A child pushes a merry go-round from rest to a final angular speed of 0.44 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 1.8 revolutions. What is the angular acceleration of the merry go-round?

To solve this problem, we need to use the formula for angular acceleration:

angular acceleration (α) = (final angular speed (ωf) - initial angular speed (ωi)) / time (t)

In this case, the final angular speed is given as 0.44 rev/s, and the initial angular speed is 0 rev/s because the merry-go-round starts from rest.

The number of revolutions the child pushes the merry-go-round is irrelevant for finding the angular acceleration. It simply gives us additional information about the distance the child pushes the merry-go-round.

Therefore, all we need to find the angular acceleration is the final angular speed and the time it takes for the merry-go-round to reach that speed.

To find the angular acceleration of the merry-go-round, we can use the kinematic equation:

ω^2 = ω₀^2 + 2αθ

Where:
ω = final angular speed = 0.44 rev/s
ω₀ = initial angular speed = 0 rev/s (since it started from rest)
α = angular acceleration (what we want to find)
θ = number of revolutions = 1.8 rev

Plugging in the values into the equation, we have:

(0.44 rev/s)^2 = (0 rev/s)^2 + 2α(1.8 rev)

Simplifying the equation further:

0.44^2 rev^2/s^2 = 2α(1.8 rev)

0.1936 rev^2/s^2 = 3.6α rev

Now we can solve for α by dividing both sides of the equation by 3.6 rev:

(0.1936 rev^2/s^2) / (3.6 rev) = α

α ≈ 0.0538 rev/s^2

Therefore, the angular acceleration of the merry-go-round is approximately 0.0538 rev/s^2.