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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.
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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}

Consider the right hand side of the identity (a) to be proved and write all terms in the sum. There will be four terms as \(i\) and \(j\) take values 1 and 2. This gives \begin{eqnarray} \frac{1}{2}S_{ij}\big(T_{ij}+T_{ji}\big) &=&\frac{1}{2}\sum_{i=1}^2\sum_{j=1}^2S_{ij}\big(T_{ij}+T_{ji}\big)\\ &=& \frac{1}{2}\sum_{i=1}^2\Big( S_{i1}\big(T_{i1}+T_{1i}\big)+ S_{i2}\big(T_{i2}+T_{2i}\big)\Big)\\ &=&\frac{1}{2} \Big( S_{11}\big(T_{11}+T_{11}\big)+ S_{12}\big(T_{12}+T_{21}\big)\Big)\\ && +\frac{1}{2} \Big( S_{21}\big(T_{21}+T_{12}\big)+ S_{22}\big(T_{22}+T_{22}\big)\Big)\\ &=& S_{11}T_{11} +S_{12}T_{12} + S_{21}T_{21}+ S_{22}T_{22}\\ &=& \sum_{i=1}^2 \sum_{j=1}^2 S_{ij}T_{ij} = S_{ij}T_{ij} \end{eqnarray} Note:- This approach will not be useful for more complicated expressions.

\begin{eqnarray} S_{ij}T_{ij} &=& S_{ji}T_{ij} \qquad \HighLight{Used symmetry property of S} \\ &=&S_{nm}T_{mn} \qquad \HighLight{renamed dummy indices}\\ &=& S_{ij}T_{ji} \qquad \HighLight{renamed dummy indices again} \label{EQ02} \end{eqnarray} Therefore the right hand side of (a) becomes \begin{eqnarray} \frac{1}{2}S_{ij}\big(T_{ij}+T_{ji}\big) &=&\frac{1}{2}S_{ij}T_{ij} +\frac{1}{2} S_{ij}T_{ji} \qquad \text{\HighLight{Now use \eqRef{EQ02} in second term}}\nonumber\\ &=& \frac{1}{2}S_{ij}T_{ij} +\frac{1}{2} S_{ij}T_{ij} = S_{ij}T_{ij}. \end{eqnarray}

\begin{eqnarray}\nonumber S_{ij}T_{ij} &=& S_{ji}T_{ji} \qquad \HighLight{Change Dummy indices}\\\nonumber S_{ij}T_{ij} &=& S_{ji}T_{ji} \qquad \HighLight{Use symmetry of S}\\ \text{add} \qquad S_{ij}T_{ij} &=& \frac{1}{2}S_{ij}\big(T_{ij}+T_{ji}\big)\nonumber \end{eqnarray}

 You are given two unit vectors \(\vec{m}=(1,1,1)\) and \(\vec{n}=(1,0,1)\). What should be the unit vector \(\hat{\ell}\) along an axis what should be the angle of rotation \(\alpha\) that will rotate vector \(\vec{n}\) and align it along the vector \(\vec{m}\).

\(\cos\alpha=\frac{\vec{m}\cdot\vec{n}}{|\vec{m}||\vec{n}|}=\frac{1}{\sqrt{3}} \)

Angle \(\alpha\) is given by \(\cos\alpha=\dfrac{\vec{m}\cdot\vec{n}}{|\vec{m}||\vec{n}|}=\dfrac{1}{\sqrt{3}} \);Clockwise rotation by \(\alpha \) about  axis along \(\hat{\ell}=\dfrac{\vec{m}\times\vec{n}}{|\vec{m}||\vec{n}|}= \frac{1}{\sqrt{2}}(1,0,-1)\);