[NOTES/ME-14011]-Rotation of a Rigid Body with One Point Fixed

1.Kinetic energy:

Let us compute the kinetic energy of a body moving with one point fixed. A general infinitesimal displacement of the body consists of rotation about some axis passing through the point. The instantaneous axis rotation will in general change from time to time. We choose a set of coordinate axes \(K\) with the fixed point as the origin. Then the velocity of a point on the body will be given by \begin{equation}{EQ01} \dd[{\mathbf {x}}_\alpha]{t} = \pmb{\omega}\times {\mathbf x}_\alpha \end{equation} where \(\pmb{\omega}\) is the angular velocity of the body. Thus we obtain the kinetic energy as \begin{eqnarray} \text{K.E.} &=& \sum_\alpha m_\alpha \dot{\vec{x}}_\alpha^{\,2}\nonumber\\ &=& \sum_\alpha m_\alpha (\pmb{\omega}\times {\mathbf x}_\alpha) \cdot(\pmb{\omega}\times {\mathbf x}_\alpha){EQ02} \end{eqnarray} We shall rearrange the expression \((\pmb{\omega}\times {\mathbf x}_\alpha) \cdot(\pmb{\omega}\times {\mathbf x}_\alpha)\) and write it in terms of components of \(\pmb{\omega}\) and \({\mathbf x}_\alpha\). Using the identity \(({\mathbf A}\times{\mathbf B})\cdot {\mathbf C}= {\mathbf A}\cdot({\mathbf B}\times {\mathbf C})\), we obtain \begin{eqnarray} (\pmb{\omega}\times {\mathbf x}_\alpha) \cdot(\pmb{\omega}\times {\mathbf x}_\alpha) &=& \pmb{\omega}\cdot\big({\mathbf x}_\alpha \times(\pmb{\omega}\times {\mathbf x}_\alpha)\big)\\\nonumber &=& \pmb{\omega}\cdot({\mathbf x}_\alpha^2\,\pmb{\omega}- {\mathbf x}_\alpha (\pmb{\omega}\cdot{\mathbf x}_\alpha))\\\nonumber &&\quad \mbox{\HighLight[LightCyan]{Used}} ~\colorbox{LightCyan}{ \({\mathbf A}\times ({\mathbf B}\times{\mathbf C}) ={\mathbf B}({\mathbf A}\cdot{\mathbf C})-{\mathbf C} ({\mathbf A}\cdot{\mathbf B}) \)} \\\nonumber &=& \{ \pmb{\omega}^2\, |{\mathbf x}_\alpha|^2- (\pmb{\omega}\cdot{\mathbf x}_\alpha)^2\}\\\nonumber &=& (\delta_{jk}\omega_j\omega_k)x_\alpha^2 -(\omega_j x_{\alpha j}) (\omega_jx_{\alpha_k}) \\\nonumber &=& \Big(\delta_{jk} |\vec{x}_\alpha|^2 - x_{\alpha j}x_{\alpha k}\Big) \omega_j \omega_k. \end{eqnarray} Substituting in \eqRef{EQ02}, we get the kinetic energy \begin{eqnarray} \text{K.E.} &=& \sum_\alpha m_\alpha \Big(\delta_{jk} |\vec{x}_\alpha|^2 - x_{\alpha j}x_{\alpha k}\Big) \omega_j \omega_k\nonumber\\\ &=& I_{jk}\, \omega_j \omega_k. \end{eqnarray} where \(I_{jk}\) are components of {\it inertia tensor}, and are defined by \begin{equation}{EQ05A} I_{jk} = \sum_\alpha m_\alpha \Big(|\vec{x}_\alpha|^2\,\delta_{jk} - x_{\alpha j}x_{\alpha k}\Big). \end{equation}

