91Ó°ÊÓ

When we talk about torque, it's essentially the rotational equivalent of linear force. Imagine you're trying to open a jar. The harder you twist (or apply force), the more torque you produce to overcome the tight lid. Torque is represented by the symbol \( \tau \), and you calculate it using the formula \( \tau = r \times F \). Here, \( r \) is the radius, the distance from the pivot point to the point where the force is applied, and \( F \) is the force applied, in this case, by the potter's hands on the clay.
In our exercise, the radius \( r \) is half the diameter of the bowl, which is 6 cm or 0.06 m. The friction force \( F \) is given as 1.5 N. So, by calculating the product of these two—\( 0.06 \times 1.5 \)—we find that the potter's hands exert a torque of 0.09 Nm on the wheel. This torque is what resists the wheel's motion, eventually bringing it to a stop.

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It's the rotational analog to mass in linear motion. If you've ever tried to roll a heavy barrel, you know how much harder it is compared to a smaller one—that's the moment of inertia at play.
It's represented by \( I \), and in our exercise, it's given as 0.11 kg \( \cdot \) m² for the wheel and the bowl combined. The larger the moment of inertia, the more torque is required to change the object's rotation. In other words, just like heavier objects need more force to move, objects with a larger moment of inertia need more torque to rotate.

• Moment of inertia depends on the mass distribution about the axis of rotation
• It's crucial for calculating angular motion alongside torque

The combination of torque and the moment of inertia determines how the wheel will decelerate over time.

Angular deceleration tells us how quickly an object slows down its rotation. It's similar to linear deceleration but, instead of meters per second squared, it's in radians per second squared.
Given the torque \( (0.09 \text{ Nm}) \) and moment of inertia \( (0.11 \text{ kg} \cdot \text{m}^2) \), we use the formula \( \tau = I \alpha \) to find angular acceleration \( \alpha \), which here acts as a deceleration. Rearranging gives \( \alpha = \frac{\tau}{I} \), or \( \alpha = \frac{0.09}{0.11} \approx 0.818 \text{ rad/s}^2 \).
This value tells us that the wheel's speed decreases by 0.818 radians per second squared. With this deceleration and the initial angular speed known, we can determine the time it takes for the wheel to come to a full stop using the kinematic equation: \( \omega = \omega_0 + \alpha t \). By setting the final speed \( \omega \) to zero, we find the stopping time as approximately 12.3 seconds.