13.12: Kinetic Energy of Rotating Rigid Body

An important observable is the kinetic energy of rotation of a rigid body. Consider a rigid body composed of $N$ particles of mass $m_{\alpha}$ where ${\alpha} = 1, 2, 3, \dots N$. If the body rotates with an instantaneous angular velocity $\boldsymbol{\omega}$ about some fixed point, with respect to the body coordinate system, and this point has an instantaneous translational velocity $\mathbf{V}$ with respect to the fixed (inertial) coordinate system, see Figure $13.3.1$, then the instantaneous velocity $\mathbf{v}_{\alpha}$ of the ${\alpha}^{th}$ particle in the fixed frame of reference is given by

$$\mathbf{v}_{\alpha} = \mathbf{V} + \mathbf{v}^{\prime\prime}_{\alpha} + \boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha} \label{13.56}$$

However, for a rigid body, the velocity of a body-fixed point with respect to the body is zero, that is $\mathbf{v}^{\prime\prime}_{\alpha} = 0$, thus

$$\mathbf{v}_{\alpha} = \mathbf{V} + \boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}$$

The total kinetic energy is given by

$$\begin{align} T & = & \sum^N_{\alpha} \frac{1}{2} m_{\alpha} \mathbf{v}_{\alpha} \cdot \mathbf{v}_{\alpha} = \sum^N_{\alpha} \frac{1}{2} m_{\alpha} (\mathbf{V} + \boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}) \cdot (\mathbf{V} + \boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}) \notag \\ & = & \frac{1}{2} \sum^N_{\alpha} m_{\alpha} V^2 + \sum^N_i m_{\alpha} \mathbf{V} \cdot \boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha} + \frac{1}{2} \sum^{N}_{\alpha} m_{\alpha} (\boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}) \cdot (\boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}) \label{13.58} \end{align}$$

This is a general expression for the kinetic energy that is valid for any choice of the origin from which the body-fixed vectors $\mathbf{r}^{\prime}_{\alpha}$ are measured. However, if the origin is chosen to be the center of mass, then, and only then, the middle term cancels. That is, since $\mathbf{V} \cdot \boldsymbol{\omega}$ is independent of the specific particle, then

$$\sum^N_{\alpha} m_{\alpha} \mathbf{V} \cdot \boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha} = \mathbf{V} \cdot \boldsymbol{\omega} \times \left( \sum^{N}_{\alpha} m_{\alpha} \mathbf{r}^{\prime}_{\alpha} \right)$$

But the definition of the center of mass is

$$\sum_{\alpha} m_{\alpha} \mathbf{r}^{\prime} = M\mathbf{R}$$

and $\mathbf{R} = 0$ in the body-fixed frame if the selected point in the body is the center of mass. Thus, when using the center of mass frame, the middle term of Equation $\ref{13.58}$ is zero. Therefore, for the center of mass frame, the kinetic energy separates into two terms in the body-fixed frame

$$T = T_{trans} + T_{rot} \label{13.61}$$

where

$$T_{trans} = \frac{1}{2} \sum^{N}_{\alpha} m_{\alpha} V^2$$

$$T_{rot} = \frac{1}{2} \sum^N_{\alpha} m_i (\boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}) \cdot (\boldsymbol{\omega} \times \mathbf{r}^{\prime}_{\alpha}) \notag$$

The vector identity

$$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{A} \times \mathbf{B}) = A^2B^2 − (\mathbf{A} \cdot \mathbf{B})^2$$

can be used to simplify $T_{rot}$

$$T_{rot} = \frac{1}{2} \sum^N_{\alpha} m_{\alpha} \left[ \omega^2 r^{\prime 2}_{\alpha} − (\boldsymbol{\omega} \cdot \mathbf{r}^{\prime}_{\alpha})^2 \right]$$

The rotational kinetic energy $T_{rot}$ can be expressed in terms of components of $\boldsymbol{\omega}$ and $\mathbf{r}^{\prime}_{\alpha}$ in the body-fixed frame. Also the following formulae are greatly simplified if $\mathbf{r}^{\prime}_{\alpha} = (x_{\alpha}, y_{\alpha}, z_{\alpha})$ in the rotating body-fixed frame is written in the form $\mathbf{r}^{\prime}_{\alpha} = (x_{\alpha,1}, x_{\alpha,2}, x_{\alpha,3})$ where the axes are defined by the numbers $1, 2, 3$ rather than $x,y,z$. In this notation the rotational kinetic energy is written as

$$T_{rot} =\frac{1}{2} \sum_{\alpha}^{N} m_{\alpha}\left[\left(\sum_{i} \omega_{i}^{2}\right) \left(\sum_{k} x_{\alpha, k}^{2}\right)-\left(\sum_{i} \omega_{i} x_{\alpha, i}\right)\left(\sum_{j} \omega_{j} x_{\alpha, j}\right)\right]$$

Assume the Kronecker delta relation

$$\omega_i = \sum^3_j \omega_j \delta_{ij}$$

where $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ if $i \neq j$.

