Short answer: No. For any choice of the inertia M and the fixed rigid thruster T, a single nonreversible input f(t) ≥ 0 cannot make the endpoint map locally surjective at 0, hence (by your own time/scale symmetries) global controllability between arbitrary angular velocities is impossible.

A simple quantitative obstruction

It is convenient to work with the angular momentum L = {M,Ω} ∈ so(4), for which Euler’s equation becomes the control-affine system L' = −[K L, L] + f(t) T, L(0) = 0, where K = {M,·}^{−1} is the positive self-adjoint linear operator that sends L to Ω, and T ∈ so(4) is the constant torque direction generated by the thruster.

Fix the Frobenius norm on matrices. There is a universal constant C0 such that for all A,B, ||[A,B]||F ≤ 2 ||A||F ||B||F. Since K is bounded on so(4) (with operator norm ||K|| depending only on M), ||[K L, L]||F ≤ 2 ||K|| ||L||F^2. (1)

Write the mild (variation-of-constants) form of the solution: L(t) = ∫_0^t f(s) T ds − ∫_0^t [K L(s), L(s)] ds. (2)

Let ε := ∫_0^{t1} f(s) ds be the total “impulse” delivered by the thruster up to time t1 (this ε is exactly the sum of the coefficients if f is a sum of Dirac deltas). Denote S := ε T. From (2) and (1), ||L(t1) − S||F ≤ ∫_0^{t1} 2 ||K|| ||L(s)||F^2 ds. (3)

A bootstrap bound gives that, for ε small enough (depending only on ||K||, ||T||F, t1), one has sup_{0≤s≤t1} ||L(s)||F ≤ 2 ||T||F ε, whence, by (3), ||L(t1) − ε T||F ≤ 8 ||K|| ||T||F^2 t1 ε^2. (4)

In words: starting from L(0)=0, the terminal angular momentum is the line segment ε T plus a uniformly quadratic correction in the total impulse ε. The same estimate holds verbatim when f is a finite sum of Dirac deltas (integrals become finite sums; the quadratic bound is preserved).

Consequences

• The reachable set from 0 in so(4) for small total impulse ε lies in an O(ε^2)-tube around the half-line {σ T : σ ≥ 0}. In particular, for ε arbitrarily small, it does not contain any open neighborhood of 0 in so(4).

• Your scaling symmetry then rules out global reachability: if one could steer 0 to any prescribed Ω1, then by scaling one would be able to steer 0 to k Ω1 for all sufficiently small k>0, contradicting the local obstruction above.

Therefore there do not exist M and T with the property stated in your question.

Remarks

• The argument does not use any 4-dimensional special structure; it works in any dimension n ≥ 2. The key features are: (i) the drift is quadratic, with no linear term at the origin, and (ii) the control enters additively along a fixed direction with a one-sided amplitude constraint f ≥ 0. These two facts force the first-order term of the endpoint map at 0 to be one-dimensional (along T), with only second-order corrections available via the drift.

• If the actuator could produce torques of both signs along T (i.e., f allowed to take positive and negative values), one can cancel the first-order term and start exploiting second-order Lie-bracket effects. For generic choices one then expects much richer accessibility. But with the one-sided thruster considered here, the first-order obstruction cannot be removed.