Tangential Component of Acceleration of Uniform Circular Motion is Zero
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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.
$\blacksquare$
Also see
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): acceleration
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): acceleration