Answer:
the angular speed of the merry-go-round is 3.77 rad/s.
Explanation:
To calculate the angular speed of the merry-go-round, we can use the formula:
ω = Δθ/Δt
where ω is the angular speed, Δθ is the change in angle, and Δt is the change in time.
The child completes 3.00 revolutions in 5.00 seconds. Therefore, the change in angle is:
Δθ = 2π × 3 = 6π
The child lands 2.00 m from the center of the merry-go-round. The radius of the merry-go-round is:
r = 2.00 m
The circumference of the merry-go-round is:
C = 2πr = 4π m
The distance traveled by the child during one revolution is equal to the circumference of the circle:
d = C = 4π m
The time taken for one revolution is:
t = Δt/Δn = 5.00 s/3.00 rev = 1.67 s/rev
Now we can calculate the speed of the child during one revolution:
v = d/t = (4π m)/(1.67 s/rev) ≈ 7.54 m/s
The angular speed of the merry-go-round can be calculated as follows:
ω = v/r = (7.54 m/s)/(2.00 m) ≈ 3.77 rad/s
Therefore, the angular speed of the merry-go-round is approximately 3.77 rad/s.