Then the kinetic energy can be written more compactly

$$\begin{align} T_{rot} & =\frac{1}{2} \sum_{\alpha}^{N} m_{\alpha}\left[\left(\sum_{i} \omega_{i}^{2}\right) \left(\sum_{k} x_{\alpha, k}^{2}\right)-\left(\sum_{i} \omega_{i} x_{\alpha, i}\right)\left(\sum_{j} \omega_{j} x_{\alpha, j}\right)\right] \notag \\ & = \frac{1}{2} \sum_{\alpha}^{N} \sum_{i, j}^{3} m_{\alpha} \left[ \left ( \omega_{i} \omega_{j} \delta_{i j}\right) \left(\sum_{k}^{3} x_{\alpha, k}^{2}\right)-\left(\omega_{i} x_{\alpha, i}\right)\left(\omega_{j} x_{\alpha, j}\right)\right] \notag \\ & = \frac{1}{2} \sum_{i, j}^{3} \omega_{i} \omega_{j}\left[\sum_{\alpha}^{N} m_{\alpha} \left[\delta_{i j} \left(\sum_{k}^{3} x_{\alpha, k}^{2}\right)-x_{\alpha, i} x_{\alpha, j}\right]\right] \end{align}$$

The term in the outer square brackets is the inertia tensor defined in equation $(13.4.1)$ for a discrete body. The inertia tensor components for a continuous body are given by equation $(13.4.2)$.

Thus the rotational component of the kinetic energy can be written in terms of the inertia tensor as

$$T_{rot} = \frac{1}{2} \sum^3_{i,j} I_{ij} \omega_i\omega_j \label{13.68}$$

Note that when the inertia tensor is diagonal, then the evaluation of the kinetic energy simplifies to

$$T_{rot} = \frac{1}{2} \sum^3_i I_{ii} \omega^2_i$$

which is the familiar relation in terms of the scalar moment of inertia $I$ discussed in elementary mechanics.

Equation $\ref{13.68}$ also can be factored in terms of the angular momentum $\mathbf{L}$.

$$T_{rot} = \frac{1}{2} \sum_{i,j} I_{ij} \omega_i \omega_j = \frac{1}{2} \sum_i \omega_i \sum_j I_{ij} \omega_j = \frac{1}{2} \sum_i \omega_i L_i \label{13.70}$$

As mentioned earlier, tensor algebra is an elegant and compact way of expressing such matrix operations. Thus it is possible to express the rotational kinetic energy as

$$T_{rot} = \frac{1}{2} \left( \omega_1 \ \omega_2 \ \omega_3 \right) \cdot \begin{pmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \end{pmatrix} \cdot \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix}$$

$$T_{rot} \equiv \mathbf{T} = \frac{1}{2} \boldsymbol{\omega} \cdot \{\mathbf{I}\} \cdot \boldsymbol{\omega}$$

where the rotational energy $\mathbf{T}$ is a scalar. Using equation $(13.11.1)$ the rotational component of the kinetic energy also can be written as

$$T_{rot} \equiv \mathbf{T} = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{L}$$

which is the same as given by $\ref{13.70}$. It is interesting to realize that even though $\mathbf{L} = \{\mathbf{I}\} \cdot \boldsymbol{\omega}$ is the inner product of a tensor and a vector, it is a vector as illustrated by the fact that the inner product $T_{rot} = \frac{1}{2} \boldsymbol{\omega}\cdot \mathbf{L} = \frac{1}{2} \boldsymbol{\omega} \cdot (\{\mathbf{I}\} \cdot \boldsymbol{\omega})$ is a scalar. Note that the translational kinetic energy $T_{trans}$ must be added to the rotational kinetic energy $T_{rot}$ to get the total kinetic energy as given by Equation $\ref{13.61}$.