Aperson exerts a tangential force of 37.7 n on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. if the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 33.2°?

 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

    I = 544.5 kg.m²

torque = force  x radius

τ = 37.7 x  2.75

τ = 103.675 N.m

angular acceleration

 α = 0.190 rad/s²

now ,

distance =

d = 0.579 rad

we know,

using equation of rotational motion

 t = 2.46 s

angular speed

 ω =  α  x t

 ω = 0.19 x 2.46

 ω = 0.467 rad/s