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: 5

Answer

$\frac{\omega_o}{\omega_i}=0.517$

Work Step by Step

Body $m=52kg$ $r=0.15m$ $I_B=\frac{MR^2}{2}=\frac{(52kg)(0.15m)^2}{2}=0.585\frac{kgm^2}{2}$ Arms outstretched $m=4kg$ $l=0.60m$ $I_A=\frac{Ml^2}{3}=\frac{(4kg)(0.60m)^2}{3}=0.72\frac{kgm^2}{2}$ Arms in $m=60kg$ $r=0.18m$ $I_{in}=\frac{MR^2}{2}=\frac{(60kg)(0.15m)^2}{2}=0.675\frac{kgm^2}{2}$ $I_{out}=0.585\frac{kgm^2}{2}+0.72\frac{kgm^2}{2}=1.305\frac{kgm^2}{2}$ $L_o=L_i$ $I_o\omega_o=I_i\omega_i$ $\frac{\omega_o}{\omega_i}=\frac{I_i}{I_o}=\frac{0.675\frac{kgm^2}{2}}{1.305\frac{kgm^2}{2}}=0.517$
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