Any question answered...

What kind of wheel obeys I = m*r^2 yet gives such a simple result as "double"?

For an ideal wheel (all mass concentrated at periphery) of
radius = r
mass = m
rolling at a forward
velocity = v
moment of inertia
I=mr²
and angular momentum
L=Iω
Angular velocity relates to forward velocity and radius
ω=v/r

Now apply an arbitrary braking
torque = τ
for a
time = t

such that
L=τt
bringing ω to zero

The
force = F
at the periphery needed to generate the torque is given by
Fr=τ

Now rearrange

L=mrv
&
L=Frt
so
Frt=mrv
and dividing each side by r gives
Ft=mv
in other words, the impulse which can be extracted from a spinning wheel is maximised at the same value as the translational momentum for the ideal condition of concentrating the whole mass of the wheel at its periphery, independent of the radius.

This gives a maximal impulse available from a wheel of exactly double the translational momentum mv; all practical wheels will provide a smaller impulse than this.

PS: I haven't been in a maths class for 30 years, so I won't be too upset if somebody proves me wrong :-)