Question

Tangential acceleration of a particle moving in a circle of radius $1m$ varies with time $t$ as (initial velocity of particle is zero). Time after which total acceleration of particle makes and angle of ${30}^o$ with radial acceleration is

• A. $4sec$
• B. $4/3sec$
• C. $2^{2/3}sec$
• D. $\sqrt{2}sec$

Answer 1

The correct option is C $2^{2/3}sec$

using equation for a straight line $(y=mx+c)$ where,

$m=\tan {60}^o=\sqrt{3}$ (slop) $C=0$ from graph

relation b/w $a_T$ and $t$ is obtained as:-
$a_T=\tan {60}^{o}t+0$

this tangential acceleration increases velocity (v) of the particless as.

$a_T=\frac{dv}{dt}\Rightarrow \frac{dv}{dt}=\sqrt{3}t$

$\Rightarrow \int dv=\int \sqrt{3}+dt\Rightarrow v=\frac{\sqrt{3}{t}^2}{2}$

so, $a_T$ a particular time, centripetal accin $\left(a_c\right)$ is given by :-

$a_c=\frac{v^2}{r}=\frac{3}{4}{t}^4(r=1)\overrightarrow{a_T}$

let a time $\overrightarrow{a}$ masses angle ${30}^o$ with $\overrightarrow{a_c}$. so it makes angles ${60}^o$ with $\overrightarrow{a_T}$ ($\overrightarrow{a_c}K\overrightarrow{a_T}$ are perpendicular)

$\Rightarrow \overrightarrow{a_T}=\sqrt{3t^2+\frac{9}{16}+8}\frac{1}{2}$ that given,

$\Rightarrow 3t^2=\frac{1}{11}(3t^2+\frac{9}{16}+8)$

$\Rightarrow t=0$ or $t=2^{2/3}$.