Question

A particle at rest starts moving with constant angular acceleration ‘α’ in circular path. At what time the magnitude of centripetal acceleration is half the tangential acceleration?

पर्याय

• `1/(sqrt(2α))`

• `1/(sqrtα)`

• `2/(sqrtα)`

• `(sqrtα)/2`

Answer 1

`bb(1/(sqrt(2α)))`

Explanation:

As the particle starts from rest, 

ω_i = 0

∴ α = `ω/t`   ...(i)

Given that, 

`a_(cp) = 1/2 xx a_T`

∴ `rω^2 = 1/2 xx αr`    ...(∵ a_cp = rω^2, a_T = ωr)`

∴ `ω^2 = α/2`

∴ `(αt)^2 = α/2`    ...[From (i)]

∴ `α^2t^2 = α/2`

∴ t = `1/(sqrt(2α))`