Calculating Angular Acceleration of a Merry-Go-Round

In many physics problems, particularly those involving rotational motion, you might encounter scenarios like a child pushing a merry-go-round from rest to a specified angular speed. Such problems can be solved using basic principles and equations of rotational motion. Let's dive deeper into the given scenario and explore the necessary calculations to find the angular acceleration of the merry-go-round.

Understanding Rotational Motion and Key Equations

When a merry-go-round (or any rotating object) experiences constant angular acceleration, several kinematic equations can be applied to describe its motion. These equations are derived from the basic relations between angular acceleration, angular velocity, and angular displacement. Two fundamental equations are:

Angular Displacement:

θ ωot frac{1}{2} αt2

Angular Velocity:

ω ωo αt

Here, θ is the angular displacement, ωo is the initial angular velocity, ω is the final angular velocity, α is the angular acceleration, and t is the time elapsed.

Given Scenario and Conversion to Radians

In the given problem, we are given the following information:

The final angular speed, ωf 0.40 rev/s The initial angular speed, ωi 0 rev/s (since the merry-go-round starts from rest) The angular displacement, θ 1.2 revolutions

To use these values in our equations, we need to convert the angular displacement from revolutions to radians. Since 1 revolution is equal to 2π radians, we calculate:

θ 1.2 rev × frac{2π rad}{1 rev} 1.2 × 2π rad 2.4π rad

Applying the Kinematic Equation for Rotational Motion

The kinematic equation that expresses the relationship between angular displacement, initial angular velocity, final angular velocity, and angular acceleration is:

ωf2 ωi2 2αθ

Substituting the given values into this equation:

(0.40 rev/s)2 (0 rev/s)2 2α(2.4π rad)

Calculating the square of the final angular velocity:

(0.40)2 0.16 (rev/s)2

Hence, the equation becomes:

0.16 (rev/s)2 2α(2.4π rad)

Now, solving for α:

0.16 4.8πα

α frac{0.16}{4.8π}

Calculating the value of α:

α ≈ frac{0.16}{15.0796} ≈ 0.0106 (rev/s)2

Conclusion

The angular acceleration of the merry-go-round is approximately 0.0106 (rev/s)2.

By using the principles of rotational kinematics and the appropriate equations, we can solve such problems effectively. Understanding these concepts is crucial for anyone studying mechanics or applied physics, and applying them correctly can help in various real-world scenarios, such as engineering and physics research.