Equation of Motion for Rigid Body Motion

Question

In a paper, eq 24 I am reading, the author mentions the equation of rigid body motion which is written as the sum of translational motion of the centre of mass, $x_G(t)$ and a rotational term about an axis through the centre of mass, viz,

$$w(t,x)=w_G(t)+R(t)\times(x-x_G(t)),$$ where $w$ is the velocity field and $w_G$ the velocity of the centre of mass and $R$, is the angular velocity(?).

Then they proceed to differentiate the equation to obtain the equation

$$\frac{dw(t,x)}{dt}=\frac{dw_G(t)}{dt}+\frac{dR(t)}{dt}\times(x-x_G(t))+R(t)\times(R(t)\times (x-x_G(t))).$$

I do not understand how the third term in this equation is obtained and what it signifies. Should the third term not be $R(t)\times w(t,x)$?

Answer 1

Equation 24 of your paper says:

$$\vec{u}=\vec{U}+\vec{\omega}\times\vec{R}$$

Here $\vec{U}$ is the position of the center of mass, $\vec{\omega}$ is the angular velocity vector, and $\vec{R}$ is the position of one point in the rigid body. Taking the derivative,

$$\frac{d\vec{u}}{dt}=\frac{d\vec{U}}{dt}+\frac{d\vec{\omega}}{dt}\times\vec{R}+\vec{\omega}\times\frac{d\vec{R}}{dt}$$

Notice that $\frac{d\vec{R}}{dt}$ is the tangential velocity, which can be written as $\vec{\omega}\times\vec{R}$. Plugging this in, the result is,

$$\frac{d\vec{u}}{dt}=\frac{d\vec{U}}{dt}+\frac{d\vec{\omega}}{dt}\times\vec{R}+\vec{\omega}\times(\vec{\omega}\times\vec{R})$$

Hopefully you can translate this notation to your own.