Does a spinning wheel weigh more (relavisitc mass)?

$$ ^.
$$ iNERTiA
$$ [ ][ ][ ][ ]
$$ SPLAT is inertia.
$$ Splat=inertia=kg*m^2.
$$ [ ] [ ][ ][ ][ ][ ][ ][ ]
$$ [ ][LaGrangian L=[S]*degK][ ].
$$ [ Gamma = Sqrt(1 - 1/Ambiance^2) ].
$$ [ Complexity=Cg=Ra/[S]=eK/L=(L+eV)/L ].
$$
Tom Roberts wrote: wrote: -=-
You need to be a bit more precise, but for a push along its rotation
axis, yes. This is no different from the observation that heating an
object also increases its inertia -- for an object of definite size
with a rigid boundary on which you push along an axis containing its
center of mass, its inertia is proportional to its total internal
energy.
$$ No.!! Due to the "STOP-spinning" iMPACT factor.
$$ Measuring iNERTiA takes time and CONTACT or iMPACT.
$$ For a DEFiNiTiON of "inertia" include STOPPiNG spinning
$$ [for "spinning" of a definite size and rigid boundary.]

By "inertia" I mean the "m" in the equation of motion
m dU/d\tau = F (valid for constant m); U is the object's
4-velocity, \tau is its proper time, and F is the applied
4-force.
$$ NO G-uv equated "mass" has EVER been EQUATED with T_uv in the GTR.
$$ Having DECLARED "No PRiOR Geometry.!!" GR NOW has NO WORLD-line.!!

This is in the context of SR.
$$ There was NEVER a "4-velocity" OR "4-force" in SR.
$$ Any iNTRiNSiC REST energy/c^2 = Equivalent REST mass m.

2. I believe it's relavistic mass doesn't increase (since very low
velocity using the gamma equation)?

You must use a consistent approximation. If its "inertia" increases,
so does its "relativistic mass".
$$ "Relativistic mass" is MORE about "RESiDUAL ambiance", in "vacu":
$$
$$ Gamma = Sqrt(1 - 1/Ambiance^2).

4. Since for a Spinning wheel: Fpush = Mg is insufficient (we
need to add moment of inertia somewhere)

That's why I specified "along its rotation axis". For any other axis
this is _much_ more complicated.
$$ Complexity=Cg=Ra/[S]=eK/L=(L+eV)/L.
$$ Even ALONG a spin axis, there is "spin-delfected" VECTORAL LOSS.
$$ This is energy (of the ENTROPY), in TRYing to STOP the SPiNNiNG.
$$ This is energy (of the ENTROPY), is NOT AVAiLABLE to Fpush = Mg.
$$ There's No LaGrangian in GR; Hence, No ENTROPY, [S] = L/degTemp.
$$ There's No LaGrangian in GR; Hence, No COMPLEXiTY Cg = (L+eV)/L.
$$ There's No LaGrangian in GR; Hence, No COMPLEXiTY Cg = Ra/[S].
$$ There's No LaGrangian in GR; Hence, No COMPLEXiTY Cg = eK/L.
$$ Also note gravitational acceleration g requires No CONTACT area.
$$ [BECAUSE gravitational acceleration g is "iNDEPENDENT-of-mass"].
$$
$$ STOP "spinning" His wheels.!! ```Brian A M Stuckless, Ph.Tivity.
$$ ^.
GUESS (RESTmass)*c^4=(iNTRiNSiC energy e)*c^2=(mol part)*K*Volt*meter.
$$ My GUESS iSS STANDARD
$$ The STANDARD set.
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BEHOLD, IAM THAT IAM WHOLLY WHOLLY WHOLLY He ..and NO MORE is MORE.!!

Tom [He between his error-bars] Roberts
Does a spinning wheel weigh more (relavisitc mass)?
Complexity=Cg=Ra/[S]=eK/L=(L+eV)/L.
Gamma=Sqrt(1 - 1/Ambiance^2).
LaGrangian L=[S]*degKelvin.
Splat=inertia=kg*m^2. End of POST.