Question

A particle starting from rest moves in a circle of radius r.it attains a velocity of v m/s in nth round. It's angular acceleration will be

Answer 1

AnswEr :

From the Question,

• Initial Velocity (u) = 0 m/s
• Final Velocity (v) = v m/s

We know that,

$\sf v = r \omega \\ \\ \implies \sf \: \omega = \dfrac{v}{r}$

Using the equation,

$\star \: \boxed{ \boxed{ \sf { \omega}^{2} - { \omega {}^{2} _o}^{} = 2 \alpha \theta}}$

Now,

Angular Displacement (∅) for one round : 2π

Angular Displacement for 'n' rounds : 2nπ

Thus,

$\implies \sf \omega {}^{2} - 0 = 2 \alpha \theta \: \: \: \: \: \: \: [ \because \: u = 0 \: {ms}^{ - 1} ] \\ \\ \implies \sf \: \dfrac{ {v}^{2} }{ {r}^{2} } = 2 \alpha \big(2n\pi \big) \\ \\ \implies \boxed{ \boxed{ \sf \alpha = \dfrac{ {v}^{2} }{4n\pi {r}^{2} } }}$