Intuition for $m (\omega \times v)$ term in unconstrained rigidbody motion I'm sorry if this is a duplicate question, but I've been Googling for hours and can't seem to find anything. I'm confused about the intuition behind the formulas for linear dynamics of an unconstrained rigid body. My textbook (Modern Robotics by Lynch and Park) gives the linear dynamics of an unconstrained rigid body in the inertial body frame $b$ (instantaneously coincident with the center of mass), with linear velocity $v_b$ and angular velocity $\omega_b$ as
$$f_b = m\dot{v}_b + m(\omega_b \times v_b)$$
I'm very comfortable with $f_b = m\dot{v}_b$, since that's just highschool physics $f = ma$. But I'm confused about the intuition behind the $m(\omega_b \times v_b)$. When I do out a simple example of a 1kg object rotating at $\omega_b = [0, 0, 1 m/s]$, traveling at $v_b = [0, 1 m/s, 0]$, with a force applied of $f_b = [0, 1N, 0]$, I end up with an acceleration of $\dot{v}_b = [-1 m/s^2, 1 m/s^2, 0]$ which doesn't make much sense to me.
So here's my question: How can a push in the positive y-axis direction at the center of mass end up creating an acceleration at a 45 degree angle just because the body is rotating? It's an inertial frame, so I thought we weren't supposed to need fictitious forces, and yet that's the only explanation I can think of.
Thanks in advance for the help!
 A: The $\vec\omega\times \vec r_b$ term appear in the rotating (non-inertial) frame.  It is there because the motion of a particle in a rotating frame expressed in the lab contains a "pure" motion on the rotating (non-inertial) frame, plus a term to account for the rotation of the non-inertial frame.
In equations, 
$\vec r'(t)=U(t) \vec r$ where $\vec r'$ is in the lab, $\vec r$ is in the rotating frame, and $U(t)$ is a time-dependent transformation that carries the orientation of rotating set of axes to the fixed ones in the lab.
Taking the time derivative of $\vec r'(t)$ thus involves two pieces by the chain rule, and one can show that 
\begin{align}
\frac{d}{dt}U(t)=U U^{-1} \dot{U}(t) 
\end{align}
and that, basically, $U^{-1}\dot{U}$ boils down to an antisymmetric matrix  $\Omega$, so that $\Omega \vec r\equiv \vec\omega \times \vec r$.
Note that a simplified version of this is provided by the Lagrangian description of a free particle 
\begin{align}
    L=\frac{1}{2}m\left(\dot x^2+\dot y^2+\dot z^2\right)\, ,
\end{align}
as viewed from a rotating coordinate system 
\begin{align}
    x'=x\cos(\theta(t))+y\sin(\theta(t))\, ,\qquad 
    y'=-x\sin(\theta(t))+y\cos(\theta(t))\, ,\qquad 
    z'=z  
\end{align}
where the angle $\theta(t)$ is some function of time.  In terms of these coordinates the Lagrangian takes the form
    \begin{align}
    L=\frac{1}{2}m\left[(\dot{x}')^2 +(\dot{y}')^2 +(\dot{z}')^2
    +2\omega(t)(x' \dot{y}'-y'\dot{x}')+\omega^2(t)(x'^2+y'^2)\right]
\end{align}
    with $\omega(t)=\dot\theta(t)$ is the angular velocity.  The equation of motion for $y(t)$ (for instance) takes the form
\begin{align}
m\ddot{y}'&= -2m\dot{x}'\omega -mx'\dot{\omega }+m\omega^2 y'\, .
\end{align}
These are obtained in the Newtonian framework by including a Coriolis force in the body-frame:
$$
\vec F_C=\hat x (2m\omega \dot{y'})-\hat y(2m\omega \dot{x'})
$$
which is basically $\vec\omega\times \vec r'$ with $\vec\omega=\omega\hat z$.  The centrifugal term in $\omega^2$ is usually neglected unless $\omega$ is large, and the Euler force is in $\dot{\omega}$ and nil if the rotation is uniform.  Of course for the free particle there is no $\dot{v}_b$ term.
