Answer:
 ω = 0.467 rad/s
Explanation:
given,
tangential force exerted by the person = 37.7 N
radius of merry-go-round = 2.75 m
mass of merry-go-round  = 144 Kg
angle =  33.2°
moment of inertia
[tex]I = \dfrac{1}{2} m R^2[/tex]
[tex]I = \dfrac{1}{2}\times 144 \times 2.75^2[/tex]
  I = 544.5 kg.m²
torque = force  x radius
τ = 37.7 x  2.75
Ï„ = 103.675 N.m
angular acceleration
[tex]\alpha= \dfrac{\tau}{I}[/tex]
[tex]\alpha= \dfrac{103.675}{544.5}[/tex]
 α = 0.190 rad/s²
now ,
distance = [tex]33.2\times \dfrca{2\pi}{360}[/tex]
d = 0.579 rad
we know,
using equation of rotational motion
[tex]d = \omega t + \dfrac{1}{2}\alpha t^2[/tex]
[tex]0.579 = \dfrac{1}{2}\times 0.190\times t^2[/tex]
 t = 2.46 s
angular speed
 ω =  α  x t
 ω = 0.19 x 2.46
 ω = 0.467 rad/s
