Relative Motion - Earth's rotation and Coriolis acceleration

Problem Statement:

A plane is traveling from the North Pole (we will assume that the earth is a sphere of radius RT) with a velocity v’ measured with respect to an Earth’s non-inertial frame of reference O’ (see figure). v’ is contained in the XY plane. The constant angular velocity of the earth is ω .
Calculate the Coriolis acceleration vector of the plane for points A, B, C and D of its trajectory. Give the results as a function of the givens.

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Solution:

Coriolis acceleration

The Coriolis acceleration is given by:

Where ω is the angular velocity of the rotating frame of reference (O’ in this problem) and v’ is the velocity of the moving particle with respect to the rotating reference frame.

As with any other cross product, the magnitude of the Coriolis acceleration is:

Where θ in the angle between ω and v’.

The direction of the Coriolis acceleration is given by the right hand rule. We are going to learn how to use it with some examples.

Point A:

As you can see in the figure, when the plane is at point A θ is 900, and therefore the magnitude of the Coriolis acceleration is:

We are going to use the right hand rule to determine the direction of the Coriolis acceleration.

First we calculate the cross product:

In the figure below you can see vectors ω and v’ at point A.

To use the right hand rule, first we have to align the right hand with the first vector (in this case ω). Then we close the hand over the second one (v’). The thumb gives the direction of the cross product.

The cross product is not commutative; therefore you will have to respect the order of the vectors to calculate it correctly.

The cross product is always orthogonal to both vectors. In this example the cross product is perpendicular to the screen and points inwards, as the thumb indicates.

Last we have to multiply by -1 (the minus sign is included in the definition of the Coriolis acceleration). This inverts the cross product direction, and therefore the Coriolis acceleration in this case will point outwards.

In the first figure you can see the unit vectors. The Coriolis acceleration at point A is parallel to k. And since we also have calculated its magnitude, the final value of the Coriolis acceleration when the plane is at point A is:


Point B: The angle between ω and v’ at this point is 180-λ, as you can see in the figure below. We have translated ω to point B to make it easier to determine the angle.

Therefore the magnitude of the Coriolis acceleration at point B is:

We are going to use the right hand rule to determine the direction of the cross vector.

The direction of the Coriolis acceleration at point B will be the same as at point A, since ω and v’ make the same plane as at point A.

The Coriolis acceleration at point B is:


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Point C: At this point the angle between ω and v’ is 180-λ, as you can see in the figure below. We have translated ω to point C to make it easier to determine the angle.


Por tanto, el módulo de la aceleración de Coriolis del avión cuando se encuentra en el punto C es:

We are going to use the right hand rule to determine the direction of the Coriolis acceleration. We begin with the cross product:

In the figure below you can see vectors ω and v’ at point C:

As in the previous cases, we are going to use the right hand rule, closing it from ω to v’. The thumb gives the direction of the cross product. In this case it is perpendicular to the screen and points outwards (parallel to k).

Last we have to multiply by -1 (the minus sign is included in the definition of the Coriolis acceleration). This inverts the cross product direction, and therefore the Coriolis acceleration in this case will point inwards (-k).

The Coriolis acceleration at point C is given by:

As you can see, the Coriolis acceleration has, for the same value of the latitude, opposite directions in the Northern and Southern hemispheres. This is why the trajectories deviate to the right in the first case and to the left in the second.


Point D: In this case ωv’ are parallel, and therefore the cross product is zero and so is the Coriolis acceleration.

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