Problem 1 An automobile traveling \(80.0 \... [FREE SOLUTION] (2024)

Chapter 12: Problem 1

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

Short Answer

Expert verified

(a) 59.2 rad/s. (b) 2.33 rad/s^2. (c) 70.7 m.

Step by step solution

Convert Units

First, convert the linear speed from km/h to m/s. Given: Speed, \(v = 80.0 \text{ km/h} = \frac{80.0 \times 1000}{3600} \text{ m/s} = 22.2 \text{ m/s}\)

Calculate the Circumference of the Tire

The diameter of the tire is given. Use the formula to calculate the circumference (C). Diameter, \(d = 75.0 \text{ cm} = 0.75 \text{ m}\) Circumference: \( C = \pi d = \pi (0.75) = 2.36 \text{ m}\)

Calculate the Rotational Speed of the Tires

Use the formula: \( \omega = \frac{v}{r}\), where radius, \(r = \frac{d}{2} = \frac{0.75}{2} = 0.375 \text{ m}\). Hence, \(\omega = \frac{22.2}{0.375} = 59.2 \text{ rad/s}\)

Calculate Total Angle Turned During Braking

Convert the total number of turns to radians. Number of turns = 30.0 One full turn in radians = \( 2\pi \text{ radians}\), Therefore, Total angle, \(\theta = 30.0 \times 2\pi = 60\pi \text{ radians}\)

Calculate Rotational Acceleration

To find the rotational acceleration, use the kinematic equation for rotational motion: \( \omega_f^2 = \omega_i^2 + 2\alpha \theta \), where final rotational speed \(\omega_f = 0\) (since the car comes to a stop). Rearranging for acceleration \(\alpha\), we get: \( 0 = (59.2)^2 + 2 \alpha (60\pi)\), Solving gives: \( \alpha = \frac{-(59.2)^2}{2(60\pi)} = -2.33 \text{ rad/s}^2\)

Calculate the Distance Moved During Braking

The distance moved during braking can be calculated using the total angle turned and the circumference of the tire: \( d = \theta \times r = (60\pi) \times (0.375)\), Hence, Distance, \( d = 70.7 \text{ m}\)

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion in Physics

Unit conversion is a fundamental skill in physics. It's necessary when dealing with measurements in different units. For example, speed is often given in kilometers per hour (km/h). To convert this to meters per second (m/s), which is more suitable for many physics equations, you multiply by \frac{1000}{3600}\ or \frac{5}{18}\ . In our exercise:

Given speed: 80.0 km/h
Conversion factor: \frac{1000}{3600}\ or \frac{5}{18}\

Thus, 80.0 km/h becomes: \(\frac{80.0 \times 1000}{3600} = 22.2 \text{ m/s} \).

Converting units correctly ensures all your measurements are compatible in calculations.

Circumference Calculation

Understanding the circumference of a circle is crucial in rotational motion problems. The circumference is the distance around the circle and is given by the formula: \ C = \pi \cdot d \.

Diameter of tire, \(d\): 0.75 meters

So the circumference of the tire is:
\(C = \pi \cdot 0.75 = 2.36 \text { m } \).

This calculation tells us how far the tire travels in one complete rotation. It's a key step in determining other parameters like rotational speed and distance traveled.

Rotational Speed Calculation

Rotational speed, denoted as \(\omega\), tells us how fast an object is rotating and is measured in radians per second (rad/s). The formula to calculate rotational speed is:
\(\omega = \frac{v}{r}\), where:

v is the linear velocity
r is the radius of the circle
The radius is half of the diameter

Given in our problem, we have:

Linear velocity (v): 22.2 m/s
Radius (r): 0.375 meters

Thus, the rotational speed \(\omega\) is:
\(\omega = \frac{22.2}{0.375} = 59.2 \text{ rad/s} \).
The rotational speed helps us understand how many radians the tire travels per second.

Angular Displacement

Angular displacement, denoted as \(\theta\), is the angle by which an object rotates. It's calculated in radians. To convert complete turns to radians, use the fact that one complete turn is equal to \(2\pi\) radians.

For our exercise, the car makes 30.0 complete turns during braking:
\(\theta = 30.0 \times 2\pi = 60 \pi \text{ radians} \).
This value shows the total angle covered by the rotating tire during the stopping process. Understanding angular displacement is crucial for calculating rotational acceleration and distance traveled.

Rotational Kinematics

Rotational kinematics deals with the motion of rotating objects using kinematic equations. The key equation in this context is:
\(\omega_f^2 = \omega_i^2 + 2\alpha\theta\),
where:

\(\omega_f\) is the final angular velocity
\(\omega_i\) is the initial angular velocity
\(\alpha\) is the angular acceleration
\(\theta\) is the angular displacement

To find the rotational acceleration \(\alpha\) when the car comes to a stop (\(\omega_f = 0\)), rearrange as:
\(0 = (59.2)^2 + 2\alpha(60\pi)\)
Solving this gives:
\(\alpha = -2.33 \text{ rad/s}^2 \).
Initial and final velocities, along with displacement, help in describing how quickly the tires stop rotating.

Distance Traveled During Braking

The distance a car travels when braking can be determined by the total angular displacement and the tire's radius. The formula is:
\(d = \theta \times r\),
where:

\(\theta\) is the angular displacement (in radians)
\(r\) is the radius of the wheel

From our exercise:
\(\theta = 60\pi\)
\(r = 0.375 \text{ m}\)
Therefore, the distance:
\(d = 60\pi \times 0.375 = 70.7 \text{ m}\).
This tells us that the car moves 70.7 meters during the braking period. The combination of the angular displacement and the tire’s radius helps calculate the overall distance.