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MHT CET 2024 (2 May, Shift 1) · Non-Uniform Circular Motion

Non-Uniform Circular Motion question from MHT CET 2024 (2 May, Shift 1)

Q8MHT CET 2024 · 2 MAY · SHIFT 1Medium

Non-Uniform Circular Motion

A particle at rest starts moving with a constant angular acceleration of $4\text{ rad/s}^2$ in a circular path. The time at which magnitudes of its centripetal acceleration and tangential acceleration will be equal, is (in second)

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Answer: (c)

From rest: $\omega = \alpha t$ , so $a_c = \alpha^2 t^2 r$ and $a_t = \alpha r$
Equal: $\alpha^2 t^2 = \alpha \Rightarrow t = \frac{1}{\sqrt{\alpha}} = \frac{1}{2}$ s
So the answer is (c) $\frac{1}{2}$ .

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