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 - Search and Learn - Page 229: 6

Answer

a) $E=5.12\times10^6J$ b) $\omega=367\frac{rad}{s}$ c) $t=46.0s$

Work Step by Step

$m=1100kg$ $r=0.75m$ $m=270kg$ $a=\frac{v_f-v_i}{2}=\frac{95\frac{m}{s}-0\frac{m}{s}}{2}=47.5\frac{m}{s^2}$ $F=ma=(1100kg)(47.5\frac{m}{s^2})=52250N$ a) $E=E_f+E_a$ $E_f=dF_f=(350km)(450N)=157500J$ $E_a=\frac{mv^2}{2}=\frac{(1100kg)(95\frac{m}{s})^2}{2}=4963750J$ $E=157500J+4963750J=5121250J$ $=5.12\times10^6J$ b) $L=I\omega$ $I=\frac{MR^2}{2}=\frac{(270kg)(0.75m)^2}{2}=75.9kgm^2$ $E_K=\frac{I\omega^2}{2}=\frac{(75.9kgm^2)(\omega)^2}{2}=5.12\times10^6J$ $\omega=367\frac{rad}{s}$ c) $150hp\times\frac{745.7W}{1hp}=1.12\times10^5W$ $t=\frac{E}{P}=\frac{5.12\times10^6J}{1.12\times10^5W}=46.0s$
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