When you take U(y) = mgy and express y as a function of some parameter x, or
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).
From: LaMontagne, Bob
Sent: Wednesday, April 04, 2012 3:11 PM
To: Forum for Physics Educators
Subject: [Phys-l] Force and potential
Deep in the sugar shock of eating too many Easter "Peeps", I came up with
the following:
Consider a bowl formed from the lower half of a sphere. In 2D I can write
the equation for the bowl as x^2 + (y-1)^2 = 1.
This gives y = 1 - sqrt(1-x^2)
If I place a marble at the inside rim (x=1, y=1) of the bowl and release it,
the marble will oscillate back and forth, repeatedly coming back to the
release point.
The potential energy is given by U=mgh=mgy = mg[1 - sqrt(1-x^2)]
Using F_x = - dU/dx , then we have F_x = -x/sqrt(1-x^2) which implies that
F_x is infinitely strong when x = 1 (the release point).