Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 22

Answer

a) $\alpha$ = -3.07012 rad/ $\sec^{2}$ b) t=12.44s c) The car has travelled 95.03 meters before stopping.

Work Step by Step

a) To find the angular acceleration, the following formula can be used: $\omega_{f}^{2}$ = $\omega_{o}^{2}$ + 2$\alpha$$\theta$. This formula is best because we either know or can calculate all the unknown variables and no unnecessary variables, such as time, are present. 1. The given speeds 55km/h and 95km/h are the translational speeds of the wheels. However, by using the formula v=$\omega$r, they can be converted into angular speeds $\omega_{o}$ and $\omega_{f}$. $\frac{55km}{h}$($\frac{1000m}{1km}$)($\frac{1h}{3600s}$) = $\frac{275m}{18s}$ $\frac{95km}{h}$($\frac{1000m}{1km}$)($\frac{1h}{3600s}$) = $\frac{475m}{18s}$ Translational speed (v) = $\omega$(radius) ($\frac{275m}{18s}$) $\div$ (0.4m) = $\frac{1375rad}{36s}$ = $\omega_{f}$ ($\frac{475m}{18s}$) $\div$ (0.4m) = $\frac{2375rad}{36s}$ = $\omega_{o}$ 2. $\theta$ is the total angle (in radians) through which the wheel has rotated as it slowed down. Since the tires make 75 revolutions and one revolution is 2$\pi$, the total angle $\theta$ = 75(2$\pi$) = 150$\pi$ 3. Then we just plug all the variables in the formula $\omega_{f}^{2}$ = $\omega_{o}^{2}$ + 2$\alpha$$\theta$ and find the angular acceleration. $\frac{1375rad}{36s}^{2}$ = $\frac{2375rad}{36s}^{2}$ + 2$\alpha$(150$\pi$) $\alpha$ = -3.07012 rad/ $\sec^{2}$ b) The answer can be determined using the formula $\omega_{f}$ = $\omega_{o}$ + $\alpha$t Stopping means that $\omega_{f}$ = 0rad/s and "how much more time" implies that $\omega_{o}$ = $\frac{1375rad}{36s}$. Plug the variables in the formula and solve for t. 0rad/s = $\frac{1375rad}{36s}$ + (-3.07012 rad/$\sec^{2}$)t t=12.44s c) We again use the formula $\omega_{f}^{2}$ = $\omega_{o}^{2}$ + 2$\alpha$$\theta$, only this time to find $\theta$. $\frac{0rad}{s}^{2}$ = $\frac{1375rad}{36s}^{2}$ + 2(-3.07012 rad/$\sec^{2}$)$\theta$ $\theta$ = 237.583 rad The relationship between the distance traveled and the angle theta is d=$\theta$(radius) d = (237.583)(0.4m) = 95.03 m The car has travelled 95.03 meters before stopping
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