Question

For the point mass m we considered in (4.2) to (4.4), the velocity is $\mathit{v}\mathbf{=}\mathit{\omega }\mathbf{\times }\mathit{r}$so the kinetic energy is$\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(\omega \times r\right)\mathbf{\nabla }\dot{}\left(\omega \times r\right)$.Show that T can be written in matrix notation as $\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\omega }}^{\mathbf{T}}\mathbf{}\mathit{I}\mathbf{}\mathit{\omega }$ where I is the inertia matrix, $\mathit{\omega }$is a column matrix, and is a row matrix with elements equal to the components of$\mathit{\omega }$

Short answer

The value of Kinetic energy is $T=\frac{1}{2}{\omega }^T/\omega$.

Step-by-step solution

Given Information

The kinetic energy,$T=\frac{1}{2}m{v}^2=\frac{1}{2}m\left(\omega \times r\right)\square \left(\omega \times r\right)$.

The velocity is, $v=\omega \times r$.

Definition of Kinetic energy

Kinetic energy is the energy that a particle has as a result of its movement.

Kinetic Energy of a rigid body consisted of N-point masses.

$T=\sum _{n=1}^n\frac{1}{2}{m}_n{v}_n^2=\sum _{n=1}^N\frac{1}{2}{m}_n{\left(\omega \times r_n\right)}^2$

Calculate square of velocity vector.

Find the square of velocity vector.

$$\begin{gathered} v^2=\left(\omega \times r\right).\left(\omega \times r\right) \\ =\left|{\left(\omega \times r\right)}_x{\omega }_{x}x \\ {\left(\omega \times r\right)}_y{\omega }_y{y}^{} \\ {\left(\omega \times r\right)}_z{\omega }_{z}z\right| \\ =\left|{\omega }_{y}z-{\omega }_z{y}_{}{\omega }_{x}x \\ {\omega }_{z}x-{\omega }_x{z}^{}{\omega }_{y}y \\ {\omega }_{x}y{\omega }_{y}x{\omega }_{z}z\right| \\ =\left|{\omega }_{y}z{\omega }_{x}x \\ {\omega }_{zx}{\omega }_y{y}^{} \\ {\omega }_{x}y{\omega }_{z}z\right|-=\left|{\omega }_{z}y{\omega }_{x}x \\ {\omega }_{x}z{\omega }_y{y}^{} \\ {\omega }_{y}x{\omega }_{z}z\right| \end{gathered}$$

Simplify further,

$$\begin{gathered} =-{\omega }_x\left|{\omega }_{z}xy \\ {\omega }_{x}yz\right|+{\omega }_y\left|{\omega }_{y}zx \\ {\omega }_{x}yz\right|-{\omega }_z\left|{\omega }_{y}zx \\ {\omega }_{z}yz\right|+{\omega }_x\left|{\omega }_{z}yx \\ {\omega }_{y}xz\right|-{\omega }_y\left|{\omega }_{z}yx \\ {\omega }_{y}xz\right|+{\omega }_z\left|{\omega }_{z}yx \\ {\omega }_{x}zz\right| \\ ={\omega }_x^2{y}^2+{\omega }_y^2+{\omega }_z^2{x}^2+{\omega }_x^2{z}^2+{\omega }_z^2{y}^2-2\left({\omega }_x{\omega }_{z}xz+{\omega }_x{\omega }_{y}xy+{\omega }_y{\omega }_{z}yz\right)\pm \left({\omega }_x{x}^2+{\omega }_y{y}^2+{\omega }_z{z}^2\right) \\ =\left({\omega }_x^2+{\omega }_y^2+{\omega }_z^2\right)\left(x^2+y^2+z^2\right)-{\left({\omega }_{x}x+{\omega }_{y}y+{\omega }_{z}z\right)}^2 \\ ={\omega }^2{r}^2-{\left(\omega ·r\right)}^2 \end{gathered}$$

So Kinetic energy is

$T=\sum _{n=1}^N\frac{1}{2}{m}_i\left({\omega }^2{r}_n^2-{\left(\omega ·r_n\right)}^2\right)$

Consider I, as matrix of tensor of inertia.

Calculate $I\omega$.

$$\begin{gathered} {\left(I\omega \right)}_i=I_k{\omega }_k \\ =\sum _{n=1}^N{m}_n\left(r_n^2{\mathcal{S}}_{ik}-r_{n,i}{r}_{n,k}\right){\omega }_k \\ =\sum _{n=1}^N{m}_n\left(r_n^2{\omega }_i-r_{n,i}\left(r_{n,k}{\omega }_k\right)\right) \\ =\sum _{n=1}^N{m}_n\left(r_n^2{\omega }_i-r_{n,i}\left(\omega .r_n\right)\right) \end{gathered}$$

Since, ${\omega }^{T}I\omega =\omega .\left(I\omega \right)$.

$$\begin{gathered} \frac{1}{2}{\omega }^{T}I\omega =\frac{1}{2}{\omega }_i{\left(I\omega \right)}_i \\ =\frac{1}{2}\sum _{n=1}^N{m}_n\left(r_n^2{\omega }_i{\omega }_i-\left(r_{n,i}{\omega }_i\right)\left(\omega ·r_n\right)\right) \\ =\frac{1}{2}\sum _{n=1}^N{m}_n\left(r_n^2{\omega }^2-{\left(\omega ·r_n\right)}^2\right) \end{gathered}$$

Equate kinetic energy with matrix of tensor of inertia.

$$\begin{gathered} T=\sum _{n=1}^N\frac{1}{2}{m}_i\left({\omega }^2{r}_n^2-{\left(\omega ·r_n\right)}^2\right) \\ =\frac{1}{2}{\omega }^{T}I\omega \end{gathered}$$

Hence, Kinetic energy is $T=\frac{1}{2}{\omega }^T\omega$.