A baseball truly at rest would have less momentum (i.e., zero) than a moving electron. But a baseball could have many, many times the momentum of an electron and still be moving so slowly as to be indistinguishable from “at rest”.
Oh, now even* I* can understand that! 
When you said ‘the faster it goes (or the heavier it is)…’ were you talking about the relativistic effects? Also, is this why we get such odd behaviour of free particles in a vacuum in the double-slit-experiment, i.e., that they are isolated from anything much bigger?
Okay, that makes sense from what ftg said. 
I have read science articles that say this is why liquid helium doesn’t freeze solid. The “quantum jiggle” is just enough to keep it liquid.
(Is this correct, or am I remembering it completely wrong?)
Can a baseball be at relative net-rest? Where its net momentum within its frame of reference is zero but the motion of its many constituent parts is each non-zero?
Good post, but I’d like to urge some caution against overstating the analogy between duration and frequency of a wave, and the uncertainty principle (not that I think you do, but people might get the wrong impression).
Basically, there’s nothing counterintuitive about the relationship between frequency and duration: a note that’s been played for a certain finite time simply doesn’t have a single frequency; and when we record that note, we have exact values for its duration and the frequency range it spans. All of its properties are thereby completely known and fixed.
The same can’t be said for a quantum particle: if it is, say, in a fairly sharp state of momentum, then it’s not possible to exactly predict its position; but upon measurement, we will always find it in a single place. So all of the particle’s properties, as they are revealed to us via measurement, are not fixed by what we know about its state.
Furthermore, uncertainty also exists between discrete variables: for, e.g., a particle’s spin, which can be either ‘up’ or ‘down’ in a given direction, there exists an uncertainty principle between different orthogonal directions of spin—perfectly knowing spin in the x-direction means you don’t know anything about spin in y-direction.
Yes, by definition. “Its own frame” means the frame where its overall momentum is zero. But we can’t tell what frame that is. (surprisingly, the math for combining quantum mechanics and special relativity still manages to all turn out just fine)
And immediately upon making that position measurement, it’s no longer in a fairly sharp momentum state.
OP has his answer, I think, so jokes are in order:
Very good. 
As a post script I think I would like to say that I have learnt that sub-atomic particles are very * elusive.*
abashed, thank you for asking this as I am in the process of reading (for fun),“Quantum Mechanics: The Theoretical Minimum” and this clears a few things up for me.
Derleth, thank You for the clear analogy!
I think we did a set of Heisenberg jokes a couple of months ago. This is the one I like.
Heisenberg is driving down the road and he gets pulled over by a cop. The cop walks up to him and says “Do you know how fast you were going?” Heisenberg says “No, but I know exactly where I am!”
This is awesome. It caused me to google, where I found this Interactive Guide to the Fourier Transform. This really helped me understand what you explained even more intuitively.
Glad to be of service swampspruce ;).