Consider a particle on the real line with:
$L=\frac{1}{2}(\partial_0q)^2 + f(q)\partial_0q$
the equation of motion is that of a free particle $\partial_0^2q=0$. In fact $\delta[f(q)\partial_0q]=0$. Is this right?
I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...
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