Problem 58 - University Physics with Modern Physics

Question

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 7.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; (b) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

Step-by-Step Solution

Answer

(a) Moment of inertia is approximately 1.34 kg·m²; (b) Friction torque is approximately -0.112 N·m; (c) Total revolutions during deceleration is about 104.

Step 1: Convert Angular Speed to Radians per Second

First, convert the given angular speed from revolutions per minute (rev/min) to radians per second (rad/s). The formula for conversion is: \[1 \text{ rev} = 2\pi \text{ rad}\]and\[1 \text{ min} = 60 \text{ s}\]So, 100 rev/min is converted as follows:\[100 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{100 \times 2\pi}{60} \approx 10.47 \text{ rad/s}\]

Step 2: Use Torque and Angular Acceleration Relationship

Calculate the angular acceleration \(\alpha\) of the wheel using the relationship between torque \(\tau\) and angular acceleration:\[\tau = I \alpha\]where \(\tau = 7.00 \text{ N} \cdot \text{m}\). Rearranging the equation:\[\alpha = \frac{\tau}{I}\]

Step 3: Use Kinematics to Find Angular Acceleration

Using the kinematic equation:\[\omega_f = \omega_i + \alpha t\]where \(\omega_f = 10.47 \text{ rad/s}\), \(\omega_i = 0\), and \(t = 2.00 \text{ s}\), solve for \(\alpha\):\[10.47 = 0 + \alpha \times 2 \implies \alpha = \frac{10.47}{2} = 5.235 \text{ rad/s}^2\]

Step 4: Calculate Moment of Inertia (I)

Now, use the value of \(\alpha\) to find the moment of inertia \(I\):\[I = \frac{\tau}{\alpha} = \frac{7.00}{5.235} \approx 1.34 \text{ kg} \cdot \text{m}^2\]

Step 5: Calculate Friction Torque

Compute the friction torque using the deceleration period. The wheel slows to rest due to a frictional torque. The kinematic equation during deceleration is:\[0 = \omega_f + \alpha_f \times t\]where \(t = 125 \text{ s}\). Use the same angular acceleration formula with negative \(\alpha_f\):\[0 = 10.47 + \alpha_f \times 125 \implies \alpha_f = -\frac{10.47}{125} \approx -0.0838 \text{ rad/s}^2\]Now, calculate the friction torque:\[\tau_f = I \times \alpha_f = 1.34 \times -0.0838 \approx -0.112 \text{ N} \cdot \text{m}\]

Step 6: Calculate Total Number of Revolutions

Calculate the total angle \(\theta\) covered during the deceleration period:\[\theta = \omega_f \times t_f + \frac{1}{2} \times \alpha_f \times t_f^2\]where \(t_f = 125 \text{ s}\):\[\theta = 10.47 \times 125 + \frac{1}{2} \times (-0.0838) \times 125^2 \approx 653.125 - 654.6875 \approx 653.125\]Convert it from radian to revolutions:\[\frac{653.125}{2\pi} \approx 104 \text{ revolutions}\]

Key Concepts

Angular Speed

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.

Moment of Inertia

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

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

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.