Question

For a particle moving along a circle of radius $8m$, the time dependent tangential acceleration is given by $a_t=k{t}^{2}m/s^2$ . Initially, if the particle is at rest and the total acceleration of the particle makes ${45}^{\circ }$ with the tangential acceleration after 2 sec, the value of $k$ (in $m/s^4$ is

Answer 1

Given that angle between total acceleration and tangential acceleration is ${45}^{\circ }$ and tangental acceleration varies with time as $a_t=k{t}^2$
$\tan \theta =\frac{a_r}{a_t}$
$\Rightarrow a_t=a_r$
[since $\theta ={45}^{\circ }$ and $\tan {45}^{\circ }=1]$
Taking $a_t=k{t}^2=\frac{dv}{dt}$
$dv=k{t}^{2}dt$
Integrating on both sides, ${\int }_0^{v}dv={\int }_0^{2}k{t}^{2}dt$
$\Rightarrow v=\frac{k{t}^3}{3}{|}_0^2=\frac{8k}{3}$
$\therefore a_r=\frac{v^2}{r}=\frac{{\left(\frac{8k}{3}\right)}^2}{8}=\frac{64k^2}{9\times 8}=\frac{8k^2}{9}$
$a_t=k{t}^2$ (at $t=2s$) $=4k$

Since $a_t=a_r(att=2sec)$,
$4k=\frac{8k^2}{9}$
$\Rightarrow k=\frac{4\times 9}{8}=4.5m/s^4$