$$E=\frac{1}{2}m\textbf{v}^2 + \frac{1}{2}I\boldsymbol{\omega}^2$$
If we assume
$$\textbf{v}=\textbf{v}(x,t),$$ $$\boldsymbol{\omega}=\boldsymbol{\omega}(x,t),$$
then
$$\frac{\partial\textbf{v}}{\partial x}=-I\frac{\partial\boldsymbol{\omega}}{\partial t},$$ and $$\frac{\partial\boldsymbol{\omega}}{\partial x}=-m\frac{\partial\textbf{v}}{\partial t},$$ we arrive at
$$\boxed{\frac{\partial^2\textbf{v}}{\partial t^2}=mI \frac{\partial^2\textbf{v}}{\partial x^2}}$$
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