91Ó°ÊÓ

Understanding how to convert angular velocity from one unit to another is crucial in rotational dynamics. Angular velocity is often given in revolutions per minute (rpm) for practical applications such as in motors or wheels. However, in physics, it is more useful to express it in radians per second (rad/s) for calculations. This is done by using the conversion:

• 1 revolution = \(2\pi\) radians,
• 1 minute = 60 seconds.

A quick conversion formula is: \(\omega = \text{rpm} \times \frac{2\pi}{60}\). In our exercise, the rotor's initial angular velocity is 3600 rpm. Using the conversion formula, this becomes \(120\pi \text{ rad/s}\). This allows us to easily integrate the angular velocity into torque, inertia, and kinetic equations.

Angular deceleration is the rate at which an object slows down its rotation. It is a key factor in determining how quickly an object will come to a stop. We can calculate angular deceleration \(\alpha\) using the formula:

• \( \alpha = \frac{\Delta \omega}{\Delta t} \),

where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time over which this change occurs. In this exercise, the rotor of the motor decelerates from an initial velocity of \(120\pi \text{ rad/s}\) to zero over a period of 252 seconds. Applying the formula, \(\alpha\) is calculated as \(\frac{-120\pi}{252} \text{ rad/s}^2\), which simplifies to \(\frac{-10\pi}{21} \text{ rad/s}^2\). This includes a negative sign, indicating that the rotor is slowing down.

Moment of inertia (I) is a measure of an object's resistance to changes in its rotation. It depends on the mass distribution relative to the axis of rotation. The larger the inertia, the harder it is to change the rotational speed. We use the relationship between torque \(\tau\), inertia \(I\), and angular deceleration \(\alpha\) as:

• \(\tau = I \cdot \alpha\).

Knowing \(\tau = 1.2 \text{ N}\cdot\text{m}\) and the calculated \(\alpha = \frac{-10\pi}{21} \text{ rad/s}^2\), we solve for \(I\) using \(I = \frac{1.2}{\left|\alpha\right|}\). This simplifies to \(I = \frac{25.2}{10\pi} \approx 0.802 \text{ kg}\cdot\text{m}^2\). Ignoring the sign, since inertia is always positive, helps determine the object’s resistance to rotational motion.

The radius of gyration \(k\) provides a simplified way of understanding how mass is distributed around the axis of rotation. It is defined as:

• \(k = \sqrt{\frac{I}{m}}\),

where \(I\) is the moment of inertia and \(m\) is the object's mass. Essentially, it shows how far from the axis of rotation the mass could be concentrated to produce the same moment of inertia. For a rotor with a mass of 25 kg and an inertia of \(0.802 \text{ kg}\cdot\text{m}^2\), substituting into the formula gives \(k = \sqrt{\frac{0.802}{25}} \approx 0.18 \text{ m}\). This means if the rotor's entire mass were concentrated at a distance of 0.18 meters from its axis, it would have the same rotational inertia as its actual distribution.