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:
$$T_{space} = T_s = \frac{1}{2}mv\cdot v = \frac{1}{2}m v \cdot (\omega \times r)$$
$$\text{by triple product property } = \frac{1}{2}m \omega \cdot (r \times v)$$
$$=\frac{1}{2}\omega \cdot m(r \times v) = \frac{1}{2} \omega \cdot L$$
$$L = I \cdot \omega$$
$$T_s = \frac{1}{2} \omega \cdot I \cdot \omega$$
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
$$F_{eff} = F – 2m(\omega \times v_{rot}) – m\omega \times (\omega \times r')$$
Integrating both sides we get
$$V' = V – \int 2m(\omega \times v_{rot}) dr – \int m\omega \times (\omega \times r) dr$$
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.
Best Answer
There are three mistakes that prevented you from arriving at the correct lagrangian.
(1) The correct form for a CM lagrangian should be $L = T_{total} - V_{total}$ instead of $L = T_{total} + V_{total}$ I think this is just a typo since later on you did use the correct lagrangian.
(2)It is not valid to assume that $T = T_{space} + T_{rot}$ since energy is not additive in this manner.
(3) There is no need to introduce $V'$. Since the origin is fixed, and since the potential is conservative, $V = V' $
Here is a hint for a correct derivation: Express the velocity of the particle in the rotating frame as a function of the velocity in fixed frame and $\omega_0$. Then plug this velocity into the standard $T = \frac{1}{2}mv^2$ formula. Expand the right hand side and you will get a three term expression that matches what you are asked for. Try it!