/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 An automobile traveling at \(80.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
Angular speed: \(59.25 \text{ rad/s}\); Angular acceleration: \(-2.34 \text{ rad/s}^2\); Distance: \(70.69 \text{ m}\).

Step by step solution

01

Convert Linear Speed to Meters per Second

The car is traveling at \(80.0 \text{ km/h}\). First, convert this speed to meters per second using the conversion: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Therefore, \(80.0 \text{ km/h} = \frac{80}{3.6} \text{ m/s} \approx 22.22 \text{ m/s}\).
02

Calculate the Angular Speed of the Tires

The diameter of the tire is given as \(75.0 \text{ cm}\), which is \(0.75 \text{ m}\). The radius \(r\) is half of the diameter: \(r = 0.375 \text{ m}\). The linear speed \(v\) is related to the angular speed \(\omega\) by the formula \(v = \omega r\). Solving for \(\omega\), we have \(\omega = \frac{v}{r} = \frac{22.22}{0.375} \approx 59.25 \text{ rad/s}\).
03

Calculate the Total Angular Displacement During Braking

The car comes to a stop after \(30.0\) complete turns, which is equivalent to an angular displacement \(\theta\). Since one complete turn equals \(2\pi\) radians, \(30.0\) turns is \(30 \times 2\pi = 60\pi \) radians.
04

Determine the Angular Acceleration

The car stops uniformly, meaning it comes to rest with a final angular speed \(\omega_f = 0\). We use the equation \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\) to find angular acceleration \(\alpha\). Substitute \(\omega_f = 0\), \(\omega_i = 59.25 \text{ rad/s}\), and \(\theta = 60\pi\): \(0 = (59.25)^2 + 2\alpha(60\pi)\). Solving for \(\alpha\), we get \(\alpha = -\frac{(59.25)^2}{2 \times 60\pi} \approx -2.34 \text{ rad/s}^2\).
05

Calculate the Distance Traveled During Braking

The distance \(d\) traveled during braking can be determined from the total angular displacement \(\theta\) using the formula \(d = \theta r\). We substitute \(r = 0.375 \text{ m}\) and \(\theta = 60\pi\):\(d = 60\pi \times 0.375 = 22.5\pi \approx 70.69 \text{ m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Speed
Angular speed, often symbolized as \( \omega \), is a measure of how quickly an object rotates or turns. It is the rate at which an object moves around a circular path. For a rotating object like a car tire, this is measured in radians per second (rad/s). To compute angular speed, we need to know the linear speed of the object and the radius of the circle it is moving along. The relationship between linear speed \( v \) and angular speed \( \omega \) is given by the equation: \[ \omega = \frac{v}{r} \] where \( r \) is the radius of the tire. This formula tells us how many radians an object covers per second, based on its linear speed and radius. For example, if a car travels at a speed of 22.22 m/s and has a tire radius of 0.375 m, its angular speed would be approximately 59.25 rad/s.Understanding this concept helps us to comprehend the rotational movement of objects and is essential for solving problems involving circular motion.
Angular Acceleration
Angular acceleration, denoted by \( \alpha \), defines how quickly the rate of rotation changes over time. It is the rotational counterpart to linear acceleration and is measured in radians per second squared (rad/s\(^2\)).When a car brakes to stop uniformly, its angular speed decreases until it reaches zero. This change in angular speed over the time it takes to stop is captured by angular acceleration. We calculate it using the formula:\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \] Here, \( \omega_f \) is the final angular speed, \( \omega_i \) is the initial angular speed, and \( \theta \) is the total angular displacement. For a stopping car tire initially spinning at 59.25 rad/s, uniformly brought to rest over 60\( \pi \) radians, the angular acceleration can be found as -2.34 rad/s\(^2\). Understanding angular acceleration is crucial for analyzing how quickly a rotating object can stop or start, which is vital in many mechanical applications.
Linear Speed Conversion
Converting linear speed from one unit to another is a common task when dealing with rotational motion. Since angular speed calculations typically require linear speed in meters per second (m/s), we often need to convert speeds given in kilometers per hour (km/h) to m/s.The conversion formula is simple: \[ 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \] Thus, to convert a speed like 80 km/h to m/s, you would compute \( \frac{80}{3.6} \approx 22.22 \, \text{m/s} \). This step is crucial to accurately use the linear speed in further angular motion equations.Being adept at converting units of speed ensures that we apply the correct values in our calculations, maintaining consistency and accuracy in results.
Angular Displacement
Angular displacement, represented by \( \theta \), is a measure of the change in the angle as an object rotates around a point or axis. Unlike linear displacement, which tracks movement along a straight path, angular displacement measures how much an object has "turned".In our tire example, we measure angular displacement in radians. One complete revolution is equivalent to \( 2\pi \) radians. Thus, if a car's tire makes 30 complete turns, the angular displacement \( \theta \) would be \( 30 \, \times \, 2\pi = 60\pi \) radians.This concept is essential in rotational dynamics, helping us determine how far a rotating object has turned. Understanding angular displacement allows us to calculate how far a car moves during braking or how much energy is required to stop it.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. 11-62). The radius of the axle is \(0.200 \mathrm{~m}\), and the rotational inertia of the wheel- axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m}\), what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\). The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m})\). After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}-(3.00 \mathrm{~m}) \hat{\mathrm{k}}\), its veloc- ity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \hat{\mathrm{j}}+(4.00 \mathrm{~N}) \hat{\mathrm{k}}\). Find (a) the accel- eration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

Is an overhead view of a thin uniform rod of length \(0.600 \mathrm{~m}\) and mass \(M\) rotating horizontally at \(80.0 \mathrm{rad} / \mathrm{s}\) counterclock- wise about an axis through its center. A particle of mass \(M / 3.00\) and traveling horizontally at speed \(40.0 \mathrm{~m} / \mathrm{s}\) hits the rod and sticks. The particle's path is perpendicular to the rod at the instant of the hit, at a distance \(d\) from the rod's center. (a) At what value of \(d\) are rod and particle stationary after the hit? (b) In which direction do rod and particle rotate if \(d\) is greater than this value?

A girl of mass \(M\) stands on the rim of a frictionless merry-goround of radius \(R\) and rotational inertia \(I\) that is not moving. She throws a rock of mass \(m\) horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(v .\) Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.