A150-kg merry-go-round in the shape of a uni- form, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. what constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?

The force required to pull the merry-go-round is 241.9 N

Explanation:

ω = ωo + a×t

where ω is the angular speed, t is time, ωo is the initial angular speed

0.7 rev / s ×(2×π radians/rev) = 4.3 radians/s

   4.3 = a×2

      a = 2.15 rad/s^2

then:

F×r = I×a

Where F is the force on the merry-go-round, r is the radius, I is the moment of inertia of the merry-go-round, and a is the angular acceleration

for a solid, horizontal disk, the moment of inertia is: 1/2×m×r^2

that is:

F×r = 1/2×m×r^2×a

F = 1/2×m×r×a

   = 1/2×(150)×(1.50)×(2.15)

  = 214.9 N

your answer is c! hope this : )

not too sure thinking all accelerations are pointed down