# Another question

What is the physics of sharp turning; sharp curves and changing direction about those curves?

Tide
Homework Helper
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.

modeman said:
What is the physics of sharp turning; sharp curves and changing direction about those curves?
Curvature:
$$\kappa = \left| {\frac{{d\vec T}}{{ds}}} \right| = \frac{{\left| {\vec T\,'\left( t \right)} \right|}}{{\left| {\vec r\,'\left( t \right)} \right|}} = \frac{{\left| {\vec r\,'\left( t \right) \times \vec r\, {''}\left( t \right)} \right|}}{{\left| {\vec r\,'\left( t \right)} \right|^3 }}$$

Sharper curves have larger values of $\kappa$.

Curvature has its physical applications; for example, let $$\vec r ( t )$$ represent the position of an object at time $t$. You know that
$$\left\{ \begin{gathered} \vec T = \frac{{\vec r\,'}} {{\left| {\vec r\,'} \right|}} = \frac{{\vec v}} {{\left| {\vec v} \right|}} \Rightarrow \vec v = \left| {\vec v} \right|\vec T \hfill \\ \kappa = \frac{{\left| {\vec T\,'} \right|}} {{\left| {\vec r\,'} \right|}} = \frac{{\left| {\vec T\,'} \right|}} {{\left| {\vec v} \right|}} \Rightarrow \left| {\vec T\,'} \right| = \kappa \left| {\vec v} \right| \hfill \\ \vec N = \frac{{T\,'}} {{\left| {T\,'} \right|}} \Rightarrow T\,' = \vec N\left| {T\,'} \right| = \kappa \left| {\vec v} \right|\vec N \hfill \\ \end{gathered} \right\}$$

And so, making the necessary substitution,
$$\begin{gathered} \vec a = \vec v \, ' = \left| {\vec v} \right| ' \vec T + \left| {\vec v} \right|\vec T' \Rightarrow \hfill \\ \vec a = \left| {\vec v} \right| '\vec T + \kappa \left| {\vec v} \right|^2 \vec N \hfill \\ \end{gathered}$$

Hope this helps

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.

BJ