Question
Explain the acceleration of a particle in UCM. State an expression for the acceleration.

Answer

A particle in uniform circular motion (UCM) moves in a circle or circular arc at constant linear speed $\mathrm{v}$. The instantaneous linear velocity $\vec{v}$ of the particle is along the tangent to the path in the sense of motion of the particle. Since $\vec{v}$ changes in direction, without change in its magnitude, there must be an acceleration that must be
1. perpendicular to $\vec{v}$
2. constant in magnitude
3. at every instant directed radially inward, i.e., towards the centre of the circular path.
Such a radially inward acceleration is called a centripetal acceleration.
$
\therefore \vec{a}=\frac{d \vec{v}}{d t}=\overrightarrow{a_{\mathrm{r}}}
$
If $\vec{\omega}$ is the constant angular velocity of the particle and $r$ is the radius of the circle,
$
\overrightarrow{a_{\mathrm{r}}}=-\omega^2 \vec{r}
$
where $\omega=|\vec{\omega}|$ and the minus sign shows that the direction of $\vec{a}_{\mathrm{T}}$ is at every instant opposite to that of the radius vector $\vec{r}$. In magnitude,
$
a_r=\omega^2 r=\frac{v^2}{r}=\omega v
$
[Note: The word centripetal comes from Latin for 'centre-seeking'.]

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