A few kids decide to go to the park to go on the merry-go-round. They are the only ones there so the merry-go-round is not moving yet. They push it for 6.00 s and it moves over an angle of 129°. What is the final angular velocity of the merry-go-round?

assume constant torque = constant angular acceleration alpha

omega = alpha * t
theta = omega initial*t +(1/2)alpha t^2

129 * 2pi/360 = 2.25 radians
so
2.25 = 0 + .5 alpha (36)
alpha = .125 rad/s^2

omega = alpha t = .125*6 = .75 rad/s

To find the final angular velocity of the merry-go-round, we can use the formula:

angular velocity (ω) = angular displacement (θ) / time (t)

Given that the angular displacement is 129° and the time is 6.00 seconds, we can substitute these values into the formula to calculate the final angular velocity.

ω = θ / t
ω = 129° / 6.00 s

However, angular velocity is typically measured in radians per second, so we need to convert the angular displacement from degrees to radians.

1 radian = 180° / π

Converting the angular displacement:

θ = (129°) * (π / 180°)

Now we can substitute the values into the formula and calculate the final angular velocity.

ω = [(129°) * (π / 180°)] / 6.00 s

Evaluating this expression will give us the final angular velocity of the merry-go-round.

To find the final angular velocity of the merry-go-round, we need to use the equation that relates angular velocity (ω), angle (θ), and time (t). The equation is:

ω = θ / t

where ω is measured in radians per second, θ is the angle moved (in radians), and t is the time taken (in seconds).

Given that the merry-go-round moved over an angle of 129° and took 6.00 seconds to do so, we can convert the angle to radians using the conversion factor:

1 radian = 180° / π

First, let's convert the angle of 129° to radians:

θ = 129° × (π / 180°)
θ = 2.25 radians

Now we can use the equation to find the angular velocity:

ω = θ / t
ω = 2.25 radians / 6.00 seconds
ω ≈ 0.375 radians/second

Therefore, the final angular velocity of the merry-go-round is approximately 0.375 radians/second.