A child pushes a merry go-round from rest to a final angular speed of 0.4 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 1.1 revolutions. What is the angular acceleration of the merry go-round?
I got .0727 but that doesn't seem to be the right answer
t = 1.1rev / 0.4rev/s = 2.75s.
a = Vf - Vo) / t,
a = (0.4 - 0) / 2.75 = 0.145 rev/s^2.
To find the angular acceleration of the merry-go-round, we can use the following formula:
ω^2 = ω0^2 + 2αθ
where ω is the final angular speed, ω0 is the initial angular speed (which is 0 in this case since the merry-go-round starts from rest), α is the angular acceleration, and θ is the angle in radians.
Given that ω = 0.4 rev/s and θ = 1.1 revolutions, we need to convert the units to radians. Since 1 revolution is equal to 2π radians, we have:
ω = 0.4 rev/s
ω = 0.4 * 2π rad/s
ω = 0.8π rad/s
θ = 1.1 revolutions
θ = 1.1 * 2π rad
θ = 2.2π rad
Now we can substitute these values into the formula:
(0.8π)^2 = (0)^2 + 2α(2.2π)
0.64π^2 = 4.4απ
Simplifying the equation by dividing both sides by π:
0.64π = 4.4α
Now, solve for α:
α = (0.64π) / 4.4
α ≈ 0.145 rad/s²
Therefore, the angular acceleration of the merry-go-round is approximately 0.145 rad/s².