Question

A particle is moving around a circular path with uniform angular speed.The radius of the circular path is r the acceleration of the particle is

Answer 1

The acceleration of the particle is a = rω^2

Explanation:

We know that

∆s = ∆θ x r

∆s / ∆t =  ∆θ /  ∆t x r

v =  ω x r    => |v| = ωr    ( Since θ = 90°)

a = dv / dt

a =  [ω x dr / dt ] + [ dω / dt x r]

ω = constant then  dω / dt = 0

a = ω x v or a = ωv

a = v / r . r = v^2 / r

a = rω^2

Hence the acceleration of the particle is a = rω^2

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