...

 

[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.

 

[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?

 

[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 v_x=0, v_y=0 and v_z=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 t_f.

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.

 

[paranormal]

The Coriolis effect is a phenomenon that occurs due to the rotation of the Earth. As the Earth rotates, objects that are moving on its surface appear to be deflected to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. This is due to the Coriolis force, which is caused by the Earth's rotation.

To solve this problem, we can use the equations provided in the homework statement. First, we need to calculate the angular velocity of the Earth, \omega. This can be done using the equation \omega = 2\pi/T, where T is the period of rotation of the Earth. The period of rotation of the Earth is approximately 24 hours, so \omega \approx 7.27 \times 10^{-5} rad/s.

Next, we can use the equation for the Coriolis acceleration, a_C, to calculate the final displacement of the object. We know that the object is dropped from rest, so its initial velocity, v, is zero. Therefore, we can simplify the equation to a_C = -2\Omega \times v = -2\Omega \times 0 = 0.

Since the Coriolis acceleration is zero, there is no deflection due to the Coriolis effect. Therefore, the final displacement of the object will be in the same direction as its initial displacement, which is downward. The magnitude of the displacement will be equal to the initial height, H, which is 40m. So, the final displacement is 40m downward.

In summary, the Coriolis effect does not have any effect on the final displacement of the object in this scenario. The object will simply fall straight down with a magnitude of 40m.