Flywheel position. The angular position of a flywheel on an engine is given as ? = c1 (t/t1) + c2 (t/t2) 2, where c1, and c2 are dimensionless constants, and t1 and t2 are characteristic times. (a) Find an expression for the flywheel’s angular velocity. (b) Find an expression for the flywheel’s angular acceleration.

Understanding angular position is crucial when studying the motion of rotating objects, like flywheels. Angular position, often represented by \( \theta \), is a measure of how much an object has rotated from a reference position. Imagine it like the hands of a clock; the position of the hands at any given time defines their angular position.
For the given exercise, the angular position of the flywheel is described by the equation:
\[ \theta = c_1 \left( \frac{t}{t_1} \right) + c_2 \left( \frac{t}{t_2} \right)^2 \] This equation tells us how the flywheel's position changes over time \( t \).
Here:

• \( c_1 \) and \( c_2 \) are constants without dimensions (they are just numbers).
• \( t_1 \) and \( t_2 \) are characteristic times.

By plugging different values of \( t \) into the equation, we can calculate the flywheel's angular position at any point in time.

Angular velocity, represented by \( \omega \), describes how fast an object is rotating. It's the rate of change of angular position with respect to time. Think of it as how quickly the minute hand of a clock moves around the dial.
Mathematically, angular velocity is the first derivative of angular position \( \theta \) with respect to time \( t \). For the given flywheel, the expression for angular velocity is:
\[ \omega = \frac{d\theta}{dt} = \frac{d}{dt} \left( c_1 \frac{t}{t_1} + c_2 \left( \frac{t}{t_2} \right)^2 \right) \]
Using the chain rule, this becomes: \[ \omega = c_1 \frac{1}{t_1} + c_2 \cdot 2 \left( \frac{t}{t_2} \right) \cdot \frac{1}{t_2} \]
Finally, we simplify the expression to get: \[ \omega = \frac{c_1}{t_1} + \frac{2c_2 t}{t_2^2} \]
This equation tells us how fast the flywheel is rotating at any given time.

Angular acceleration, represented by \( \alpha \), describes how quickly the angular velocity is changing. It's like how the speed of a car changes when you press the gas pedal.
In our exercise, angular acceleration is the time derivative of angular velocity \( \omega \). From our earlier calculation, we know that: \[ \omega = \frac{c_1}{t_1} + \frac{2c_2 t}{t_2^2} \]
Differentiating this with respect to time \( t \), we get: \[ \alpha = \frac{d\omega}{dt} = \frac{d}{dt} \left( \frac{c_1}{t_1} + \frac{2c_2 t}{t_2^2} \right) \]
Only the second term \( \frac{2c_2 t}{t_2^2} \) involves time, so: \[ \alpha = \frac{d}{dt} \left( \frac{2c_2 t}{t_2^2} \right) = \frac{2c_2}{t_2^2} \]
This tells us how quickly the angular velocity of the flywheel is changing. By understanding this, we can predict the flywheel's future motion more precisely.