91Ó°ÊÓ

Torque, often represented by the Greek letter \( \tau \), plays a crucial role in rotational dynamics. It essentially measures how much a force acting at a distance from a pivot point, or axis, causes an object to rotate. For our example, the force varies with time and can be calculated directly from the given equation. To find the torque at a specific time, say \( t = 3.0 \) seconds, we first compute the force at this time using the formula:

\( F(t) = (0.50 \text{ N/s}) t + (0.30 \text{ N/s}^2) t^2 \).

Plugging in \( t = 3.0 \) seconds, the resultant force is:\[ F(3.0 \text{ s}) = 4.2 \text{ N} \].

Given the radius \( r = 0.1 \text{ m} \), torque is then calculated as:\[ \tau = F \, \times \, r = 4.2 \text{ N} \times 0.1 \text{ m} = 0.42 \text{ Nm} \].

Understanding this concept highlights how varying magnitudes and directions of force can influence an object's rotation.

Rotational acceleration, denoted by the symbol \( \alpha \), describes how rapidly an object's rotational speed changes with time. For our exercise, the key to solving rotational acceleration lies in the relationship between torque and the object's moment of inertia—the rotational equivalent of mass. The moment of inertia given is \( I = 1.0 \times 10^{-3} \text{ kg} \text{ m}^2 \).

Using the formula:

\[ \alpha = \frac{\tau}{I} \]

where \( \tau \) is the torque (0.42 \text{ Nm} from the previous section), we find the rotational acceleration to be:

\[ \alpha = \frac{0.42}{1.0 \times 10^{-3}} = 420 \text{ rad/s}^2 \].

This value shows how quickly the pulley’s rotational speed increases due to the applied force. This insight allows us to understand how different torques, moments of inertia, and radius sizes impact the rate at which an object can spin up or slow down.

Rotational speed, symbolized by \( \omega \), indicates how fast an object revolves around an axis. Starting from rest, our goal is to determine its speed after a certain period. Knowing \( \omega_0 = 0 \text{ (initial rotational speed)} \), rotational acceleration \( \alpha = 420 \text{ rad/s}^2 \), and time \( t = 3.0 \text{ s} \), we can find the final rotational speed using the equation:

\[ \omega = \omega_0 + \alpha t \]

Substituting the provided values, we get:

\[ \omega = 0 + (420 \text{ rad/s}^2 \times 3.0 \text{ s}) = 1260 \text{ rad/s} \].

This tells us the pulley spins at a remarkably high rate of 1260 radians per second after 3 seconds. By considering how rotational speed evolves under constant and varying accelerations, students gain a clearer view on practical applications, from wheels speeding up to planets in motion.