Search any question & find its solution
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
Options:
Solution:
2750 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2} \sec$
$\begin{aligned} & \alpha=4 \mathrm{rad} / \mathrm{s}^2 \\ & \text { Centripetal (radial) acceleration, } \mathrm{a}_{\mathrm{r}}=\mathrm{r} \omega^2 \\ & \text { Tangential acceleration, } \mathrm{a}_{\mathrm{t}}=\mathrm{r} \alpha \\ & \text { If } \mathrm{a}_{\mathrm{r}}=\mathrm{a}_{\mathrm{t}} \text {, then } \mathrm{r} \omega^2=\mathrm{r} \alpha \\ & \omega=\sqrt{\alpha}=2 \mathrm{rad} / \mathrm{s} \\ & \omega=\omega_0+\alpha \mathrm{t}=0+\alpha \mathrm{t}=\alpha \mathrm{t} \\ & \text { or } \mathrm{t}=\frac{\omega}{\alpha}=\frac{1}{2} \sec \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.