Tangential Component of Centrifugal Acceleration

[SebastianRM]

The thing is in page p.347 Taylor, it is said that the component is:

g_tan = Omega^2*Rsin(theta)cos(theta) However the angle between the centrifugal Force and the axis normal to the direction of the grav Force is actually 90 - theta, I am not really getting where I am going wrong understanding this.

 

[kuruman]

I thought I explained that to you here.
https://www.physicsforums.com/threads/what-is-the-tangential-component-taylor-p-347.965665/
If there is something you still did not understand, you should have responded there instead of starting a new thread. In any case, post a drawing of what you think is the case. I suspect you have misidentified something.

 

[Andrew Mason]

The angle in question is between the perpendicular to the Earth's axis (the direction of the centrifugal force component) and a tangent to the surface. That angle is θ which you can see by geometry. θ is the complement of the angle of latitude, L (θ = 90-L).

The magnitude of the centrifugal acceleration (see equation 9.43) is: a_cf = Ω²Rsinθ. So the tangential component is a_cfcosθ. At the equator, where θ is 90° (L=0), the centrifugal force is maximum (sinθ = 1) but the tangential component is 0 because it is all in the radial direction (cosθ = 0). Towards the pole, the direction is almost tangential (cosθ=1) but the magnitude of the centrifugal force approaches 0 (sinθ = 0).

AM

 

[SebastianRM]

I thought I explained that to you here.
https://www.physicsforums.com/threads/what-is-the-tangential-component-taylor-p-347.965665/
If there is something you still did not understand, you should have responded there instead of starting a new thread. In any case, post a drawing of what you think is the case. I suspect you have misidentified something.

Sorry about that, I could not track the post I had done already. Here is the sketch of how I am working it out on my mind.

https://imgur.com/spbCzfS
Since he says: 'the tangential component of g (the component normal to the true grav force)'

 

[Andrew Mason]

Sorry about that, I could not track the post I had done already. Here is the sketch of how I am working it out on my mind.

https://imgur.com/spbCzfS
Since he says: 'the tangential component of g (the component normal to the true grav force)'

In the drawing you have provided, you have shown two angles of 90-θ making up a right angle! The angle between the F_cf vector and the tangent is 90 - (90-θ) = θ! (the angle of the tangent to the radial vector being necessarily a right angle).

AM

 

[kuruman]

To supplement post #5 by @Andrew Mason, it is known from geometry that two angles that have their sides mutually perpendicular are equal. In your diagram, the z-axis (along Ω) is perpendicular to F_cf and the radial vector is perpendicular to the tangential component (not shown). Therefore the angle that you show as θ is equal to the angle formed by F_cf and the tangential direction.

 

[SebastianRM]

Thank you so much guys! I see it now!