# Tangential acceleration in circular motion?

1,603

For any path in 3D or 2D the velocity and acceleration vector is decomposed as

$$\vec{v} = v \vec{e} \\ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n}$$

where $\vec{e}$ is the tangent vector, $\vec{n}$ is a normal vector, and $\rho$ the radius of curvature. So when keeping the path circular ($\rho=\mbox{const.}$) the value of the radial acceleration increased with $v^2$. This is commonly referred to as centrifugal acceleration.

There is always a force of magnitude $N$ along $\vec{n}$ which keeps the object along the path, so for a particle of mass $m$ you have

$$N = m \frac{v^2}{\rho}$$

In the case the tangential component is not zero then the applied force is decomposed as

$$\vec{F} = T \vec{e} + N \vec{n} = m \dot{v} \vec{e} + \frac{m v^2}{\rho} \vec{n}$$

$$T = m \dot{v} \\ N = m \frac{v^2}{\rho}$$

In the end I do not see what the problem is you are asking about. Maybe you can rephrase it based on the above convention.

Share:
1,603

Author by

### dfg

Updated on December 05, 2020