RMS of absolute-valued sinusoid A signal $$x(t)=\left|\sin(2\pi(1000)t)\right|$$ ($\left|x\right|$ is the absolute value function) is fed to an ideal low-pass filter with cutoff frequency of $1500\,\mathrm{Hz}$ to produce an output $y(t)$, what is the RMS voltage of $y(t)?$

My initial belief is that it will be zero since the absolute value doubles the frequency of the signal and so the frequency of $y(t)$ will be $2000\,\mathrm{Hz}$ which is beyond the cut-off frequency of the ideal low pass filter.

However, upon reflecting on the Fourier transform of $y(t)$, I believe it will have a DC component with zero frequency which will give it an RMS voltage. Moreover, $y(t)$ is a fully rectified signal which, when passed through low pass filters in reality (i.e. capacitors) will yield a smooth DC signal. 

How do I find the RMS voltage of the resulting $y(t)?$ Is it really nonzero? Will it be just the same as the RMS voltage of $x(t)?$
Your reasoning is correct. The signal $x(t)=|\sin(2\pi f_0t)|$ is periodic with period $T/2$, where $T=1/f_0$. Consequently, it has components at frequencies $2k/T$, $k=0,1,\ldots$, i.e., at $0$Hz, $2000$Hz, $4000$Hz, etc. Since the ideal low pass filter has a cut-off frequency of 1500Hz, all harmonics are completely suppressed, and only the DC component appears at the output.

I'm sure that you know how to compute the DC component of $x(t)$ (which, if positive, equals its RMS value).

Your concerns about actually implemented filters and capacitors etc. are irrelevant to the question, which explicitly states that the low pass filter  can be assumed to be ideal. But even a non-ideal low pass filter of appropriate order would suppress all harmonics sufficiently, such that they would be buried in noise.