91Ó°ÊÓ

Angular acceleration is a measure of how quickly the angular velocity of an object changes with respect to time.
It is similar to linear acceleration, but it applies to rotating objects. In this problem, the merry-go-round starts from rest, meaning its initial angular velocity is zero.
The formula for angular acceleration \( \alpha \) is given by:

• \( \alpha = \frac{\omega_f - \omega_i}{t} \)

Here, \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity, and \( t \) is the time period over which the change occurs.
By substituting the given values into the formula, you can calculate the angular acceleration. With the values \( 0.68 \, \text{rad/s} \) for \( \omega_f \) and \( 24 \, \text{s} \) for \( t \), we find that \( \alpha \) is approximately \( 0.0283 \, \text{rad/s}^2 \).
This tells us how fast the merry-go-round is speeding up as it rotates.

Moment of inertia is a crucial concept in rotational dynamics, as it plays the same role in angular motion as mass does in linear motion.
It determines how difficult it is to change the rotational state of an object. For a uniform disk, such as our merry-go-round, the moment of inertia \( I \) can be calculated using:

• \( I = \frac{1}{2} m r^2 \)

In this formula, \( m \) represents the mass of the object, and \( r \) is its radius.
This equation shows that moment of inertia depends on both the mass and how far the mass is distributed from the axis of rotation.
Substituting the merry-go-round's mass \( 31,000 \, \text{kg} \) and radius \( 7.0 \, \text{m} \) into the formula, we find \( I = 756,500 \, \text{kg} \, \text{m}^2 \).
A higher moment of inertia means the object resists angular acceleration more strongly.

Torque is the rotational equivalent of force in linear motion, and it influences how an object rotates.
It is calculated as the product of an object's moment of inertia and its angular acceleration:

• \( \tau = I \alpha \)

This equation highlights that the greater the moment of inertia and angular acceleration, the more torque is required to achieve the desired rotational motion.
For the merry-go-round, with its moment of inertia calculated as \( 756,500 \, \text{kg} \, \text{m}^2 \), and angular acceleration as \( 0.0283 \, \text{rad/s}^2 \), we find the torque \( \tau \) to be approximately \( 21,400 \, \text{N} \, \text{m} \).
This means a net torque of 21,400 \( \text{N} \, \text{m} \) is needed to bring the merry-go-round to its specified angular speed within the given time.