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Bias variance tradeoff seems to behave like the uncertainty principle, is it just another name for the same principle?

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No, the uncertainty principle describes a property that is specific to electrons. That electrons don't display their wave and particle properties simultaneously. Here from Wikibooks:

The Heisenberg uncertainty principle states that the exact position and momentum of an electron cannot be simultaneously determined. This is because electrons simply don't have a definite position, and direction of motion, at the same time!

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This is due to the fact that electrons cannot exhibit both their wave and particle properties at the same time when being observed to interact with their surroundings. The momentum of an electron is proportional to its velocity, but based on its wave properties; its position is based on its particle position in space.

The bias-variance tradeoff describes a property of predictive models. That error from bias and variance usually exist in a tug-o-war where you cannot decrease one without increasing the other and that you have to find the perfect middle ground. You can observe it here from this Medium:

bias-variance curve

I agree that there are some similarities. There is a sort of tradeoff between measuring the momentum and position of electrons. But they are not the same thing.

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    $\begingroup$ The uncertainty principle is true of all fundamental particles. In fact it is true of all objects, but usually not a practical issue for larger objects. To augment the second part of your answer, it also takes a precise mathematical form as a limit to measurement $\Delta \chi \Delta \rho \ge \frac{h}{4\pi}$ with specific physical units, whilst bias/variance has no such form (and you can have zero bias, zero variance models in theory) $\endgroup$
    – Neil Slater
    Apr 23, 2019 at 6:24
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    $\begingroup$ I suspected that it applied to more than electrons but didn't want to spend to much time reading up on the rabbit hole that is quantum mechanics. So thank you for expanding and answering my questions, @NeilSlater! :) $\endgroup$
    – Simon Larsson
    Apr 23, 2019 at 7:05

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