[NOTES/ME-08005]-Pseudo forces in a Rotating Frame

The position of a point as seen in an inertial frame will not appear in our discussions below. Hence we will now drop the superscript $\prime$ and use $\vec{x}$ to denote the position of the particle as seen from a rotating frame and write the EOM as \begin{eqnarray} \left.\DD[\vec{x}]{t}\right|_\text{rf} =\frac{\vec{F}}{m} - 2 \vec{\omega}\times \vec{v}- \vec{\omega}\times(\vec{\omega}\times\vec{x}). \end{eqnarray} Consider a point $P$ on the earth at latitude $\lambda$. Choose the $X_1$ axis towards east, $X_2$ towards north and $X_3$ axis vertically upwards as shown in Fig.\ref{me-fig-060} below. The angular velocity vector of the earth lies in a plane containing the axis of spin of the earth and the $X_3$ axis. It has horizontal component $\omega \cos\lambda$ along the $X_2$ axis and a vertical $\omega \sin \lambda$. \\ \FigBelow{30,10}{100}{80}{me-fig-060}{}\\ The angular velocity vector is therefore given by \begin{equation} \vec{\omega} = (0,\omega \cos\lambda, \omega \sin \lambda).  \end{equation} Therefore, \begin{eqnarray} \vec{\omega} \times \vec{v} &=& \begin{vmatrix} \hat{e}_1 & \hat{e}_2 & \hat{e}_3 \\ 0 & \omega \cos\lambda &\omega \sin \lambda\\ v_1 & v_2 & v_3 \end{vmatrix} \\ &=&(\cos\lambda v_3- \sin\lambda v_2)\omega\hat{e}_1 +v_1 \omega \sin\lambda \hat{e}_2 - v_1\omega \cos\lambda \hat{e}_3 \end{eqnarray} For rotation of Focault pendulum's plane of oscillation, we need to consider only the first and second components of the equation of motion \begin{eqnarray} \left.\DD[\vec{x}]{t}\right|_\text{rf} =\frac{\vec{F}}{m} - 2 \vec{\omega}\times \vec{v}- \vec{\omega}\times(\vec{\omega}\times\vec{x}). \end{eqnarray} Neglecting the vertical motion of the pendulum and \(\omega^2\) terms, the equations of motion for \(x_1, x_2\) coordinates take the form \begin{eqnarray} \ddot{x}_1 &=& - 2v_2\omega \sin \lambda = - 2\dot{x}_2 \omega \sin \lambda \\ \ddot{x}_2 &=& 2v_1\omega \sin \lambda = 2\dot{x}_1 \omega \sin \lambda \end{eqnarray} Multiply {EQ06}) by $x_2$ and ({EQ07}) by $x_1$ to get \begin{eqnarray} x_2 \ddot{x}_1 &=& - 2x_2 \dot{x}_2 \omega \sin \lambda ,\\ x_1 \ddot{x}_2 &=&2 x_1 \dot{x}_1 \omega \sin \lambda{EQ19}. \end{eqnarray} Subtracting the above two equations we get \begin{eqnarray} x_2 \ddot{x}_1- x_1 \ddot{x}_2 &=& -2 ( x_2 \dot{x}_2+x_1 \dot{x}_1 )\omega \sin \lambda \\ \text{or} \qquad \dd{t}(x_2 \dot{x}_1- x_1 \dot{x}_2) &=& -\omega \sin \lambda \dd{t}(x_1^2+x_2^2). \end{eqnarray} Integrating w.r.t. time we get \begin{eqnarray} (x_2 \dot{x}_1- x_1 \dot{x}_2) &=& - (x_1^2+x_2^2)\omega \sin \lambda + \text{Const} \end{eqnarray} Introducing polar coordinates $\rho, \phi$ by means of equation \begin{equation} x_1 = \rho \cos \phi, \quad x_2 = \rho\sin\phi,\end{equation} we note that \(\dot{\rho}\) terms in the left hand side cancels and we are left \begin{equation} x_2 \dot{x}_1- x_1 \dot{x}_2 = \rho^2 \dd[\phi]{t}, \quad \text{and} \quad x_1^2+x_2^2=\rho^2 .\\ \end{equation} This allows us to write {EQ11} in the form \begin{eqnarray} \dd[\phi]{t}= -\omega\sin\lambda . \end{eqnarray} where we have used initial condition \(\phi=0\) at time \(t=0\). Therefore, the plane of the pendulum rotates with angular velocity $\omega \sin\lambda$.