Tangential Component of Acceleration of Uniform Circular Motion is Zero

Theorem
Let $P$ be a particle moving in space in uniform circular motion.

Let $\mathbf a$ denote the acceleration of $P$.

Then the tangential component of $\mathbf a$ is zero.

Proof
From Acceleration of Point in Plane in Intrinsic Coordinates, the motion of $P$ can be expressed in intrinsic coordinates as:
 * $\mathbf a = \dfrac {\d \map v t} {\d t} \mathbf s + \dfrac {\paren {\map v t}^2} \rho \bspsi$

where:
 * $\mathbf s$ denotes the unit vector along the tangential direction of $P$
 * $\bspsi$ denotes the unit vector toward the center of curvature of the motion of $P$
 * $\map v t$ is the speed of $P$ at the time $t$
 * $\rho$ is the radius of curvature of the motion of $P$ at time $t$.

By definition, the tangential component of $\mathbf a$ is defined as $\dfrac {\d v} {\d t}$.

Let $\map v t$ denote the speed of $P$ at time $t$.

By definition of uniform circular motion, $\map v t$ is constant.

Hence by Derivative of Constant it follows that:
 * $\dfrac {\d \map v t} {\d t} = 0$

Hence the result.

Also see

 * Definition:Tangential Component of Acceleration