How Does Colatitude Affect the Direction of Coriolis Force?

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Colatitude affects the direction of the Coriolis force, which is influenced by the Earth's rotation. The discussion highlights confusion regarding the interpretation of "due north" at a specific colatitude, with some participants suggesting that the direction should be radial toward the Earth's center of mass. The conversation also touches on the need to understand the vector direction of the Earth's angular velocity (denoted as ##\vec \Omega##) and its orthogonal components. Participants are encouraged to clarify these vector relationships to resolve the confusion. Understanding these concepts is crucial for accurately determining the Coriolis force's direction.
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Homework Statement
I am stuck at getting the right direction of the coriolis force.. I can get the force amplitude by formula tho...
Relevant Equations
Coriolis force, centrifugal force
1650169278243.png


So I don't understand "due north from a position at colatitude ##\theta## " , whether how I translate it...

I keep getting that direction should be radial...(toward Earth CM)

This is my work:

1650169455634.png


Thank you so much!
 
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drop_out_kid said:
Homework Statement:: I am stuck at getting the right direction of the coriolis force.. I can get the force amplitude by formula tho...
Relevant Equations:: Coriolis force, centrifugal force

View attachment 300106

So I don't understand "due north from a position at colatitude ##\theta## " , whether how I translate it...

I keep getting that direction should be radial...(toward Earth CM)

This is my work:

View attachment 300107

Thank you so much!
What is the direction of the vector ##\vec \Omega##? What directions are orthogonal to both that and due north?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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