91Ó°ÊÓ

Torque is an essential concept in physics, particularly when dealing with rotational motion. Imagine torque as the rotational equivalent of force in linear motion. When applying a tangential force on the rim of an object, such as a merry-go-round, you generate torque, causing the object to rotate.
To calculate torque in this exercise, we use the formula:

• \( \tau = r \times F \)

where \( \tau \) is the torque, \( r \) is the radius, and \( F \) is the force applied.
In our case:

• The radius \( r = 2.74 \) meters
• The tangential force \( F = 36.1 \) Newtons
• Thus, \( \tau = 2.74 \times 36.1 = 98.914 \) Nm

Understanding how torque is calculated helps us appreciate how different amounts and points of force affect rotational movement.
Like pushing a door closer to its hinge versus further away makes it easier to open.

The moment of inertia is crucial when analyzing how objects rotate. It's akin to mass in linear motion, providing an object's resistance to changes in rotation. For any rotating object, its moment of inertia depends not only on its mass but also on how that mass is distributed relative to the axis of rotation.
In this exercise, we are working with a disk-shaped merry-go-round, so we use the formula:

• \( I = \frac{1}{2} m r^2 \)

where \( m \) is the mass and \( r \) is the radius of the disk.
Let's break it down:

• Mass \( m = 167 \) kg
• Radius \( r = 2.74 \) meters
• Moment of inertia \( I = \frac{1}{2} \times 167 \times (2.74)^2 = 627.907 \) kg m\(^2\)

Understanding the moment of inertia helps explain why it's harder to stop or start rotating objects with more mass concentrated far from the axis.
The distribution significantly affects how an object behaves when subjected to external forces.

Angular motion describes how objects move when they spin or rotate. It's a counterpart to linear motion, involving terms like angular speed and acceleration.
After understanding torque and moment of inertia, we can delve into angular motion using angular acceleration \( \alpha \), derived from torque and moment of inertia.

• \( \alpha = \frac{\tau}{I} \)

With \( \tau = 98.914 \) Nm and \( I = 627.907 \) kg m\(^2\):

• \( \alpha = \frac{98.914}{627.907} = 0.1575 \) rad/s\(^2\)

When calculating final angular speed after an angular displacement \( \theta \), use the equation:

• \( \omega^2 = \omega_0^2 + 2\alpha\theta \)

Starting from rest (\( \omega_0 = 0 \)), with \( \alpha = 0.1575 \) rad/s\(^2\) and \( \theta = 0.5672 \) radians:

• \( \omega^2 = 2 \times 0.1575 \times 0.5672 \)
• \( \omega^2 = 0.1786 \)
• \( \omega = \sqrt{0.1786} = 0.4227 \) rad/s

This calculation illustrates how various factors contribute to the rotational speed of a spinning object, providing insight into its dynamic motion and behaviors.