Question

The speed of a particle moving in a circle of radius $r=2m$ varies with time $t$ as $v=t^2$ where $t$ is in second and $v$ in m/s. Find the radial, tangential and net acceleration at $t=2s$.

Answer 1

Linear speed of particle at t=2s is

$v={\left(2\right)}^2=4m/s$

$\therefore$ Radial acceleration

$a_r=\frac{v^2}{r}=\frac{{\left(4\right)}^2}{2}=8m/s^2$

The tangential acceleration is $a_t=\frac{dv}{dt}=2t$

$\therefore$ Tangential acceleration at t=2s is
$a_t=\left(2\right)\left(2\right)=4m/s^2$

$\therefore$ Net acceleration of particle t=2s is

$a=\sqrt{{\left(a_r\right)}^2+{\left(a_t\right)}^2}=\sqrt{{\left(8\right)}^2+{\left(4\right)}^2}$
or $a=\sqrt{80}m/s^2$

Note: On any curved path (not necessarily a circular one) the acceleration of the particle has two components $a_{t}and{a}_n$ in two mutually perpendicular directions. Component of $\overrightarrow{a}along\overrightarrow{v}$ is $a_t$ and perpendicular to $\overrightarrow{v}is{a}_n.$ Thus,
$\left|\overrightarrow{a}\right|=\sqrt{a_t^2+a_n^2}$