Question

An automobile traveling \(78.3 \mathrm{~km} / \mathrm{h}\) has tires of \(77.0-\mathrm{cm}\) diameter. (a) What is the angular speed of the tires about the axle? (b) If the car is brought to a stop uniformly in \(28.6\) turns of the tires (no skidding), what is the angular acceleration of the wheels? (c) How far does the car advance during this braking period?

Short answer

The angular speed of the tires is \(56.4935 \mathrm{~rad/s}\). The angular acceleration is \(-111.6287 \mathrm{~rad/s}^2\). During braking, the car advances \(69.1937 \mathrm{m}\).

Step-by-step solution

Calculate the Angular Speed

First, find the speed in \(\mathrm{m/s}\) by multiplying with a conversion factor: \(78.3 \mathrm{~km/h} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}} = 21.75 \mathrm{~m/s}\). Then, calculate the radius of the tires in \(\mathrm{meters}\): \(\frac{77.0 \mathrm{cm}}{2}\times \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.385 \mathrm{m}\). Now, compute the angular speed using the formula \(\omega = \frac{v}{r}\), where \(v\) is the linear speed of the car and \(r\) is the radius of the tires. Substituting these values gives \(\omega = \frac{21.75 \mathrm{m/s}}{0.385 \mathrm{m}} = 56.4935 \mathrm{~rad/s}\)

Determine the Angular Acceleration

The car is brought to a stop, so the final angular speed will be \(0 \mathrm{~rad/s}\). To find the time it took to stop, we compute the time for one rotation (or period) by taking the inverse of the angular speed \(\frac{1}{56.4935 \mathrm{~rad/s}} = 0.0177 \mathrm{s}\). Then multiply by the total number of turns to get the stopping time \(0.0177 \mathrm{s} \times 28.6 = 0.5058 \mathrm{s}\). Now using the formula for angular acceleration, \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular speed and \(\Delta t\) is the change in time. We get \(\alpha = \frac{-56.4935 \mathrm{~rad/s}}{0.5058 \mathrm{s}} = -111.6287 \mathrm{~rad/s}^2\)

Calculate the Distance Advanced During Braking

Multiply the angular displacement by radius to get the linear displacement. The angular displacement is given by \(28.6 \times 2\pi = 179.5949 \mathrm{rad}\). Therefore, the displacement is \(179.5949 \mathrm{rad} \times 0.385 \mathrm{m} = 69.1937 \mathrm{m}\)

Key concepts

Angular Speed

Angular speed refers to how quickly an object rotates about an axis. It's expressed in radians per second (rad/s). In this problem, to find angular speed, you first need the linear speed of the car, given in km/h, which we convert to meters per second. Here's how it works:

• Convert the speed: Multiply the speed in km/h by \(\frac{1000}{3600}\), which changes it to m/s.

Once we have the linear speed, we use the formula:
\[ \omega = \frac{v}{r} \]where \(v\) is the linear speed and \(r\) is the radius of the tire. Divide linear speed by the radius to find the angular speed. This formula reflects the proportionality between linear speed and angular speed, considering the distance from the axis is constant (the tire's radius). Ensuring the conversion from cm to m for the radius is crucial for accuracy.

Angular Acceleration

Angular acceleration describes how the angular velocity of an object changes over time, measured in radians per second squared (rad/s²). When a car stops, its wheels experience angular deceleration. Essential steps include:

• Determine the initial and final angular speeds (Initial is the calculated speed; final speed is 0 as the car stops).
• Find the total time taken for this change. First, calculate the time for one tire rotation, and multiply this duration by the total number of rotations until the car stops.

With time and change in speed known, use the angular acceleration formula:
\[\alpha = \frac{\Delta \omega}{\Delta t}\]Understanding this idea helps in calculating how quickly the car is losing its speed, providing insights into the car's braking capabilities and safety dynamics. The negative sign in our solution infers deceleration.

Linear Displacement

Linear displacement measures the distance traversed by an object along a straight line. In this scenario, as the car stops, we desire to evaluate how far it progresses. This uses the concept of angular displacement, calculated as the number of wheel rotations multiplied by 2π (complete rotation in radians). Steps include:

• Calculate angular displacement in radians: Number of turns × 2π.
• Apply the formula: Linear displacement = Angular displacement × Radius of the tire.

This relationship illustrates the direct conversion from angular motion to linear motion, showing the link between rotational measures and straight-line distance, an essential aspect of understanding how vehicle dynamics translate from angular to linear realms on a practical level.

Conversion of Units

Unit conversion is vital for integrating various dimensions. Dealing with different metrics demands fluency in conversions.
To convert speed:

• From km/h to m/s: Multiply the km/h value by \(\frac{1000}{3600}\) to transition from kilometers per hour to meters per second.

For measuring dimensions like tire diameters or radii:

• Convert cm to m: As 1 m equals 100 cm, multiply or divide accordingly.

Grasp these conversions to seamlessly transition between different physical quantities, which is a critical skill in physics and real-world applications for consistency and accuracy.

Braking Dynamics

Braking dynamics explore how forces decelerate a vehicle. It's intertwined with angular deceleration and displacement.

• Understanding braking starts with calculating angular deceleration as the wheels stop. We then determine stopping distance through linear displacement.

Within these dynamics, you assess how a vehicle transitions from motion to rest in a controlled manner, involving the transformation from kinetic energy to heat (through friction). The uniform stop assumes consistent deceleration across the entire braking duration, clarifying vehicle safety calculations and the importance in engineering safe braking systems.