Lagrangian in rotating reference frame

I have a homework question I am working on, and I’m stuck on the last term in this proof. The problem states:

Question: Consider a primed set of axes coincident in origin with an inertial set of axes but rotating with respect to the inertial frame with fixed angular velocity $omega_0$. If a system of mass points is subject to forces derived from a conservative potential $V$ depending only on the distance to the origin, show that the Lagrangian for the system in terms of coordinates relative to the primed set can be written as

$$mathcal{L} = T’ + omega cdot L’ + frac{1}{2} omega cdot I’ cdot omega_0 – V$$

where primes indicate the quantities evaluated relative to the primsed set of axes.

Important Notes: Here $L’$ is the angular momentum in the rotating frame and $V$ is the potential in the non-rotating frame.

Attempt at Solution:

We know that the Lagrangian is given by
$$mathcal{L} = T_{tot} – V_{tot}$$

I will proceed by only considering one particle (this is a rigid body problem, thus if we prove this is true for one particle it is true for a distribution of particles). The kinetic energy is equal to,

$$T_{tot} = T_{rot}+T_{space}$$

where “space” refers to the space coordinates in the inertial frame. We get $T_{space}$ by the following manipulation:

$$T_{space} = T_s = frac{1}{2}mvcdot v = frac{1}{2}m v cdot (omega times r)$$
$$text{by triple product property } = frac{1}{2}m omega cdot (r times v)$$
$$=frac{1}{2}omega cdot m(r times v) = frac{1}{2} omega cdot L$$
$$L = I cdot omega$$
$$T_s = frac{1}{2} omega cdot I cdot omega$$

Thus I obtain one of the four terms. Now I consider the kinetic energy and potential energy in the rotating reference frame. The problem wants me to leave the kinetic energy in the reference frame in terms of $T’$, so I already get that term (now I have 2/4 terms). The final thing to consider is the potential energy in the rotating reference frame.

$$mathcal{L} = T’ – V’ + frac{1}{2} omega cdot I cdot omega$$

We know that in a rotating reference frame, the effective force is given by

$$F_{eff} = F – 2m(omega times v_{rot}) – momega times (omega times r’)$$
Integrating both sides we get

$$V’ = V – int 2m(omega times v_{rot}) dr – int momega times (omega times r) dr$$

This is where I start getting confused (hopefully I’m thinking of this step correctly). $omega$ is constant, but $v_r$ shouldn’t be since there are forces acting on the particle. Therefore, we would expect the velocity to change. So I don’t know how to integrate this first term. I could try putting the velocity in terms of the angular momentum (which should be a constant because it is a conservative force), but then I get a natural log in my answer, which of course isn’t right.

The second term I think I can integrate. In cylindrical coordinates, if we define $vec{omega} = omega hat{z}$ then $omega times (omega times r) = -omega^2 r$. Integrating I get $-frac{1}{2} m omega^2 r^2$ I’m not really sure what to even do with this term.

Any help where I am going wrong would be much appreciated! Thanks.