/*! 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 3 The tires on a new compact car h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 3

The tires on a new compact car have a diameter of \(2.0 \mathrm{ft}\) and are warranted for 60000 miles. (a) Determine the angle (in radians) through which one of these tires will rotate during the warranty period. (b) How many revolutions of the tire are equivalent to your answer in part (a)?

Short Answer

Expert verified
The tire rotates through \(316800000\pi\) radians or makes \(158400000\) revolutions during the warranty period.

Step by step solution

01

Calculate the circumference of the tire

The diameter of the tire is 2 feet. Therefore, its radius is 1 foot. The circumference of the tire can be found using formula for circumference of a circle which is \(2\pi r \). Hence, the circumference of the tire is \(2\pi \times 1 = 2\pi\) feet.
02

Calculate the total rotation angle

The warranty for the tires is 60000 miles. Convert this distance into feet: \(1\) mile \(=5280\) feet, hence \(60000\) miles \(=60000 \times 5280\) feet. The rotation angle in radians is equal to the total distance traveled divided by the radius of the tire. So rotation angle \(=\frac{60000 \times 5280}{1}=316800000 \pi\) radians.
03

Calculate number of revolutions

The rotation angle in number of full revolutions is obtained by dividing the total rotation angle by \(2\pi\). Hence the number of revolutions \(=\frac{316800000\pi}{2\pi}=158400000\) revolutions.

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 91影视!

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 satellite of Mars, called Phoebus, has an orbital radius of \(9.4 \times 10^{6} \mathrm{~m}\) and a period of \(2.8 \times 10^{4} \mathrm{~s}\). Assuming the orbit is circular, determine the mass of Mars.

An electric motor rotating a workshop grinding wheel at a rate of \(1.00 \times 10^{2} \mathrm{rev} / \mathrm{min}\) is switched off. Assume the wheel has a constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in part (a)?

The pilot of an airplane executes a constantspeed loop-the-loop maneuver in a vertical circle as in Figure \(7.15 \mathrm{~b}\). The speed of the airplane is \(2.00 \times\) \(10^{2} \mathrm{~m} / \mathrm{s}\), and the radius of the circle is \(3.20 \times 10^{3} \mathrm{~m}\). (a) What is the pilot's apparent weight at the lowest point of the circle if his true weight is \(712 \mathrm{~N}\) ? (b) What is his apparent weight at the highest point of the circle? (c) Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body. Under what conditions does this occur? (d) What speed would have resulted in the pilot experiencing weightlessness at the top of the loop?

A piece of mud is initially at point \(A\) on the rim of a bicycle wheel of radius \(R\) rotating clockwise about a horizontal axis at a constant angular speed \(\omega\) (Fig. P7.8). The mud dislodges from point \(A\) when the wheel diameter through \(A\) is horizontal. The mud then rises vertically and returns to point \(A\). (a) Find a symbolic expression in terms of \(R, \omega\), and \(g\) for the total time the mud is in the air and returns to point \(A\). (b) If the wheel makes one complete revolution in the time it takes the mud to return to point \(A\), find an expression for the angular speed of the bicycle wheel \(\omega\) in terms of \(\pi, g\), and \(R\).

The diameters of the main rotor and tail rotor of a single-engine helicopter are \(7.60 \mathrm{~m}\) and \(1.02 \mathrm{~m}\), respectively. The respective rotational speeds are \(450 \mathrm{rev} / \mathrm{min}\) and \(4138 \mathrm{rev} / \mathrm{min}\). Calculate the speeds of the tips of both rotors. Compare these speeds with the speed of sound, \(343 \mathrm{~m} / \mathrm{s}\).
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Circular Motion and Gravitation

Read Explanation

Oscillations

Read Explanation

Electricity and Magnetism

Read Explanation

Physics of Motion

Read Explanation

Dynamics

Read Explanation
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.