A particle at rest starts moving with constant angular acceleration ' α ' in a circular path of radius 'r'. At certain instant, the magnitude of centripetal acceleration is (1/3)rd the tangential acceleration. The relation between linear speed (V) and angular acceleration ( α) is (a) V = (rα)/(3) (b) V = α√(r/2) (c) V = r√((α)/(3)) (d) V = √((α r)/(3))

Answer: (c). at = rα, ac = V²/r Given ac = at/3: V²/r = (rα)/(3) V² = (r²α)/(3) ⇒ V = r√((α)/(3)) So the answer is (c).

A particle at rest starts moving with constant angular acceleration '… | MHT CET 2026 (15 Apr, Shift 2)

Q5MHT CET 2026 · 15 APR · SHIFT 2Medium

Non-Uniform Circular Motion

A particle at rest starts moving with constant angular acceleration ' $\alpha$ ' in a circular path of radius 'r'. At certain instant, the magnitude of centripetal acceleration is $\left(\frac{1}{3}\right)^{rd}$ the tangential acceleration. The relation between linear speed (V) and angular acceleration ( $\alpha$ ) is

Show answer and solution

Answer: (c)

$a_t = r\alpha$ , $a_c = \frac{V^2}{r}$
Given $a_c = \frac{a_t}{3}$ : $\frac{V^2}{r} = \frac{r\alpha}{3}$
$V^2 = \frac{r^2\alpha}{3} \Rightarrow V = r\sqrt{\frac{\alpha}{3}}$
So the answer is (c).

Report a mistake

Non-Uniform Circular Motion question from MHT CET 2026 (15 Apr, Shift 2)

More questions on Non-Uniform Circular Motion