Direction of the Coriolis force

[John Mcrain]

Homework Statement

Determine direction of Coriolis force

Relevant Equations

-2mΩ x v

How can I determine direction of Coriolis force for these cases?1)Ship travel at 45th north parallel to the east (course 90°)2)Ship travel at north hemisphere at course 45°3)Ship travel at equator at east (course 90°)

My attempt:

1) Coriolis force is point up,vertical to the Earth surface
2)I really don't have idea where is direction
3)Coriolis force point up,vertical to the surface

But problem is still don't know how to apply right hand rule to my cases,because fingers can't be set in directions of velocity and omega what my cases has ?

 

[jambaugh]

I suggest you set up a coordinate system with z in the direction of the North pole and the ship in the x-z plane. Then work out the corresponding vectors directly and take the cross product by components.

 

[Herman Trivilino]

But problem is still don't know how to apply right hand rule to my cases,because fingers can't be set in directions of velocity and omega what my cases has ?

Sure they can! You point your fingers in the direction of $\vec{\Omega}$ and then "curl" them in the direction of $\vec{v}$.

 

[John Mcrain]

Sure they can! You point your fingers in the direction of $\vec{\Omega}$ and then "curl" them in the direction of $\vec{v}$.

If object velocity(index finger) is travel to the north,how can I set thumb parallel to axis of rotation?

 

[haruspex]

If object velocity(index finger) is travel to the north,how can I set thumb parallel to axis of rotation?

These right hand rules do tend to assume the arguments to the cross product operation are orthogonal.
Pick one of the arguments, the spin axis say, and ignore any component of the other which is parallel to it. If traveling North at the equator, that means throwing away the whole velocity vector, leaving a zero Coriolis force (correct). More generally, just use the component of the velocity that lies in the plane of latitude.

 

[John Mcrain]

More generally, just use the component of the velocity that lies in the plane of latitude.

So basically I must "convert" everything at 2D view,where plane latitude represent turning table and use just component of the velocity that lies in the plane of latitude.
-So if object travel to north at northen hemishpere ,is same as object travel toward center of counterclock wise turnning table,coriolis is prependiculart to objct velocity to the right
-if person travel to the south at northen hemisphere is same as object travel outward at counterclock wise turning table,coriolis direction is prependicular to object velocity to the right
-if object travel to west is same as object travel in circle in clockwise directions when turning table rotate counterclockwise,coriolis direction is inward toward center
etc

 

[PeroK]

So basically I must "convert" everything at 2D view,where plane latitude represent turning table and use just component of the velocity that lies in the plane of latitude.
-So if object travel to north at northen hemishpere ,is same as object travel toward center of turnning table
-if person travel to the south at northen hemisphere is same as object travel radialy outward at turning table
-if object travel to west is same as object travel in circle in opposite direction as truning table rotation
etc

At the end of the day we have:
$$\hat x \times \hat y = \hat z$$
And, using the "physics" convention with $\theta$ being taken from the z-axis:
$$\hat r \times \hat \theta = \hat \phi$$
And we have $\pm$ for even/odd permutations thereof.

To calculate the cross product of two vectors, I find the safest approach is to use the determinant:

https://math.stackexchange.com/questions/1476391/what-is-the-cross-product-in-spherical-coordinates

 

[Herman Trivilino]

If object velocity(index finger) is travel to the north,how can I set thumb parallel to axis of rotation?

You're not applying the right hand rule correctly. You point the fingers of your right hand in the direction of $\vec{\Omega}$ and then curl those fingers until they point in the direction of $\vec{v}$. Your thumb will then point in the direction of the force, that is, in the the direction of $\vec{\Omega}\times\vec{v}$.

 

[haruspex]

You're not applying the right hand rule correctly. You point the fingers of your right hand in the direction of $\vec{\Omega}$ and then curl those fingers until they point in the direction of $\vec{v}$. Your thumb will then point in the direction of the force, that is, in the the direction of $\vec{\Omega}\times\vec{v}$.

I am not able to visualise that procedure.
As far as I am aware, the thumb-and-curled-fingers case of the rule is used only for mapping between a rotation (curled fingers) and the vector that represents it (thumb), or, similarly, for field lines from an electric current, etc.
Assuming the Coriolis force is given in the form of a cross product, and we have figured out the rotation vector, we then need the other case of the rule, involving thumb, index and middle fingers at right angles.
See e.g. https://en.m.wikipedia.org/wiki/Right-hand_rule.

If you have a link that makes your version more intelligible please post it.

 

[Herman Trivilino]

If you have a link that makes your version more intelligible please post it.

 

[haruspex]

Ok, thanks for clarifying, but it doesn't make the other way of applying the rule wrong. They both work, and the three finger version is, in my experience, the usual one.
Indeed, the video is not quite what you described since it rotates the palm to the second vector rather than curling the fingers. The curled fingers model makes it quite close to what I wrote about mapping rotation vectors: its the rotation from first vector to second vector, by the shorter route.

 

[John Mcrain]

You're not applying the right hand rule correctly. You point the fingers of your right hand in the direction of $\vec{\Omega}$ and then curl those fingers until they point in the direction of $\vec{v}$. Your thumb will then point in the direction of the force, that is, in the the direction of $\vec{\Omega}\times\vec{v}$.

You forget to mention,I must use only at velocity component which lay in latidude plane, this component is perpendicular to the axis of rotation .
Without that you can't do nothing..

 

[Herman Trivilino]

You forget to mention,I must use only at velocity component which lay in latidude plane, this component is perpendicular to the axis of rotation .
Without that you can't do nothing..

That is not correct. The direction of $\vec{\Omega}\times\vec{v}$ can be determined from the known direction of $\vec{\Omega}$ and the given directions of $\vec{v}$ only.

 

[John Mcrain]

That is not correct. The direction of $\vec{\Omega}\times\vec{v}$ can be determined from the known direction of $\vec{\Omega}$ and the given directions of $\vec{v}$ only.

I can visualise how.