NO Uncertainties for particles in their own frames! Well I had this thought experiment in which a particle observes itself, and something like the following is observed.
Taking in mind the uncertainty principle all particles even stopped at 0K jiggle about, but nothing stops us from going into there frame and observing from there. We do just that, we go in the frame of a particle and keep it at the origin, since we are in the frame of the particle it does not move, its entire universe seems to jiggle as from outside frame we would say the particle is jiggling.
So how the principle seems to have failed here? We most definitely know that our particle is motionless at the origin when we see from its own frame, we do not care about the entire universe jiggling about and we notice that the uncertainty principle becomes invalid! Why is such a thing happening? Am I doing something wrong? Or is this a known loophole to the principle?
 A: A 'particle' (in the way that you are thinking about it) is a classical concept. In special relativity, it is possible to go to the rest frame of a particle, but you cannot hope to apply the uncertainty principle.
In quantum mechanics the existence of a 'particle' is described by a state $| \psi \rangle$ in the Hilbert space. The classical concepts of energy, momentum and position are then described in terms of linear operators ${\hat H}$, ${\hat p}$ and ${\hat x}$. It is possible to move into the rest frame of a particle here. (There are issues about normalizability of the such a state, but those can be swept under the rug for the level of rigour here.) However, in quantum mechanics the 'rest frame' is the statement that the state satisfies
$$
{\hat H} | \psi \rangle = m | \psi \rangle,~~ {\hat p} | \psi \rangle = 0
$$
The position of the particle is then described by the expection of the ${\hat x}$ operator, $x_0 = \langle \psi | {\hat x} | \psi \rangle$. This quantity is going to be a precise number. However, the interpretation of this is not that the particle exists at $x_0$. Rather, if one were to measure the position of the particle, it is most likely to be found at $x_0$. More generally, multiple experiments measuring the position of such a particle will yield many different results. The statistical variance of the multiple experimental measurements will satisfy the Uncertainty relation. In this case of course, since $\Delta p = 0$, the variance in $x$ is infinite.
Remember that the uncertainty relation is not a statement about the particle jiggling about and thereby providing uncertainties in measurements. Rather, it is a statement about the fundamental breakdown of the concept of a particle, as we are used to thinking about it. 
A: Let's think a little bit about what it would mean to place ones self in the rest frame of a particle. That is, I want to outline a procedure for making this so rather than simply assuming it.


*

*I boost to the particle's frame 
I put my lab on a spacecraft, boost to interstellar space and wait of a proton to wander by. When one does I measure it's momentum and then design a chase program to bring me into the same frame.  
That measurement entails some uncertainty, or if it does not I must randomize the particles position any in frame, including it's own. 

*I stop a particle in the lab 
Who wants to bother with all that troublesome spaceflight stuff anyway? I'll just stop a particle in the lab and then I'll be in it's rest frame.  
So, I set up some particle detectors around the lab and when the right one comes along (presumably from a beam I designed, but that is a irrelevant detail) I interact with it somehow to bring it to a stop. I know two approaches to this problem.  
First, I could measure it's momentum and hit it with a carefully calibrated hammer (say a tuned photon) to bring it to a stop. That involves both the measurement uncertainty and the momentum uncertainty of the photon.
Second, I could bind it into some kind of trap and then slowly cool the trap. That makes the particle bound and imposes a ground-state energy and consequent momentum uncertainty from the standing wave boundary condition.
So, no matter what you think you have achieved by assuming that you know the rest frame of the particle, there is no operational way to achieve this trick.
A: In particle physics we continually go to the center of mass system for the interactions under study, because the mathematical expressions are simpler and Lorenz invariance assures that the results will be the same in whatever system one studies the interaction.
Suppose we have an electron in a bubble chamber. 

This is a pion decaying into a muon and an electron, with missing two neutrinos, but we are interested in the electron. It leaves a track like a particle should, and this track are tiny bubbles from soft interactions/ionization  of the electron so it keeps losing energy to ionization and that is why it makes a spiral in the magnetic field that permeates the chamber. We can get its momentum at the point of the muon decay with good accuracy by considering the curvature in the beginning. This introduces a measurement error that is much larger than the heisenberg uncertainty principle (HUP) of the delta(x) of the muon-to-electron point with the delta(p) of our measurement, so even though we go to the center of mass of the electron , the measurement uncertainty will propagate on all the variables in the system, the momentum of the neutrinos and the muon in the center of mass system of the electron. All values will have a +/-delta to them, thus position and momentum are not known at the same time, because of measurement errors . The position of the electron has an error and thus even if we think we are at the electron center of mass there will be a measurement uncertainty at the decay point of the decaying muon and electron neutrino, and the momentum of the electron will be 0+/-delta(p) .
Suppose that we could measure all the bubbles to a very small accuracy and the curvature of the electrons to such a small accuracy. The HUP tells us that this is not possible because the wave nature of the probability distribution for the electron will limit the measured  values with a delta(x) and a delta(p). Thus in the center of mass of the electron the decay point will be uncertain and the 0 momentum of the electron will have a delta spread.
This would be true for a hypothetical free electron too at rest, the zero in x will have +/- delta(x) as well as the zero in momentum +/- delta(p) from the  HUP uncertainty.
