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$\newcommand{\Label}[1]{\label{#1}}\newcommand{\Prime}{{^\prime}}\newcommand{\pp}[2][]{\frac{\partial#1}{\partial #2}}\newcommand{\PP}[2][]{\frac{\partial^2#1}{\partial #2^2}}\newcommand{\dd}[2][]{\frac{d#1}{d #2}}\newcommand{\DD}[2][]{\frac{d^2#1}{d #2^2}}$
A body falling under the earth's gravitational field experiences a side ways deflection due to Coriolis force. An expression for this deflection is obtained and numerical values are estimated.
The position 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}).\Label{EQ12} \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 \Figref{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. Remembering that the earth turns anticlockwise if viewed from the Northern hemisphere, the angular velocity 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). \Label{EQ13} \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} \Label{EQ14}\\ &=&(\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\Label{EQ15} \end{eqnarray}
Neglecting the vertical motion of the pendulum, the EOM for the motion in the $X_1, X_2$ plane are given by
\begin{eqnarray} \ddot{x}_1 &=& 2v_2\omega \sin \lambda = 2\dot{x}_2 \omega \sin \lambda \Label{EQ16}\\ \ddot{x}_2 &=& - 2v_1\omega \sin \lambda = -2\dot{x}_1 \omega \sin \lambda\Label{EQ17} \end{eqnarray}
Multiply \eqRef{EQ16} by $x_2$ and (\EqRef{EQ17}) by $x_1$ to
get
\begin{eqnarray} x_2 \ddot{x}_1 &=& 2x_2 \dot{x}_2 \omega \sin \lambda \Label{EQ18},\\ x_1 \ddot{x}_2 &=& -2x_1 \dot{x}_1 \omega \sin \lambda\Label{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 \omega+x_1 \dot{x}_1 )\omega \sin \lambda \Label{EQ20} \\ \text{or} \qquad \dd{t}(x_2 \dot{x}_1- x_1 \dot{x}_2) &=& \dd{t}(x_1^2+x_2^2)\omega \sin \lambda \Label{EQ21} \end{eqnarray} Introducing polar coordinates $\rho, \phi$ by means of equation
\\begin{equation} x_1 = \rho \cos \phi, \quad x_2 = \rho\sin\phi,\Label{EQ22} \end{equation}
we note that \begin{equation} x_2 \dot{x}_1- x_1 \dot{x}_2 =- \rho^2 \dd[\phi]{t} \Label{EQ23}\\ \end{equation} which allows us to write \EqRef{EQ21} in the form
\begin{eqnarray} - \dd[(\rho^2\phi)]{t} = \Big( \dd[\rho^2]{t}\Big) \omega \sin \lambda,\Label{EQ24} \\ \therefore \qquad \rho^2\dd[\phi]{t} = - \rho^2 \omega \sin\lambda,\Label{EQ25} \\ \Rightarrow \qquad \dd[\phi]{t}= - \omega\sin\lambda . \Label{EQ26} \end{eqnarray}
Therefore, the plane of the pendulum rotates with angular velocity $\omega \sin\lambda$. The negative sign in \(\dot{\phi}\) means that the plane of oscillation turns clockwise in the northern sphere. As seen from the inertial frame, the plane of oscillation of the Focault pendulum remains fixed.
