A particle starts moving at t = 0 in a circle of radius `R = 2m` with constant angular acceleration of `a - 3 "rad/sec"^(2)`. Initial angular speed of the particle is 1 rad/sec. At time `t_(0)` the angle between the acceleration vector and the velocity vector of the particle is `37^(@)`.
What is the value of `t_(0)` ?

Options

A. 1s
B. `(1)/(2)s`
C. `(1)/(3)s`
D. `(1)/(6)s`

Answer

D

Explanation

`a_(t) = a cos 37^(@)`

`a_(c ) = a sin 37^(@)`
`a_(t) = alpha R = 3 xx 2 = 6 m//s^(2)`
`6 = a (4)/(5) " " a = (15)/(2) = 7.5 m//s^(2)`
`a_(c ) = omega^(2) R = a sin 37^(@) = (15)/(2) (3)/(5) = (9)/(2)`
`omega^(2) = omega + alpha t`
`(3)/(2) = 1 + 3 t`
`t = (1)/(6)s`