Coriolis Torque On a Spinning Object

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Homework Help Overview

The discussion revolves around the Coriolis force and its effect on a spinning object, specifically a horizontal hoop of mass m and radius r spinning with angular velocity w at a certain colatitude. Participants are exploring how to derive the torque produced by the Coriolis force due to the Earth's rotation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Coriolis force and its calculation, with attempts to express the torque in terms of the hoop's parameters. Questions arise regarding the treatment of different mass elements of the hoop and their respective velocity vectors.

Discussion Status

Some participants have provided guidance on introducing a coordinate system to better express the position and velocity vectors of mass elements. There is an ongoing exploration of the correct formulation of these vectors and the implications for the torque calculation.

Contextual Notes

There is mention of confusion regarding the direction of the torque and the need to clarify the relationship between the angular velocity and the Coriolis force. Participants are also considering the implications of treating each mass element independently.

Dextrine
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Homework Statement


The Coriolis force can produce a torque on a spinning object. To illustrate this, consider a horizontal hoop of mass m and radius r spinning with angular velocity w about its veritcal axis at colatitude theta. Show that the Coriolis force due to the Earth's rotation produces a torque of magnitude mwWr^2sin[theta] directed to the west, where W is the Earth's angular velocity.

Homework Equations


Coriolis force = 2mr' X W
Torque = F X d

The Attempt at a Solution


r'=rXw

Fcor= 2m(rXw) X W
= 2mrwW

Torque = Fcor X r
= 2mr^2wWsin[theta]

I'm supposed to get half of this value somehow, and the direction is not at all clear and from my diagram is constantly changing...
 
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Dextrine said:

Homework Equations


Coriolis force = 2mr' X W[/B]
This equation is for a particle of mass m moving with linear velocity ##\dot{\mathbf{r}}##. Each mass element of the hoop should be treated as a particle with its own velocity vector. (Different mass elements have different position and velocity vectors.)
 
TSny said:
This equation is for a particle of mass m moving with linear velocity ##\dot{\mathbf{r}}##. Each mass element of the hoop should be treated as a particle with its own velocity vector. (Different mass elements have different position and velocity vectors.)
I thought that rewriting r' as w X r with both w and r being vectors, would take care of this? If not, how would I go about treating each point independently?
 
You'll need to introduce a coordinate system and express the position and velocity vectors of a mass element with respect to the coordinate system. For example, you could introduce Cartesian axes with origin at the center of the hoop and with the z axis along the axis of rotation of the hoop, the x-axis pointing east, and the y-axis pointing north.

A mass element of the hoop will be at some angle ##\phi## to the x-axis. How would you express the position and velocity vectors of this mass element?
 

Attachments

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Last edited:
TSny said:
You'll need to introduce a coordinate system and express the position and velocity vectors of a mass element with respect to the coordinate system. For example, you could introduce Cartesian axes with origin at the center of the hoop and with the z axis along the axis of rotation of the hoop, the x-axis pointing east, and the y-axis pointing north.

A mass element of the hoop will be at some angle ##\phi## to the x-axis. How would you express the position and velocity vectors of this mass element?

[r cos[phi] x direction + r sin[phi] y direction] X w z direction
 
Dextrine said:
[r cos[phi] x direction + r sin[phi] y direction] X w z direction
I'm guessing that this is an expression for the velocity vector of the mass element located at angle ##\phi##. Does the expression have the correct overall sign? Can you simplify by carrying out the cross product?

Note that math symbols are available by clicking on the ##\Sigma## tab.
 
Thank you for that, I couldn't remember how i had made symbols before. Anyway, I understand it now, thanks a lot!
 

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