Problem 2 - Fundamentals of Physics

Question

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(75.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

Step-by-Step Solution

Answer

The angular speed is 59.25 rad/s, the angular acceleration is -9.34 rad/s², and the car moves 70.7 meters during braking.

Step 1: Convert linear speed to m/s

The car's speed is given in km/h. To convert it to m/s, use the conversion factor: 1 km/h = 0.27778 m/s. Thus, \( 80.0 \text{ km/h} = 80.0 \times 0.27778 = 22.22 \text{ m/s} \).

Step 2: Calculate the angular speed

The relationship between linear speed \( v \) and angular speed \( \omega \) is \( v = r \omega \), where \( r \) is the radius in meters. Convert the diameter of the tire to meters: \( r = \frac{75.0 \text{ cm}}{2} = 0.375 \text{ m} \). Now solve for \( \omega \): \( \omega = \frac{v}{r} = \frac{22.22}{0.375} \approx 59.25 \text{ rad/s} \).

Step 3: Calculate the final angular speed at stop

When the car comes to a stop, the final angular speed \( \omega_f \) is 0 rad/s.

Step 4: Calculate the initial angular displacement

The car stops after 30 complete turns, so the total angular displacement \( \Delta \theta \) is \( 30 \times 2\pi = 60\pi \text{ rad} \).

Step 5: Calculate the angular acceleration

Use the kinematic equation \( \omega_f^2 = \omega_i^2 + 2\alpha \Delta\theta \), where \( \omega_f = 0 \text{ rad/s} \), \( \omega_i = 59.25 \text{ rad/s} \), and \( \Delta\theta = 60\pi \text{ rad} \). Solving for \( \alpha \), \( \alpha = -\frac{\omega_i^2}{2\Delta\theta} = -\frac{(59.25)^2}{2 \times 60\pi} \approx -9.34 \text{ rad/s}^2 \).

Step 6: Calculate the distance moved during braking

Since the wheels roll without slipping, the linear distance \( d \) is given by \( d = r \Delta\theta \). Using \( r = 0.375 \text{ m} \) and \( \Delta\theta = 60\pi \text{ rad} \), we find \( d = 0.375 \times 60\pi \approx 70.7 \text{ meters} \).

Key Concepts

Angular Speed

Angular speed is a measure of how fast something rotates or revolves relative to another point, such as in a circular path. In our exercise, angular speed describes how quickly the tires of a car spin around their axles when the car is moving. It's expressed in radians per second (rad/s).

To find angular speed, we use the relationship between linear speed and angular speed:

• The formula is given by\( v = r \omega \), where \( v \) is linear speed, \( r \) is the radius of the path, and \( \omega \) is angular speed.
• This formula shows that angular speed is derived from dividing the linear speed by the radius, \( \omega = \frac{v}{r} \).
• In our problem, we calculate the angular speed of the tires by first converting the given tire dimension from centimeters to meters and adjusting the car's speed from kilometers per hour into meters per second.

Understanding angular speed is crucial in numerous applications, such as determining how effectively a vehicle can maintain stability while driving, or how a machine part appropriately transfers motion and force.

Angular Acceleration

Angular acceleration is a measure of how the angular speed of an object changes with time. It tells us how quickly a spinning object, like our car tires, is speeding up or slowing down its rotation.

In the exercise, we find the magnitude of the angular acceleration required for the car to uniformly come to a stop over a certain number of tire rotations:

• When end speed (\( \omega_f \)) is zero, angular acceleration can be found using the kinematic equation \( \omega_f^2 = \omega_i^2 + 2\alpha \Delta\theta \).
• Here, initial angular speed (\( \omega_i \)), final angular speed (\( \omega_f \)), total angular displacement (\( \Delta\theta \)), and other associated values are input to find the angular acceleration (\( \alpha \)).
• Negative angular acceleration indicates a decrease in angular speed, as seen in braking.

Angular acceleration is vital for analyzing systems where rotation speed changes, like braking systems, motors, or amusement park rides, enabling precise control over their functioning.

Kinematic Equations

Kinematic equations are fundamental tools for solving problems related to motion, including linear and angular motion. They connect key quantities like displacement, velocity, acceleration, and time.

In the context of angular motion, kinematic equations help decipher how rotational quantities relate to each other:

• These equations can solve for different variables such as angular speed, angular acceleration, or displacement when specific conditions are known.
• The kinematic equation used in our problem, \( \omega_f^2 = \omega_i^2 + 2\alpha \Delta\theta \), relates the initial and final angular speeds, angular acceleration, and total angular displacement.
• This particular equation provided insight into the braking process, showing how the car's speed was reduced to a stop.

Understanding and applying these equations is crucial in physics and engineering, where accurately predicting motion and mechanics can lead to safer and more efficient designs and operations.