Question

A body of mass $1\text{kg}$ starts moving from rest at $t=0$ in a circular path of radius $8\text{m}$. Its kinetic energy varies with time as $k=2t^2\text{J}$, then the magnitude of centripetal acceleration (in ${\text{m/s}}^2$) at $t=2\text{s}$ is:

Answer 1

Given, Kinetic energy $k=2t^2$
On differentiating, we get
$\frac{dk}{dt}=4t$
$\Rightarrow dk=4tdt=F.ds$
$\Rightarrow 4tdt=a_{T}ds$ (since mass is 1 kg)
Integrating,
$\underset{0}{\overset{t}{\int }}4tdt=\underset{0}{\overset{v}{\int }}vdv$
$\Rightarrow \frac{4t^2}{2}+c=\frac{v^2}{2}$
Using the given condition $v=0$ at $t=0$, we get $c=0$
$\Rightarrow v=2t$
$\therefore$ At $t=2\text{s}v=4\text{m/s}$
So, centripetal acceleration
$a_c=\frac{v^2}{R}=\frac{4^2}{8}=2{\text{m/s}}^2$