A particle starting from rest moves along the circumference of a circle of radius r with angular acceleration α. The magnitude of the average velocity, in the time it completes the small angular displacement θ is (a) r((2)/(αθ))² (b) r((αθ)/(2))² (c) r((αθ)/(2)) (d) r((αθ)/(2))1/2

Answer: (d). From rest: θ = 1/2α t² ⇒ t = √((2θ)/(α)) Small θ: displacement ≈ rθ vavg = (rθ)/(t) = r((αθ)/(2))1/2 So the answer is (d).

A particle starting from rest moves along the circumference of a… | MHT CET 2020 (16 Oct, Shift 2)

Q19MHT CET 2020 · 16 OCT · SHIFT 2Medium

Non-Uniform Circular Motion

A particle starting from rest moves along the circumference of a circle of radius $r$ with angular acceleration $\alpha$ . The magnitude of the average velocity, in the time it completes the small angular displacement $\theta$ is

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Answer: (d)

From rest: $\theta = \frac{1}{2}\alpha t^2 \Rightarrow t = \sqrt{\frac{2\theta}{\alpha}}$
Small $\theta$ : displacement $\approx r\theta$
$v_{avg} = \frac{r\theta}{t} = r\left(\frac{\alpha\theta}{2}\right)^{1/2}$
So the answer is (d).

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Non-Uniform Circular Motion question from MHT CET 2020 (16 Oct, Shift 2)

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