/*! 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 46 As you drive down the highway, t... [FREE SOLUTION] | 91Ó°ÊÓ

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As you drive down the highway, the top of your tires are moving with a speed \(v\). What is the reading on your speedometer?

Short Answer

Expert verified
The speedometer reads \( \frac{v}{2} \).

Step by step solution

01

Understanding the Problem

When you drive, your car's tires rotate around their axes. The top of the tire moves twice as fast as the car itself because the topmost point of the tire consists of both the forward motion of the car and the tire's rotation.
02

Identifying Speed Contributions

At the top of the tire, speed is the sum of the speed due to the car moving forward and the speed due to the tire rotating in relation to the center, meaning it is equal to twice the car's speed.
03

Setting Up the Equation

Let the car's speed be denoted as \( v_c \). The speed of the top of the tire \( v_t \) is given by \( v_t = 2v_c \). Thus, if \( v \) is the speed at the top of the tire, then \( v = 2v_c \).
04

Solving for the Speedometer Reading

Given \( v = 2v_c \), the speedometer, which displays the speed of the car itself, reads \( v_c = \frac{v}{2} \). It does not account for the doubling effect observed at the top of the tire.

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Key Concepts

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

Rotational motion
Understanding rotational motion is crucial when studying the movement of objects like car tires. When a car moves forward, its tires rotate around their axes. This rotation is an example of rotational motion, where each point on the tire moves in a circular path around the tire’s center.

The tire's rotation creates different speeds at different points on the tire. For instance, the base of the tire, where it meets the road, is momentarily at rest as the tire rolls forward. Other points on the tire have different speeds depending on their position.

At the top of the tire, the rotational motion adds to the car's forward speed. Understanding this combination of movements helps explain why the top of a tire moves faster than the rest of the tire. This concept is critical in analyzing the dynamics of rotating objects and understanding how rotational motion affects the overall motion of an object.
Speed calculation
Calculating the speed of different points on a rotating object helps us understand how those points contribute to the object's overall motion. For a car tire, the speed calculation involves both the car’s linear speed and the tire’s rotational speed.

To find the speed at the top of the tire, you add the speed of the car moving forward and the speed due to the tire's rotation. Since the topmost point of the tire moves in the same direction as the car, these speeds combine.

Let's denote the speed of the car as \( v_c \). The speed of the top of the tire, \( v_t \), is the sum of the car's speed and the tire's rotational effect, giving us \( v_t = 2v_c \). Thus, if the speedometer reads \( v_c \), the speed at the top of the tire is actually twice this value.

This understanding is vital for interpreting speedometer readings accurately, acknowledging that they measure the car's linear speed, not the speed at various points on the tire.
Circular motion
Circular motion is a fundamental concept when we discuss the dynamics of rotating objects such as car tires. When a tire rotates, every point along the tire traces a circular path. This is particularly evident at the tire’s outer edge.

In circular motion, the velocity of a point is always tangent to the circle at that point. This explains why the top of the tire has the highest speed—it is both moving forward with the car and rotating, contributing to this increased speed.

To conceptualize this, imagine a toy car with a visible wheel. As it rolls, track any single point on the toy wheel. You will notice that while it moves forward, it also traces circular paths around the central axis of the wheel. This dual movement helps clarify why understanding the patterns of circular motion is vital in comprehending how objects like tires move and interact with other forces.

Thus, grasping circular motion provides insights into everyday applications, like why certain points on a rotating tire move faster than others, ultimately improving your understanding of dynamics.

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Most popular questions from this chapter

Find the rate at which the rotational kinetic energy of the Earth is decreasing. The Earth has a moment of inertia of \(0.331 M_{\mathrm{E}} R_{\mathrm{E}}^{2}\) where \(R_{E}=6.38 \times 10^{6} \mathrm{m}\) and \(M_{\mathrm{E}}=5.97 \times 10^{24} \mathrm{kg},\) and its rotational period increases by \(2.3 \mathrm{ms}\) with each passing century. Give your answer in watts.

A compact disk, which has a diameter of \(12.0 \mathrm{cm},\) speeds up uniformly from 0.00 to 4.00 rev \(/ \mathrm{s}\) in \(3.00 \mathrm{s}\). What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is (a) 2.00 rev \(/\) s and (b) \(3.00 \mathrm{rev} / \mathrm{s} ?\)

Angular Acceleration of the Crab Nebula The pulsar in the Crab nebula (Problem 9) was created by a supernova explosion that was observed on Earth in A.D. 1054 . Its current period of rotation \((33.0 \mathrm{ms})\) is observed to be increasing by \(1.26 \times 10^{-5}\) seconds per year. (a) What is the angular acceleration of the pulsar in rad/s \({ }^{2} ?\) (b) Assuming the angular acceleration of the pulsar to be constant, how many years will it take for the pulsar to slow to a stop? (c) Under the same assumption, what was the period of the pulsar when it was created?

Weightless on the Equator In Quito, Ecuador, near the equator, you weigh about half a pound less than in Barrow, Alaska, near the pole. Find the rotational period of the Earth that would make you feel weightless at the equator. (With this rotational period, your centripetal acceleration would be equal to the acceleration due to gravity, g.)

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proton. In the lowest-energy orbit the distance from the proton to the electron is \(5.29 \times 10^{-11} \mathrm{m},\) and the linear speed of the electron is \(2.18 \times 10^{6} \mathrm{m} / \mathrm{s}\), (a) What is the angular speed of the electron? (b) How many orbits about the proton does it make each second? (c) What is the electron's centripetal acceleration?

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