Pseudo force and Potential energy

Question

The pseudo force equation is

$$\vec F_p=-2\,m\,(\vec \omega\times \vec v)-m\,(\vec\omega\times (\vec\omega\times \vec R))$$

where $~\vec v=\dot{\vec{R}}$ and $~\vec\omega=\text{const.}$

the pseudo force is depending on the generalized coordinates $~\vec q~$ and the generalized velocities $~\dot{\vec{q}}$

If we applied the Euler-Lagrange to obtain the equations of motion, the pseudo force is external force , but instead we can use potential energy with this Lagrange function

$$\mathcal{L}(~\vec q~,~\dot{\vec{q}})=\frac m2\,\vec{v}\cdot \vec{v}-U_p\\ U_p=m\,(\vec\omega\times\vec v)\cdot \vec R-\frac m2 (\vec\omega\times\vec R)\cdot (\vec\omega\times\vec R)$$

my question is what is the mathematical derivation to obtain the pseudo force potential $~U_p$?

Answer 1

I don't think there is a general method to find $U_p$, but it must satisfy (using $x,y,z$ as generalized coordinates) $$\vec{F}_p=\frac{d}{dt}\left(\vec{\nabla}_{\vec{v}}U_p\right)-\vec\nabla U_p$$or, in components, $$\left(\vec{F_{p}}\right)_i=\frac{d}{dt}\left(\frac{\partial U_p}{\partial v_i}\right)-\frac{\partial U_p}{\partial x_i}.\tag{1}$$

You could try a solution in the form (repeated indices are summed from $1$ to $3$)$$U_p=A_{ijk}~\omega_i~v_j~x_k~+~B_{ijkl}~\omega_i~\omega_j~x_k~x_l~+...\tag{2}$$ where $x_i$ are the components of the vector $\vec{R}.$

Note that since the force does not depend on the acceleration, the generalized potential may depend at most linearly on the velocity. You can add terms to $(2)$ using arguments like this.

More notes:

• The components of the force are (repeated indices are summed) $$\left(\vec{F_{p}}\right)_i=-2m~\epsilon_{ijk}~\omega_j~v_k-m~\epsilon_{ijk}~\omega_j~\epsilon_{kab}~\omega_a~x_b$$
• By construction, $B_{ijkl}=B_{jikl}=B_{jilk}$.