A particle at rest starts moving with a constant angular acceleration of $4 \mathrm{rad} / \mathrm{s}^2$ in…
A particle at rest starts moving with a constant angular acceleration of $4 \mathrm{rad} / \mathrm{s}^2$ in a circular path. The time at which magnitudes of its centripetal acceleration and tangential acceleration will be equal, is (in second)
$\frac{1}{4}$
$\frac{1}{3}$
$\frac{1}{2}$
$\frac{2}{3}$
Solution
Given that, $\alpha=\frac{4 \, \text{rad}}{s^2}$
Centripetal (radial) acceleration, $a_{r}=r \omega^2$
Tangential acceleration, $a_t=r \alpha$
If $a_r=a_t$, then $r \omega^2=r \alpha$
$\begin{aligned}
\therefore \, & \omega^2=\alpha=4 \\
\therefore \, & \omega=\sqrt{4}=2 \, \text{rad} / \text{s} \\
\therefore \, & \text{But, } \omega=\omega_0+\alpha t=0+\alpha t=\alpha t \\
\therefore \, & 2=4 t \\
\therefore \, & t=\frac{1}{2} \, s
\end{aligned}$