Answer:
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
