Coriolis Effect for a level surface on Earth

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Homework Help Overview

The problem involves analyzing the motion of a puck sliding on a frictionless ice surface, specifically examining how it behaves in the Earth's rotating frame due to the Coriolis effect. The original poster attempts to demonstrate that the puck moves in a circular path and is tasked with determining the radius and angular frequency of this motion, under the assumption that the puck's circle is small compared to the Earth's radius.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the Coriolis force and the puck's velocity, with one participant expressing concern about the completeness of their reasoning. Questions arise regarding the conditions necessary for circular motion, particularly the implications of the puck's circular path being small relative to the Earth's radius.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the Coriolis force being perpendicular to the puck's velocity. Some guidance has been offered regarding the conditions required for circular motion, and participants are actively questioning and clarifying assumptions related to the problem.

Contextual Notes

Participants are considering the assumption that the puck's circle is small compared to the radius of the Earth, which may influence the nature of the forces acting on the puck and the resulting motion.

Habeebe
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Homework Statement


A puck slides with speed v on a frictionless ice that is level in the sense that the surface is perpendicular to geff at all points. Show that the puck moves in a circle as viewed in the Earth's rotating frame. Determine the radius of the circle and the angular frequency of the motion. Assume that the puck's circle is small compared to the radius of the earth.


Homework Equations



\vec{F}_{cor}=2m\vec{v}\times\Omega

The Attempt at a Solution



I have a solution, I just want to make sure that my logic is correct. I'm worried that I'm missing something subtle.

Assuming that geff the normal force and the z component of the Coriolis force all balance (it's sitting on top of the ice at all times). Then the only thing I have to worry about is the Coriolis force.

\vec{F}_{cor}=2m\vec{v}\times\Omega dot product by v on both sides
\vec{F}_{cor}\cdot\vec{v} = 2m(\vec{v}\times\Omega)\cdot\vec{v}
(\vec{v}\times\Omega)\cdot\vec{v} = 0 therefore:
\vec{F}_{cor}\cdot\vec{v}=0 therefore, \vec{F}_{cor}\perp\vec{v}, \forall\vec{v}

Since the force is perpendicular to the velocity, it must go in a circle.
 
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You didn't have to go through all those contortions to arrive at the fact that force is perpendicular to velocity. The cross product of two vectors that are not parallel or antiparallel to one another is a third vector that is perpendicular to both vectors involved in the cross product.

That said, that force is perpendicular to velocity alone does not make for circular motion. You need some other condition. Hint: You were told to "assume that the puck's circle is small compared to the radius of the earth."
 
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The puck's circle is small implies v stays in a plane (in its reference frame). So I need to say that the motion is confined to a plane for a consistently perpendicular force to make a circle, right?
 
That's part of it. Do you get a circle if the force isn't constant?
 
Ohhhhhh. I'm pretty embarrassed I missed that part. Thanks!
 

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