[QUE/ME-14001] ME-PROBLEM

A point of rigid body is pivoted in way that the body can rotate freely about a horizontal axis passing thorough the point of suspension. Show that the frequency of small oscillations is given by \[\omega^2= \frac{Mgh}{Mh^2 + I_1n_1^2+I_2n_2^2+I_3n_3^2}\] where \(I_1, I_2, I_3\) are principal moments of inertia about the centre of mass and \(h\) is distance of centre of mass from the point of suspension. Also \(\hat{n}=(n_1,n_2,n_3)\) is unit vector along the axis of rotation and the components of \(\hat{n}\) are taken w.r.t. the principal axes relative to the centre of mass of the body.
%FigBelow{10,-47}{40}{0}{me-fig-14006}

This is a problem about rotation of a rigid body about a fixed axis. In this case the axis of rotation is horizontal line passes through the point of suspension. The equation of motion is obtained by equating the rate of change of angular momentum to the torque. The angular momentum is \(I\dot{\theta}\), where \(I\) is the moment of inertia about the axis of rotation. The torque is given by \(Mgh\sin\theta\) and is in the clockwise direction of decreasing \(\theta\). Therefore the equation of motion becomes \begin{equation} \dd[L]{t} = - Mgh \sin\theta\\ \end{equation} Substituting \(L=I\dot{\theta}\) and making small amplitude approximation we get \begin{eqnarray} I \ddot{\theta} + Mgh \theta=0\\ \therefore \qquad \qquad \ddot{\theta} + \frac{Mgh}{I} \theta=0. \end{eqnarray} Thus the angular frequency of oscillations is \(\omega^2=\dfrac{Mgh}{I}\). Using the parallel axes theorem the moment of inertia can be written as \begin{equation} I= I_0 + Mh^2 \end{equation} where \(I_0\) is the moment of inertia about a parallel axis passing through the centre of mass. In terms of the principal moments of inertia the value of \(I_0\) is given by \begin{equation} I =I_1n_1^2 + I_2n_2^2 + I_3n_3^2. \end{equation} Thus we get the final answer \begin{equation} \omega^2=\frac{Mgh}{I_0 + I_1n_1^2 + I_2n_2^2 + I_3n_3^2}. \end{equation}