Unless you have a specific unstated issue in mind then the answer is that it's the same physics as for gentle turning, curves and changing direction, i.e. Newton's Laws of Motion.
#3
bomba923
760
0
modeman said:
What is the physics of sharp turning; sharp curves and changing direction about those curves?
Curvature:
[tex] \kappa = \left| {\frac{{d\vec T}}{{ds}}} \right| = \frac{{\left| {\vec T\,'\left( t \right)} \right|}}{{\left| {\vec r\,'\left( t \right)} \right|}} = \frac{{\left| {\vec r\,'\left( t \right) \times \vec r\, {''}\left( t \right)} \right|}}{{\left| {\vec r\,'\left( t \right)} \right|^3 }} [/tex]
Sharper curves have larger values of [itex] \kappa [/itex].
Curvature has its physical applications; for example, let [tex] \vec r ( t ) [/tex] represent the position of an object at time [itex] t [/itex]. You know that
[tex] \left\{ \begin{gathered}
\vec T = \frac{{\vec r\,'}}
{{\left| {\vec r\,'} \right|}} = \frac{{\vec v}}
{{\left| {\vec v} \right|}} \Rightarrow \vec v = \left| {\vec v} \right|\vec T \hfill \\
\kappa = \frac{{\left| {\vec T\,'} \right|}}
{{\left| {\vec r\,'} \right|}} = \frac{{\left| {\vec T\,'} \right|}}
{{\left| {\vec v} \right|}} \Rightarrow \left| {\vec T\,'} \right| = \kappa \left| {\vec v} \right| \hfill \\
\vec N = \frac{{T\,'}}
{{\left| {T\,'} \right|}} \Rightarrow T\,' = \vec N\left| {T\,'} \right| = \kappa \left| {\vec v} \right|\vec N \hfill \\
\end{gathered} \right\} [/tex]
And so, making the necessary substitution,
[tex] \begin{gathered}
\vec a = \vec v \, ' = \left| {\vec v} \right| ' \vec T + \left| {\vec v} \right|\vec T' \Rightarrow \hfill \\
\vec a = \left| {\vec v} \right| '\vec T + \kappa \left| {\vec v} \right|^2 \vec N \hfill \\ \end{gathered} [/tex]
Hope this helps
#4
Dr.Brain
538
2
This is studied under 'TNB Physics' which constitutes the T=Unit Tangent vector N= Principle vector and B Vector . The curvature given by 'k' is defined by rate of change of unit normal vector per unit length.