Alright. This is a problem of angular acceleration. Let's solve it step-by-step.
Step 1: Understand the given
We are given that the initial angular speed (omega_0) is 0 rev/s because the merry-go-round initially was at rest. The final angular speed (omega) has been observed to be 0.570 rev/s. The merry-go-round has been displaced by (theta) 2.00 revolutions. We are required to find the angular acceleration, which is the rate of change of angular speed.
Step 2: Convert to common units
To proceed further, we need to convert the given quantities into a common unit. Here, we need to convert the angular speed and displacement from revolutions to radians.
1 revolution is equal to 2π radians. Therefore,
* omega_0 = 0 rev/s = 0 * 2π = 0 rad/s
* omega = 0.570 rev/s = 0.570 * 2π = 3.581 rad/s
* theta = 2.00 rev = 2.00 * 2π = 12.566 rad
Step 3: Apply the Formula
Now we apply the standard formula for angular acceleration (alpha), i.e., angular acceleration is the change in angular speed divided by the time taken.
But we do not have the time here. No worries. We can still find alpha using the formula for final angular speed and the formula for displacement in angular motion. These two equations are obtained from the rules of kinematic motion.
* omega = omega_0 + alpha*t and
* theta = omega_0*t + 0.5*alpha*t^2
By rearranging the first equation, we get t = (omega - omega_0) / alpha.
Substituting this in the second equation, we get theta = 0.5*alpha*((omega - omega_0) / alpha)^2
This simplifies to alpha = (omega^2 - omega_0^2) / (2*theta)
Substituting the known values in the equation, we get:
alpha = ( (3.581)^2 - (0)^2) ) / (2 * 12.566) = 0.510 rad/s²
Hence, the angular acceleration of the merry-go-round is 0.510 rad/s².