Solution Example II

Let us define the z-axis equal to to the direction of the angular velocity vector ${\bf\omega}$. The variables in the coordinate system K and those in the co-rotational system K' are related by the rotation matrix as

$$\left( \begin{array}{c} x \\ y \\ \end{array}\right) = \... ...\left( \begin{array}{c} x^{'} \\ y^{'} \\ \end{array}\right)$$ (10)

where $\theta$ is an angle between $x$-axis in K and $x^{'}$-axis in K'. Note that

$$\dot{\theta}\equiv {d\theta\over dt} = \omega$$ (11)

and

$${d^2\theta\over dt^2} = {d\omega\over dt}\equiv \dot{\omega}$$ (12)

Then the time differential operation applied in the equation (10) gives

$$\left( \begin{array}{c} {d^2x\over dt^2} \\ {d^2y\over dt^22} \\ \end{array}\right) = \left( \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta ... ...( \begin{array}{c} {dx^{'}\over dt} \\ {dy^{'}\over dt} \\ \end{array}\right)$$

$$-\dot{\omega} \left( \begin{array}{c} y \\ -x \\ \end{array}\right) -\omega^2 \left( \begin{array}{c} x \\ y \\ \end{array}\right).$$ (13)

Here we occasionally used the relation (10) whenever applicable. Multiplication of the inversed rotation matrix from left-side to the equation above then leads to

$$\left( \begin{array}{c} {d^2x^{'}\over dt^2} \\ {d^2y^{'}\over dt^2} \\ \end{array}\right) = \left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta ... ...( \begin{array}{c} {dx^{'}\over dt} \\ {dy^{'}\over dt} \\ \end{array}\right)$$

$$+ \dot{\omega} \left( \begin{array}{cc} \cos\theta & \sin\theta \\ ... ... \\ \end{array}\right) \left( \begin{array}{c} x \\ y \\ \end{array}\right).$$

Using the Newtonian equation of motion

$$\left( \begin{array}{c} m{d^2x\over dt^2} \\ m{d^2y\over dt... ...t) = \left( \begin{array}{c} F_x \\ F_y \end{array}\right)$$ (14)

where $(F_x, F_y)$ is the external force of the x and y component, respectively, and

$$\left( \begin{array}{c} F_x \\ F_y \\ \end{array}\right) ... ...( \begin{array}{c} F_x^{'} \\ F_y^{'} \\ \end{array}\right),$$ (15)

we finally obtain

$$m{d^2x^{'}\over dt^2} = F_x^{'}+2m\omega {dy^{'}\over dt} +m\omega^2x^{'}+m\dot{\omega}y^{'}$$ (16)

$$m{d^2y^{'}\over dt^2} = F_y^{'}-2m\omega {dx^{'}\over dt} + m\omega^2y^{'}-m\dot{\omega}x^{'}$$ (17)

One finds the apparent force term by $\dot{\omega}$.