I think what you are asking is how to decompose an acceleration vector $\vec{a}$ into tangential and normal components and find the center of rotation of the normal (centrifugal) component.

• A velocity vector $\vec{v}$ is always tangent to the path, with a tangent vector $\hat{e}$ and magnitude (speed) $v$ $$ \vec{v} = v \,\hat{e} $$
• An acceleration vector $\vec{a}$ has both a tangent component with magnitude $\dot{v}$ and a normal component along $\hat{n}$ with magnitude $v^2/r$ $$ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} $$ where $r$ is the radius of curvature of the path.

See the linked Wikipedia article on the calculation of the radius of curvature from the path definition:

$$ r = \frac{ \left(1+ \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right)^{3/2} }{ \frac{{\rm d}^2 y}{{\rm d}x^2} } $$

 

$$ r = \frac{ \left( \dot{x}^2 + \dot{y}^2 \right)^{3/2} }{ \dot{y} \ddot{x} - \ddot{y} \dot{x} } $$

I think what you are asking is how to decompose an acceleration vector $\vec{a}$ into tangential and normal components and find the center of rotation of the normal (centrifugal) component.

• A velocity vector $\vec{v}$ is always tangent to the path, with a tangent vector $\hat{e}$ and magnitude (speed) $v$ $$ \vec{v} = v \,\hat{e} $$
• An acceleration vector $\vec{a}$ has both a tangent component with magnitude $\dot{v}$ and a normal component along $\hat{n}$ with magnitude $v^2/r$ $$ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} $$ where $r$ is the radius of curvature of the path.

See the linked Wikipedia article on the calculation of the radius of curvature from the path definition:

$$ r = \frac{ \left(1+ \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right)^{3/2} }{ \frac{{\rm d}^2 y}{{\rm d}x^2} } $$

$$ r = \frac{ \left( \dot{x}^2 + \dot{y}^2 \right)^{3/2} }{ \dot{y} \ddot{x} - \ddot{y} \dot{x} } $$