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

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.

i already answered it by 34/s