Respuesta :
Answer:
a) θ = 11 rad, θ = 1.75 rev., b)  w = 9 rad / s, c)  θ = 7.17 rev
Explanation:
This is a rotation kinematics exercise
     θ = θ₀ + w₀ t + ½ α t²
They indicate the initial angular velocity w₀ = 2.00 rad / s, the angular acceleration α = 3.50 rad / s² and that at the initial instant θ₀ = 0
a) let's find the rotated angle
     θ = 0 + 2.00 2.00 +1/2 3.5 2²
     θ = 11 rad
let's reduce 2Ï€ rad = 1 rev
    θ = 11 rad (1 rev / 2π rad)
    θ = 1.75 rev.
b) angular velocity
     w = w₀ + α t
     w = 2.00 + 3.50 2
     w = 9 rad / s
c) the angular displacement to reach this speed
     w² = w₀² + 2 α θ
    Â
in this case they indicate that w = 2 Â 9 = 18 rad / s
     θ = [tex]\frac{w^2 - w_o^2}{2 \alpha }[/tex]
     θ = [tex]\frac{18^2 - 2^2 }{2 \ 3.5 }[/tex]
     θ = 45.7 rad
let's reduce to rev
     θ = 45.7 rad (1rev / 2π rad)
     θ = 7.17 rev
