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Chapter 8: Problem 30

An automobile traveling at \(97 \mathrm{~km} / \mathrm{h}\) has wheels of diameter \(76 \mathrm{~cm} .(a)\) Find the angular speed of the wheels about the axle. \((b)\) The car is brought to a stop uniformly in 30 turns of the wheels. Calculate the angular acceleration. \((c)\) How far does the car advance during this braking period?

Short Answer

Expert verified
The angular speed of the wheels is \(709.06 \, \text{rad/s}\). The angular acceleration is \(-3.76 \, \text{rad/s}^2\). The car travels about \(71.12 \, \text{m}\) during the stopping period.

Step by step solution

01

Find the angular speed

The linear speed \(v\) can be converted to angular speed \(\omega\) through the relation: \(\omega = \frac{v}{r}\), where \(r\) is the radius of the wheel. Convert speed to cm/s: \(v = 97 \, \text{km/h} * \frac{1000 \, \text{m}}{1 \, \text{km}} * \frac{100 \, \text{cm}}{1 \, \text{m}} * \frac{1 \, \text{h}}{3600 \, \text{s}} = 26944.44 \, \text{cm/s}\). Afterward, take \(r = \frac{76 \, \text{cm}}{2} = 38 \, \text{cm}\). Substituting the values, we have: \(\omega = \frac{26944.44 \, \text{cm/s}}{38 \, \text{cm}} = 709.06 \, \text{rad/s}\).
02

Calculate the angular acceleration

Given that the car came to a stop uniformly in 30 turns of the wheels, we can use \(\alpha = \frac{-\omega}{\theta}\) to find the angular acceleration \(\alpha\), where \(\theta\) is the total angle in radians (which is \(30\) turns \(* 2\pi\)). Hence, \(\theta = 30 * 2\pi = 60\pi \, \text{rad}\). Substituting the values, we have: \(\alpha = \frac{-709.06 \, \text{rad/s}}{60\pi \, \text{rad}} = -3.76 \, \text{rad/s}^2\).
03

Find the distance covered during braking

Finally, the distance \(d\) covered during this period can be calculated from the relation: \(d = r \times \theta\). Substituting the given radius and calculated theta: \(d = 38 \, \text{cm} * 60\pi \, \text{rad} = 7111.8 \, \text{cm or} \, 71.12 \, \text{m}\).

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Most popular questions from this chapter

A certain wheel turns through 90 rev in \(15 \mathrm{~s}\), its angular speed at the end of the period being \(10 \mathrm{rev} / \mathrm{s} .(a)\) What was the angular speed of the wheel at the beginning of the 15 -s interval, assuming constant angular acceleration? \((b)\) How much time had elapsed between the time the wheel was at rest and the beginning of the 15 -s interval?

A rigid object rotating about the \(z\) axis is slowing down at \(2.66 \mathrm{rad} / \mathrm{s}^{2} .\) Consider a particle located at \(\overrightarrow{\mathbf{r}}=(1.83 \mathrm{~m}) \hat{\mathbf{j}}+\) \((1.26 \mathrm{~m}) \hat{\mathbf{k}}\). At the instant that \(\overrightarrow{\boldsymbol{\omega}}=(14.3 \mathrm{rad} / \mathrm{s}) \hat{\mathbf{k}}\), find \((a)\) the velocity of the particle and \((b)\) its acceleration. \((c)\) What is the radius of the circular path of the particle?

What is the angular speed of a car rounding a circular turn of radius \(110 \mathrm{~m}\) at \(52.4 \mathrm{~km} / \mathrm{h}\) ?

If an airplane propeller of radius \(5.0 \mathrm{ft}(=1.5 \mathrm{~m})\) rotates at 2000 rev/min and the airplane is propelled at a ground speed of \(300 \mathrm{mi} / \mathrm{h}(=480 \mathrm{~km} / \mathrm{h})\), what is the speed of a point on the tip of the propeller, as seen by \((a)\) the pilot and \((b)\) an observer on the ground? Assume that the plane's velocity is parallel to the propeller's axis of rotation.

A rigid body, starting at rest, rotates about a fixed axis with constant angular acceleration \(\alpha\). Consider a particle a distance \(r\) from the axis. Express \((a)\) the radial acceleration and \((b)\) the tangential acceleration of this particle in terms of \(\alpha, r\), and the time \(t .(c)\) If the resultant acceleration of the particle at some instant makes an angle of \(57.0^{\circ}\) with the tangential acceleration, through what total angle has the body rotated from \(t=0\) to that instant?
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