Question

Tangential acceleration of a particle moving in a circle of radius $1\text{m}$ varies with time t as shown in figure (initial velocity of particle is zero). Time after which total acceleration of the particle makes an angle of ${30}^{\circ }$ with radial acceleration is:

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

Answer 1

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

Let total acceleration be $a$.
From the equation for a straight line:
$(y=mx+C)$
$m=\text{slope}=\tan {60}^{\circ }=\sqrt{3}$ and $\text{intercept}\left(C\right)=0$

Relation between $a_t$ and $t$ can be obtained from using $(y=mx+C)$ as :

$a_t=\tan {60}^{\circ }t+0=\sqrt{3}$

$\Rightarrow$ $a_t$ acts in tangential direction to increase the tangential speed $v$
$\therefore a_t=\frac{dv}{dt}\Rightarrow \frac{dv}{dt}=\sqrt{3}t$
$\Rightarrow {\int }_0^{v}dv={\int }_0^t\sqrt{3}tdt\Rightarrow v=\frac{\sqrt{3}{t}^2}{2}$

Now, centripetal accceleration $\left(a_c\right)$ is given as
$a_c=\frac{v^2}{r}=\frac{3}{4}{t}^4(\because r=1\text{m})$

since, total acceleration $a$ makes an angle of ${30}^{\circ }$ with $a_c$,


Hence, $\tan 30=\frac{a_t}{a_c}=\frac{\sqrt{3}t}{\frac{3}{4}{t}^4}$
$\frac{1}{\sqrt{3}}=\frac{4\sqrt{3}}{3t^3}$
$\therefore t=4^{1/3}=2^{\frac{2}{3}}\text{sec}$