Question

For a particle performing uniform circular motion $\overrightarrow{v}=\overrightarrow{w}\times \overrightarrow{r}$ obtain an expression for linear acceleration of the particle performing non-uniform circular motion.

Answer 1

Let the particle be moving around a circular path of constant radius $r$. If it speeds up or slows down, its motion is non-uniform and both the angular speed ($\omega$) and linear speed $\left(v\right)$ change with time. Thus at any instant, $\omega$, $v$ and $r$ are related as:
Hence, the angular acceleration of the particle is given by
$\alpha =\frac{d\omega }{dt}$
and the tangential acceleration $a_t$ is given by
$a_t=\frac{dv}{dt}=\frac{d\left(\omega r\right)}{dt}=r\frac{d\omega }{dt}$ [because $r$ is a constant]
Thus, $a_t=\alpha r$