Pendulum attached to a rotating vertical disk

AI Thread Summary
The discussion revolves around the invariance of the pendulum mass coordinates when using a different Cartesian coordinate system, specifically one rotated counterclockwise by 90 degrees. The user confirms that they derived the same coordinates for the pendulum mass in both systems. A response clarifies the coordinate system's orientation, affirming that the x-axis is positive to the right and the y-axis is positive up. The equations provided illustrate the relationship between the coordinates and the pendulum's motion. The conversation highlights the consistency of coordinate transformations in physics.
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Homework Statement
I try to derive coordinates for a mass on a rotating pendulum using a different coordinate system
Relevant Equations
$$(x,y)$$
For this problem,
nBGOVfUox0x9n0dkz5EMPFanasHH6gtnK2EIj5snnUjWHO3YJ0.png

I correctly got the same coordinates for the pendulum mass using another coordinate system. The coordinate system I used was the other coordinate system rotated counterclockwise by 90 degrees. Why is the pendulum mass coordinates invariant in my cartesian coordinate system (x̄,ȳ)?

Thanks for any help!
 
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ChiralSuperfields said:
Homework Statement: I try to derive coordinates for a mass on a rotating pendulum using a different coordinate system
Relevant Equations: $$(x,y)$$

For this problem,
View attachment 343207
I correctly got the same coordinates for the pendulum mass using another coordinate system. The coordinate system I used was the other coordinate system rotated counterclockwise by 90 degrees. Why is the pendulum mass coordinates invariant in my cartesian coordinate system (x̄,ȳ)?

Thanks for any help!
Do you mean that your system was x positive to the right and y positive up?
In that system it would be ##x=a\cos(\omega t)+b\sin(\theta)##, ##y=a\sin(\omega t)-b\cos(\theta)##.
 
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haruspex said:
Do you mean that your system was x positive to the right and y positive up?
In that system it would be ##x=a\cos(\omega t)+b\sin(\theta)##, ##y=a\sin(\omega t)-b\cos(\theta)##.
Thank you for your reply @haruspex!

Yes you are correct!

Thanks!
 
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