Rotating Coordinates Homework: Finding Force

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SUMMARY

The discussion centers on calculating the force required to maintain a mass m in a straight line within a 2D rotating coordinate system defined by the equations x=Xcos(ωt)+Ysin(ωt) and y=-Xsin(ωt)+Ycos(ωt). The derived forces are F_x=-mω²x and F_y=-mω²y, indicating that the force behaves similarly to a spring force, as it is proportional to the displacement in the rotating frame. The equations of motion were derived through double differentiation of the position equations with respect to time.

PREREQUISITES
  • Understanding of 2D coordinate transformations
  • Familiarity with angular frequency (ω) and its implications in motion
  • Knowledge of Newton's second law (F=m d²x/dt²)
  • Proficiency in calculus, specifically differentiation
NEXT STEPS
  • Study the implications of rotating reference frames in classical mechanics
  • Explore the concept of pseudo-forces in non-inertial frames
  • Learn about harmonic motion and its relation to spring forces
  • Investigate the applications of angular frequency in oscillatory systems
USEFUL FOR

This discussion is beneficial for physics students, mechanical engineers, and anyone interested in the dynamics of rotating systems and forces in non-inertial frames.

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Homework Statement


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

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


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


The Attempt at a Solution


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

So we have:
[tex]F_x=-m \omega^2x[/tex]
[tex]F_y=-m \omega^2y[/tex]

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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