91Ó°ÊÓ

Angular speed refers to how quickly an object rotates or revolves relative to another point. In this exercise, we focus on converting angular speed from revolutions per minute (rev/min) to radians per second (rad/s) because radians are the standard unit for angular measurements in physics. The conversion is crucial because it allows us to apply consistent units throughout our calculations.

To convert from rev/min to rad/s, remember that one revolution is equal to two \pi \, and there are 60 seconds in a minute:

• Use the formula: 100 rev/min \( \times \frac{2\pi}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \), which simplifies to approximately 10.47 rad/s.

Understanding these units and how to convert them is fundamental in problems involving rotational motion.

The moment of inertia \(I\) is a measure of an object's resistance to changes in its rotational motion, akin to mass in linear motion. It depends on how the mass is distributed relative to the rotational axis. In our problem, the bicycle wheel's moment of inertia is essential to understand its rotational dynamics.

Using the relationship between torque \(\tau\), moment of inertia \(I\), and angular acceleration \(\alpha\):

• \(\tau = I \alpha\) allows us to rearrange to find \(I = \frac{\tau}{\alpha}\).

With the applied torque and calculated angular acceleration, we can calculate the moment of inertia as approximately 1.34 kg·m².
This value tells us how the wheel's mass is affecting its rotational acceleration. Large moment of inertia signifies that more torque is required to achieve the same angular acceleration as a smaller inertia object.

Friction torque arises when friction opposes the rotational motion, gradually causing a rotating object to slow down and eventually stop. In this scenario, friction torque is the opposing force that halted the bicycle wheel's motion over 125 seconds.

The calculation demands understanding how deceleration is related to angular velocity and time:

• Recognize that the angular acceleration \(\alpha_f\) during deceleration is negative because it opposes initial motion.
• Using \(\alpha_f = -\frac{10.47}{125} \approx -0.0838\text{ rad/s}^2\).
• Apply the formula \(\tau_f = I \alpha_f\) to find friction torque of approximately \(-0.112 \text{ N} \cdot \text{m}\).

This negative value highlights how frictional forces work against rotational motion, eventually bringing the wheel to rest.

Rotational kinematics involves analyzing the motion of rotating objects using angular quantities. It extends the concepts of linear kinematics to rotational systems, allowing us to find angles, angular speeds, and accelerations.

In the given problem, once the bicycle wheel starts decelerating due to friction, we use kinematic equations to determine its rotational behavior:

• First, calculate the total angle \(\theta\) during deceleration using: \(\theta = \omega_f \times t_f + \frac{1}{2} \alpha_f \times t_f^2\).
• Substitute the given values for the final angular speed, deceleration time, and calculated angular acceleration.

The computed angle helps us find the number of revolutions made during the 125-second interval. The total revolutions output (approximately 104 revolutions) gives a clear measure of the wheel's rotation before it stopped.