Question

Obtain an expression for acceleration of a particle performing uniform circular motion.

Answer 1

Let a particle is performing circular motion in a circular path of radius r, particle makes $\theta$ with centre of circular path.
Let,angular velocity is $\omega$,angular acceleration is $\alpha$.

we know, $v=r\omega$

so, $\omega=\frac{v}{r}$

angular acceleration is the rate of change of angular velocity. e.g., $\alpha=\frac{\Delta{\omega}}{\Delta{t}}$

or, $\alpha=\frac{d\omega}{dt}$

or, $\alpha=\frac{\left(d\frac{v}{r}\right)}{dt}$

or, $\alpha=\frac{dv}{dt}\frac{1}{r}$

or, $\alpha=\frac{a}{r}$

or, $a=r\alpha$.....(1) formula

now check figure,
here, $\delta{\theta}=\frac{RQ}{BR}$

= $\frac{\delta{v}}{v}$

$\delta{v}=v\delta{\theta}$

acceleration, $a=\frac{\delta{v}}{\delta{t}}$

$a=\frac{v\delta{\theta}}{\delta{t}}$

$a=v\omega$

we know, $v=r\omega$

$a=v\frac{v}{r}=\frac{v^2}{r}$