Equation for the relation between linear acceleration and angular acceleration:
Deduce the equation from the General Method:
We know that angular acceleration is
$\alpha=\frac{\Delta \omega}{\Delta t} \qquad (1)$
$\alpha=\frac{\Delta (v/r)}{\Delta t} \qquad \left( \because \omega=\frac{v}{r} \right)$
$\alpha=\frac{1}{r} \frac{\Delta v}{\Delta t}$
If $\Delta t \rightarrow 0$, then the above equation can be written as
$\alpha=\frac{1}{r} \: \underset{\Delta t \rightarrow 0}{Lim} \frac{\Delta v}{\Delta t}$
Where
$\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta v}{\Delta t} \rightarrow$ Instantaneous Acceleration $(a)$
$\alpha=\frac{a}{r}$
Deduce the equation from the Differential Method
We know that angular acceleration is
$\alpha=\frac{d \omega}{dt} \qquad (1)$
$\alpha=\frac{d (v/r)}{dt} \qquad \left( \because \omega=\frac{v}{r} \right)$
$\alpha=\frac{1}{r} \frac{dv}{dt} \qquad (2)$
Where
$\frac{dv}{dt}$ = Instantaneous Acceleration $(a)$
Now equation $(2)$ can be written as
$\alpha=\frac{a}{r}$