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Question: Answered & Verified by Expert
A particle at rest starts moving with a constant angular acceleration of $4 \mathrm{rad} / \mathrm{s}^2$ in a circular path. At what time the magnitude of its centripetal acceleration and tangential acceleration will be equal
PhysicsRotational MotionMHT CETMHT CET 2021 (20 Sep Shift 1)
Options:
  • A $\frac{1}{4} \mathrm{sec}$
  • B $\frac{2}{3} \sec$
  • C $\frac{1}{2} \sec$
  • D $\frac{1}{3} \sec$
Solution:
2750 Upvotes Verified Answer
The correct answer is: $\frac{1}{2} \sec$
$\begin{aligned} & \alpha=4 \mathrm{rad} / \mathrm{s}^2 \\ & \text { Centripetal (radial) acceleration, } \mathrm{a}_{\mathrm{r}}=\mathrm{r} \omega^2 \\ & \text { Tangential acceleration, } \mathrm{a}_{\mathrm{t}}=\mathrm{r} \alpha \\ & \text { If } \mathrm{a}_{\mathrm{r}}=\mathrm{a}_{\mathrm{t}} \text {, then } \mathrm{r} \omega^2=\mathrm{r} \alpha \\ & \omega=\sqrt{\alpha}=2 \mathrm{rad} / \mathrm{s} \\ & \omega=\omega_0+\alpha \mathrm{t}=0+\alpha \mathrm{t}=\alpha \mathrm{t} \\ & \text { or } \mathrm{t}=\frac{\omega}{\alpha}=\frac{1}{2} \sec \end{aligned}$

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