2. Angular momentum

The angular momentum of the rigid body relative, to a given set of axes, is defined as \begin{eqnarray} \vec{L}= \sum_\alpha m_\alpha\left( \vec{x}_\alpha \times \dd[\vec{x}_\alpha]{t}\right). \end{eqnarray} Using \eqRef{EQ01} we get \begin{eqnarray} \vec{L} &=& \sum_\alpha m_\alpha \vec{x}_\alpha \times (\vec{\omega}\times\vec{x}_\alpha )\nonumber\\ &=& \sum_\alpha m_\alpha \Big(|x_\alpha|^2 \vec{\omega} - (\vec{x}_\alpha\cdot\vec{\omega}\vec{x}_\alpha\Big) \end{eqnarray} Writing \(|\vec{x}_\alpha|^2= \delta_{jk}x_{\alpha j}x_{\alpha k}\) and \((\vec{x}_\alpha\cdot\vec{\omega}) =x_{\alpha k \omega_k }\), we get \begin{eqnarray} \vec{L} &=& \sum_\alpha m_\alpha \Big(|\vec{x}_\alpha|^2 \vec{\omega} - (x_{\alpha k} \omega_k )\vec{x}_\alpha\Big)\ \therefore\qquad\qquad L_j &=& \sum_\alpha m_\alpha \Big(|\vec{x}_\alpha|^2 \omega_j - (x_{\alpha k} \omega_k )x_{\alpha j}\Big)\ &=& \sum_\alpha m_\alpha \Big(|\vec{x}_\alpha|^2 (\delta_{jk}\omega_k) - (x_{\alpha k} \omega_k )x_{\alpha j}\Big)\ &=& I_{jk} \omega_k \end{eqnarray} where \(I_{jk}\) is the inertia tensor defined by \eqRef{EQ05A}.

3.The moment of inertia tensor

The expressions for kinetic energy and angular momentum are conveniently written in terms of the inertia tensor. The inertia tensor is a nine component object \(I_{jk}\) defined by \begin{equation}{EQ05B} I_{jk} = \sum_\alpha m_\alpha \Big(|\vec{x}_\alpha|^2\,\delta_{jk} - x_{\alpha j}x_{\alpha k}\Big). \end{equation} The following matrix notation is very convenient in writing many equations. \begin{eqnarray} \underline{\Ibb} &=& \begin{vmatrix} I_{11}& I_{12} & I_{13}\\ I_{21} & I_{22} & I_{23}\\ I_{31} & I_{32} & I_{33} \end{vmatrix},\qquad \qquad\underline{\Lbb} =\begin{pmatrix} L_1\\L_2\\L_3 \end{pmatrix}, \end{eqnarray} the expression for kinetic energy and angular momentum can be written in a compact and useful form as \begin{equation} T= \omega^T\underline{\Ibb}\, \underline{\sf \omega}; \qquad \underline{\Lbb}=\underline{\Ibb}\, \underline{\sf \omega}. \end{equation} These relationships are also written symbolically as \begin{equation} T= \overrightarrow{\omega} \cdot\overleftrightarrow{I}\cdot\vec{\omega},\qquad \vec{L} = \overleftrightarrow{I}\cdot\vec{\omega}. \end{equation}

4.Inertia tensor relative to the centre of mass

A body is said to be free to rotate if no point of the body is fixed. In this case, it turns out to be useful to choose the centre of mass as the origin of the coordinate system. This choice has the advantage that the total kinetic energy can be decomposed into a sum of two parts referring to translation of the centre of mass and and rotational motion about the centre of mass. The expression for the angular momentum also becomes a sum of two parts referring to translational and rotational motions. The expressions for the kinetic energy and angular momentum in terms of moment of inertia tensor remain unchanged, if all the components are computed w.r.t.the system of axes having origin at the centre of mass of the rigid body. The relation between components of moment of inertia tensor relative to any point and relative to the centre of mass of the rigid body is given by parallel axes theorem and will taken up next.

5.Parallel axes theorem} \input{me-lec-14007}

6.Transformation of inertia tensor under rotations

Under a rotation of axes, the components of the inertia tensor will change and the Transformation property can be worked out using the definition. \begin{equation}{EQ05C} I_{jk} = \sum_\alpha m_\alpha \Big(|\vec{x}_\alpha|^2\,\delta_{jk} - x_{\alpha j}x_{\alpha k}\Big). \end{equation} The components of each position vector \(vec{x}_\alpha\) with respect to two systems of axes, are related by a rotation matrix \(R\) \begin{equation} x_{\alpha j} = R_{jm} x_{\alpha m}. \end{equation} Therefore the components of inertia tensor, \(I_{jk}\) transform as follows. \begin{equation} I_{jk}{'} = R_{jm}R_{kn} I_{mn}. \end{equation} An object transforming in this fashion is called a second rank tensor.