Mathematics

Deriving radius of curvature using basic physics

Definition– Radius of curvature is the radius of the circular arc which best approximates the curve at that point or you can say the best fit circle for the curve at a given point.

Mathematically, its function is defined as

R(x) = \frac {\mid 1 + (f'(x))^2 \mid^{\frac 32}}{\mid f''(x) \mid}

In physics we know that R_c = \frac {v^2}{a_{\perp}} = \frac {mv^2}{F_{\perp}}

where F_{\perp} is the component of force perpendicular to the velocity at a given point.

R_c = \frac {m\mid \vec{v} \mid^2}{\mid \vec{F_{\perp}} \mid}

and \mid \vec{F_{\perp}} \mid = \mid \vec{F} sin\theta \mid = \mid \hat{v} \times \vec{F} \mid

where \hat{v} is the unit vector along \vec{v}.

We can write \vec{r} = x\hat{i} + y\hat{j}

\vec{v} = (\frac {\partial x}{\partial t})\hat{i} + (\frac {\partial y}{\partial t})\hat{j}

\vec{F} = m(\frac {\partial^2 x}{\partial t^2})\hat{i} + m(\frac {\partial^2 y}{\partial t^2})\hat{j}

where m is the mass of the particle.

\hat{v} = \frac {\vec{v}}{\mid \vec{v} \mid} = \frac{(\frac {\partial x}{\partial t})\hat{i} + (\frac {\partial y}{\partial t})\hat{j}}{{({(\frac {\partial x}{\partial t})}^2 + {(\frac {\partial y}{\partial t})}^2})^{\frac 12}}

\mid \vec{v} \mid^2 = (\frac {\partial x}{\partial t})^2 + (\frac {\partial y}{\partial t})^2

\hat{v} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac {\partial x}{\partial t} & \frac {\partial y}{\partial t} & 0 \\ m\frac {\partial^2 x}{\partial t^2} & \frac {\partial^2 y}{\partial t^2} & 0 \end{vmatrix} \times \frac {1}{{({(\frac {\partial x}{\partial t})}^2 + {(\frac {\partial y}{\partial t})}^2})^{\frac 12}}

Therefore,

\mid \hat{v} \times \vec{F} \mid = \frac {m\mid (\frac {\partial^2 y}{\partial t^2})(\frac {\partial x}{\partial t}) - (\frac {\partial^2 x}{\partial t^2})(\frac {\partial y}{\partial t}) \mid}{{({(\frac {\partial x}{\partial t})}^2 + {(\frac {\partial y}{\partial t})}^2})^{\frac 12}}

\mid \hat{v} \times \vec{F} \mid = \frac {m\mid (\frac {\partial x}{\partial t})^2 \frac {d(\frac {\partial y}{\partial t} / \frac {\partial x}{\partial t})}{dt} \mid}{{({(\frac {\partial x}{\partial t})}^2 + {(\frac {\partial y}{\partial t})}^2})^{\frac 12}}

r_c = \frac {mv^2}{F_{\perp}} = \frac {m((\frac {\partial x}{\partial t})^2 + (\frac {\partial y}{\partial t})^2)^{\frac 32}}{m\mid (\frac {\partial x}{\partial t})^2 \frac {d(\frac {\partial y}{\partial t} / \frac {\partial x}{\partial t})}{dt} \mid} = \frac { (1 + (f'(x))^2 )^{\frac 32}}{\mid f''(x) \mid}

Leave a Reply

Your email address will not be published.