91Ó°ÊÓ

In the context of rotational motion, the rotational position function describes the angular position of a rotating object over time. It is denoted by \( \theta \), which represents the angle in radians. For the given exercise, the rotational position of the flywheel is given by:
\[ \theta = (a \, \text{rad/s}) \cdot t + (b \, \text{rad/s}^3) \cdot t^3 - (c \, \text{rad/s}^4) \cdot t^4 \]
Here, \ a \, \ b \, and \ c \, are constants. This function allows us to determine the angle at which the flywheel is rotating at any given time \ t \, by substituting the value of \ t \ into the equation. The constants \ a \, \ b \, and \ c \, characterize specific aspects of the flywheel's motion, such as its initial rate of rotation and the manner in which its rotation changes over time.

Rotational velocity, also known as angular velocity, is a measure of how rapidly the rotational position \( \theta \) of an object changes with time. It is represented by the symbol \omega \ and is calculated as the first derivative of the rotational position function concerning time. In simpler terms, it's how fast something is spinning.
To find the rotational velocity for the flywheel's position function:\[ \omega = \frac{d \theta}{dt} \]
Given the position function:\[ \theta = (a \, \text{rad/s}) t + (b \, \text{rad/s}^3) t^3 - (c \, \text{rad/s}^4) t^4 \]
The rotational velocity can be derived as:

• The derivative of \(a \, \text{rad/s} \cdot t \) is \ a \, \text{rad/s} \
• The derivative of \(b \, \text{rad/s}^3 \cdot t^3 \) is \ 3b \, \text{rad/s}^3 \cdot t^2 \
• The derivative of \( -c \, \text{rad/s}^4 \cdot t^4 \) is \ -4c \, \text{rad/s}^4 \cdot t^3 \

Combining these results, we get:
\[ \omega = a \, \text{rad/s} + 3b \, \text{rad/s}^3 \cdot t^2 - 4c \, \text{rad/s}^4 \cdot t^3 \]

Rotational acceleration, or angular acceleration, measures how quickly the rotational velocity \( \omega \) changes with time. It is denoted by the symbol \alpha \ and is found by taking the first derivative of the rotational velocity concerning time. Essentially, this tells us how the speed of the flywheel's rotation is changing at any moment.
To find rotational acceleration for the derived rotational velocity function:\[ \alpha = \frac{d \omega}{dt} \]
Given the rotational velocity:\[ \omega = a \, \text{rad/s} + 3b \, \text{rad/s}^3 \cdot t^2 - 4c \, \text{rad/s}^4 \cdot t^3 \]
The rotational acceleration is calculated as:

• The derivative of \(3b \, \text{rad/s}^3 \cdot t^2 \) is \ 6b \, \text{rad/s}^3 \cdot t \
• The derivative of \(-4c \, \text{rad/s}^4 \cdot t^3 \) is \ -12c \, \text{rad/s}^4 \cdot t^2 \

Combining these results, we get:\[ \alpha = 6b \, \text{rad/s}^3 \cdot t - 12c \, \text{rad/s}^4 \cdot t^2 \]Understanding the relationship between rotational position, velocity, and acceleration is crucial for studying how objects move in circular paths.