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 time it completes the small angular displacement ' θ ' is (a) (r²)/(2αθ) (b) (r)/(2αθ) (c) (rαθ)/(2) (d) r/√(2)√(αθ)

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

A particle starting from rest moves along the circumference of a… | MHT CET 2024 (16 May, Shift 1)

Q10MHT CET 2024 · 16 MAY · SHIFT 1Medium

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 time it completes the small angular displacement ' $\theta$ ' is

Show answer and solution

Answer: (d)

From rest: $\theta = \frac{1}{2}\alpha t^2 \Rightarrow t = \sqrt{\frac{2\theta}{\alpha}}$
For small $\theta$ , displacement $\approx$ arc $= r\theta$
$v_{avg} = \frac{r\theta}{t} = r\theta\sqrt{\frac{\alpha}{2\theta}} = \frac{r}{\sqrt{2}}\sqrt{\alpha\theta}$
So the answer is (d).

Report a mistake

Non-Uniform Circular Motion question from MHT CET 2024 (16 May, Shift 1)

More questions on Non-Uniform Circular Motion