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Chapter 9: Problem 41

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

Expert verified
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

01

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 \(\omega = \frac{21.75 \mathrm{m/s}}{0.385 \mathrm{m}} = 56.4935 \mathrm{~rad/s}\)
02

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 \(\alpha = \frac{-56.4935 \mathrm{~rad/s}}{0.5058 \mathrm{s}} = -111.6287 \mathrm{~rad/s}^2\)
03

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 \(179.5949 \mathrm{rad} \times 0.385 \mathrm{m} = 69.1937 \mathrm{m}\)

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Most popular questions from this chapter

Let \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} .\) Let \(\overrightarrow{\mathbf{c}}=\) \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\). ( \(a\) ) Find \(\mathbf{c}\), expressed in unit vector notation. ( \(b\) ) Find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\).

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