University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 295: 9.19


(a) $\omega=1.99\times10^{-7}\space rad/s$ (b) $\omega=7.27\times10^{-5}\space rad/s$ (c) $v_{tan}=2.98\times10^4\space m/s$ (d) $v_{tan}=463\space m/s$ (e) $\alpha_{rad}=0.0337\space m/s^2$ $\space\space\space\space\space\alpha_{tan}=0\space m/s^2$

Work Step by Step

(a) $\omega=\frac{\Delta\theta}{\Delta t}$ Earth orbits the Sun once every 365.3 days, there are 86400 secs in 1 day and $\Delta\theta=2\pi \space rad$ $\omega=\frac{2\pi \space rad}{365.3\space days \times 86400 \space s}=1.99\times10^{-7}\space rad/s$ (b) $\omega=\frac{\Delta\theta}{\Delta t}$ It takes 24 hours for earth to spin once around its axis so 1 spin is $\Delta\theta=2\pi \space rad$ and $\Delta t=86400s$ $\omega=\frac{2\pi \space rad}{86400s}=7.27\times10^{-5}\space rad/s$ (c) Earth's orbital radius is given in Appendix F as $1.5\times10^{11}\space m$ $v_{tan}=r\omega=(1.5\times10^{11}\space m)(1.99\times10^{-7} \space rad/s)=2.98\times10^4\space m/s$ (d) Earth's radius is is given in Appendix F as $6.37\times10^6\space m$ $v_{tan}=r\omega=(6.37\times10^6\space m)(7.27\times10^{-5}\space rad/s)=463\space m/s$ (e) $\alpha_{rad}=r\omega^2=(6.37\times10^6\space m)(7.27\times10^{-5}\space rad/s)^2=0.0337\space m/s^2$ Since the Earth is not accelerating in its axis spin its angular acceleration $\alpha$ is 0 therefore, $\alpha_{tan}=r\alpha=0\space m/s^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.