91Ó°ÊÓ

Angular velocity is a measure of how fast something is rotating. It's akin to linear speed, but for circular motion. When we talk about angular velocity, we typically denote it by the symbol \( \omega \) and measure it in radians per second (rad/s). In this context, angular velocity helps us understand how quickly the motor shaft is spinning at any given moment.

To find the angular velocity from an angular displacement equation like \( \theta(t) = (250 \text{ rad/s}) t - (20.0 \text{ rad/s}^2)t^2 - (1.50 \text{ rad/s}^3)t^3 \), you have to take the derivative with respect to time \( t \). By differentiating, we get:
\[ \omega(t) = \frac{d}{dt}(250t - 20t^2 - 1.5t^3) = 250 - 40t - 4.5t^2 \]
This equation tells us exactly what the angular velocity is at any point in time \( t \). You'll notice it changes over time, indicating the motor accelerates and decelerates.

Just as angular velocity measures the speed of rotation, angular acceleration tells us how that speed changes over time. It's like linear acceleration but for things that spin. Represented by the symbol \( \alpha \), angular acceleration is measured in radians per second squared (rad/s²).

To find angular acceleration, we differentiate the angular velocity equation. From the previous section, we have:
\[ \omega(t) = 250 - 40t - 4.5t^2 \]
Taking the derivative gives us the angular acceleration:
\[ \alpha(t) = \frac{d}{dt}(250 - 40t - 4.5t^2) = -40 - 9t \]
In this case, the angular acceleration is a linear function of \( t \), meaning it also changes over time. When the motor shaft reaches zero angular velocity, you substitute this \( t \) value (8.89 s in this problem) into the equation to find \( \alpha \). This calculation shows the rate of change of the shaft's spinning at that precise moment.

Displacement in angular motion refers to the change in the angular position of the rotating object. In our problem, the angular displacement of the motor shaft is given by the equation:
\[ \theta(t) = (250 \text{ rad/s}) t - (20.0 \text{ rad/s}^2)t^2 - (1.50 \text{ rad/s}^3)t^3 \]
This formula gives us \( \theta \), which is the angle through which the motor shaft has turned over time. To find out how many revolutions the motor shaft makes by the time the angular velocity reaches zero, compute \( \theta \) at this specific time \( t = 8.89 \) s.

Once you have \( \theta(8.89) \), convert radians to revolutions, since 1 revolution equals \( 2\pi \) radians. So, you use:
\[ \text{Revolutions} = \frac{\theta(t)}{2\pi} \]
This gives you the total number of full turns the shaft has made up to the point in question.

Quadratic equations are a fundamental part of solving problems involving motion, including angular motion. A quadratic equation is generally expressed in the form \( at^2 + bt + c = 0 \). In this exercise, we use such an equation to find when the angular velocity of the motor shaft becomes zero.

The specific quadratic equation derived from the angular velocity function is:
\[ 250 - 40t - 4.5t^2 = 0 \]
To solve this, apply the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = -4.5 \), \( b = -40 \), and \( c = 250 \) are the coefficients. Calculating gives us the time \( t \) when the angular velocity reaches zero.

Solving quadratic equations is critical in finding specific points like zero velocity in rotational motion, making it a valuable tool in physics.