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When you take U(y) = mgy and express y as a function of some parameter x, orThis ignores the angular velocity contribution, a la Galileo, of course.
t or whatever other parameter,
you can still use this representation of U to produce dU/dy by using the
chain rule ==>
dU(y)/dy = dU(x)/dx * dx(y)/dy . This will recover -F_y (and ONLY -F_y ;
U does not represent any other force).
It doesn’t matter what the parameter x (or t or whatever) is or what it
represents physically, it is here only a mathematical parameter – a bridge
to y.
Try it with your y(x), or with my y(t).