Potential energy with constraints moving body

Question

I know that for conservative forces $\vec{F}=-\nabla{U}$. Let's consider the case of gravitational potential energy, I know that $U=mgy$. Just to check: $\vec{F}=-\nabla{U}=(0,-mg)$: perfect! Now, let's suppose that the body is only allowed to move on the line $y=x$. The potential energy is as before $U=mgy$, but now I could also write it as $U=mgx$.

Then I want the force, let's use the second equivalent expression: $\vec{F}=-\nabla{U}=(-mg,0)$: definitely not! What happened?

I conclude that I must be careful when deriving potential energy. How?

Other similar problems: I force the body to move along a specific constraint and I manage to express the potential as function of say z. Am I allowed to conclude that $F_x=0$ because derivative of a function that not depends on x is zero?

What is the point behind these problems?

Answer 1

The potential energy function is not a function of the path across which a particle moves along. Yes, you could use $U=mgx$ to describe the potential energy of the particle-field system when the particle is at some position along the curve $y=x$, but the potential energy is still defined over all space as $U=mgy$.

It's kind of analogous to the calculus 1 error of plugging in numbers before taking a derivative. The derivative of $y=x^2$ with respect to x at x = 2 is $y'(2)=2\cdot(2)=4$, not $(y(2))'=(4)'=0$

Answer 2

I think you are not clear where the potential of a field comes from. A field is conservative if work only depends on its start and end point, and not on the path. If we take the gravitational field as an example, we can see that it depends only on its distance from the centre (1/r^2), so we can say that it is conservative.

You should also be careful about the coordinate system you choose. If you move in the xy plane and take into account the whole earth, you have to decompose the gravitational force into vectors, so I will use a spherical system(you can use the Cartesian system if you are on a "differential" surface of the earth). If you move around the angle phi or theta, the gravitational force is a constant, since it only changes with the distance to its centre, so mg"theta"=mg"phi" will always be a constant. I don't know if that helped you solve your question.

Answer 3

I'm assuming that your coordinate system has $-y$ pointing toward the center of the earth, and that $x$ is perpendicular to $y$. You don't make that precisely clear. I have to say that I don't understand what you are after with these questions.

Part 1: $U\neq mgx$ Why do you think that $U=mgx$?

Part 2: I'm not sure I understand what you are asking. But if the potential energy depends only on $z$, then $\partial U/\partial x = 0$, so indeed $F_x=0$.