Rotating Coordinates Homework: Finding Force

AI Thread Summary
The discussion revolves around finding the force required to keep a mass m moving in a straight line within a 2D rotating coordinate system defined by specific equations. The user derives the second derivatives of the coordinates, concluding that the forces in the x and y directions are proportional to the negative of the coordinates multiplied by the mass and the square of the angular frequency. This leads to the expressions F_x = -mω²x and F_y = -mω²y, suggesting a relationship akin to a spring force. There is a query about whether the solution is complete, indicating a potential oversight in the response. The analysis highlights the complexities of motion in rotating frames and the resulting forces.
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Homework Statement


A 2D rotating coordinate system (x,y) is defined by:
x=Xcos\omega t+Ysin \omega t
y=-Xsin\omega t+Y cos \omega t

Where (X,Y) is the coordinate of the inertial frame and omega is some angular frequency. What is the force required to keep a mass m moving in a "straight" line (x,y)=(ut,0) where u is a constant?


Homework Equations


F=m\frac{d^2x}{dt^2}
and the given equations of the new coordinates.


The Attempt at a Solution


Let me take the derivative of the given equations twice:
\frac{d(Xcos\omega t+Ysin \omega t)}{dt}=-X\omega sin \omega t + Y \omega cos \omega t=\omega y
\frac{d^2x}{dt^2}=\omega \frac{dy}{dt}=\omega \frac{d(-Xsin\omega t+Y cos \omega t)}{dt}= \omega \left ( -\omega X cos \omega t -\omega Y sin \omega t \right)= -\omega^2 x
Similarly:
\frac{dy}{dt}=-\omega x
\frac{d^2y}{dt^2}=-\omega^2 y

So we have:
F_x=-m \omega^2x
F_y=-m \omega^2y

Um... is this some kind of a spring force?
 
Last edited:
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...did you forget to finish typing out your attempt at the solution?
 
diazona said:
...did you forget to finish typing out your attempt at the solution?

?? Is my Latex showing?
 
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