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Textbook answer says $ $ $ $ $U_{min}=-2mga$.

So U is negative, given that m, g, a are all positive (unless, I'm mistaken).
How can this be? I thought energy was a non-negative scalar quantity?

Also, does that mean $ $ $ $ $U_{max}=-amg\cdot0=0$ $ $?

If so, then again, how? The mass is clearly hanging in the air when both angles are $90°$ so the potential energy shouldn't be $0$.

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  • $\begingroup$ Very useful to downvote without leaving a comment... $\endgroup$
    – Stephen
    Commented Nov 25, 2017 at 18:04

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You have some deep misconceptions about the nature of energy, and specifically, potential energy. Recall (hopefully) that only differences in energy matter and that scaling all the energies in a problem up or down by some constant factor will not change the dynamics of the problem. This fact allows us to define the 0 point of (in this case, gravitational) potential energy anywhere we like. And consequently, we can indeed have negative values for potential energies with no problem at all. Negative values in potential energy mean we are deeper in the potential well than the 0 point, while a positive value would mean we are in a shallower part of the potential well (than the 0 point). If we choose to define the 0 point of potential energy at the height of the pivot point of the pendulum, then by definition when we are at a point that is lower than the pivot point, we will be in regions of negative potential energy, so it makes perfect sense that the potential energy at the lowest point would be $U=-2mga$. By definition, then, it is also perfectly sensible that when the pendulum bob is at the height of the pivot point ($\theta=90^\circ, \quad \phi=0^\circ$) that the potential energy is in fact 0. With these definitions now, the 0 of potential energy is no longer the minimum of potential energy and so "the bob hanging there in mid-air" is not an equilibrium point (as you would expect).

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  • $\begingroup$ Thank you. Physics is not my thing as you guessed. I'm a maths student, and this exercise popped up in my "Functions of several variables" unit. So, the only thing I know about energy dates from high school... Hopefully, I'll learn more about this in Unit 9 as was hinted in the question. Still, your explanation is very clear. Cheers. $\endgroup$
    – Stephen
    Commented Nov 25, 2017 at 18:23

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