My work-energy analyses in my last note are okay, but incomplete. I
think one can actually distinguish four subcases:
A. work = pseudo-translational-work, mechanical energy does not
include rotations
W-E thm => -f*d = 0.5*m*v^2 - mgh
B. work = pseudo-translational-work + pseudo-rotational-work,
mechanical energy includes rotations
W-E thm => -f*d + f*R*s = 0.5*m*v^2 + 0.5*I*w^2 - mgh
Moral: if you include rotational energy side on the RHS of the W-E
thm, you have to include rotational work on the LHS.
C. work = thermodynamic work, mechanical energy does not include rotations
1st law => 0 = 0.5*m*v^2 - mgh + delta (E_int)
The wheel gains rotational internal energy. This is not what is
usually done in thermo, but can be treated perfectly consistently.
D. work = thermodynamic work, mechanical energy includes rotations
1st law => 0 = 0.5*m*v^2 + 0.5*I*w^2 - mgh
Moral: you have to define clearly what you mean by "internal energy."
Different authors use different definitions. Caveat emptor.
In case it's not obvious in the above, d = translational distance
along plane, s = total rotational distance of rim of wheel = R*d, h =
vertical height of incline = d*sin(theta).
It is analyses A and D that I think are most useful for students and
that I called 1 and 2 in my last message. Carl