A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V0 m/s in the nth round. Its angular acceleration will be :-

Question

A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V0 m/s in the nth round. Its angular acceleration will be :- (A) (∨0/n)rad / s2 (B) (∨20/2π nr2)rad / s2 (C) (∨20/4π nr2)rad / s2 (D) (∨40/4π nr)rad / s2

Tardigrade Admin · Accepted Answer

θ = (2π n), ω0 = 0, ω = V0/r α = (ω2 - ω02/2θ) = ((V0/r)2 - 0/2(2π n)) = (V02/4π nr2)

Q. A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V₀ m/s in the $n^{t h}$ round. Its angular acceleration will be :-

$\frac{\lor _{0}}{n} r a d / s^{2}$

$\frac{\lor _{0}^{2}}{2 πn r ^{2}} r a d / s^{2}$

$\frac{\lor _{0}^{2}}{4 πn r ^{2}} r a d / s^{2}$

$\frac{\lor _{0}^{4}}{4 πn r} r a d / s^{2}$

$θ = (2 πn), ω_{0} = 0, ω = V_{0} / r$
$α = \frac{ω ^{2} - ω _{0}^{2}}{2 θ} = \frac{( V _{0} / r ) ^{2} - 0}{2 ( 2 πn )}$
$= \frac{V _{0}^{2}}{4 πn r ^{2}}$

Q. A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V0 m/s in the nth round. Its angular acceleration will be :-

n∨0​​rad/s2

2πnr2∨02​​rad/s2

4πnr2∨02​​rad/s2

4πnr∨04​​rad/s2