# kinetic energy of uniform circular motion under radial force F

Lets say we have a body rotating around a point and a radial force F (with respect to the point).

Its kinetic energy will be $\frac{1}{2} mv^2$ where $v=\sqrt{ar}$ and $a=\frac{F}{m}$

so that

$$E_k=\frac{1}{2}m\Bigg(\sqrt{\frac{F}{m}r}\Bigg)^2=\frac{1}{2}{F}{r}$$

Why does it look like the the kinetic energy is independent of mass and velocity?

• The force is perpendicular to the velocity and so no work is done on the mass and so it’s kinetic energy stays constant. – Farcher Jan 2 '18 at 17:57

$F$ in $E_k = \frac{Fr}{2}$ is not just a generic force, but the exact force required to keep the object in uniform circular motion, often referred to as $F_c$. What that required force is depends on: (1) the object's mass (2) the rate it's orbiting and (3) the radius of circular motion. In other words, $F = F_c = m\omega^2r$.
A more correct notation would be $E_k = \frac{1}{2} F_c(m, \omega, r) \cdot r$