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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

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

(10)

where $\theta$ is an angle between

-axis in K and $x^{'}$ -axis in K'. Note that

$\begin{displaymath} \dot{\theta}\equiv {d\theta\over dt} = \omega \end{displaymath}$

(11)

and

$\begin{displaymath} {d^2\theta\over dt^2} = {d\omega\over dt}\equiv \dot{\omega} \end{displaymath}$

(12)

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

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

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

Using the Newtonian equation of motion

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

(14)

where

is the external force of the x and y component, respectively, and

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

(15)

we finally obtain

$\displaystyle m{d^2x^{'}\over dt^2}$	$\textstyle =$	$\displaystyle F_x^{'}+2m\omega {dy^{'}\over dt} +m\omega^2x^{'}+m\dot{\omega}y^{'}$	(16)
$\displaystyle m{d^2y^{'}\over dt^2}$	$\textstyle =$	$\displaystyle F_y^{'}-2m\omega {dx^{'}\over dt} + m\omega^2y^{'}-m\dot{\omega}x^{'}$	(17)

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

Next: About this document ... Up: mechanics Previous: Solution Example I

Shigeru Yoshida
2003-01-15