Question

A particle is moving with constant angular acceleration$\left(\alpha \right)$in a circular path of radius $\sqrt{3}m$.At $t=0$,it was at rest and at $t=1s$, the magnitude of its acceleration becomes $\sqrt{6}m/s^2$,then $\alpha \left(\text{in}rad/s^2\right)$ is

Answer 1

Formula used: $a=\sqrt{a_c^2+a_t^2},$

$a_c={\omega }^{2}r,a_t=ar$

Given:$r=\sqrt{m}m,t=1s,a=\sqrt{6}m/s^2$

$\sqrt{6}=\sqrt{{\left({\omega }^{2}r\right)}^2+{\left(ar\right)}^2}$

As we know initial speed is zero, so angular speed at $t=1s$ will be

$\omega =at=arad/s$

$\sqrt{6}=\sqrt{{({\alpha }^2\times \sqrt{3})}^2+{(\alpha \times \sqrt{3})}^2}$

$6=3{\alpha }^4+3{\alpha }^{2}or3{\alpha }^4+3{\alpha }^2-6=0$

Above is quadratic equation in $\alpha$

So,

$\alpha =\frac{-b_{-}^{+}\sqrt{b^2-4ac}}{2a}$

$=\frac{-3_{-}^{+}\sqrt{9+4\times \times 3\times 6}}{6}\text{or}=\frac{-3-\sqrt{9+4\times \times 3\times 6}}{6}$

$\alpha =1,-2$

so, the correct answer is $1rad/s^2$
FInal Answer: $1$