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 - Page 226: 67

Answer

(a) $0.52\frac{rad}{s}$ (b ) $370J, 2.0\times 10^2J$

Work Step by Step

(a)According to law of conservation of angular momentum $L_i=L_f$ $\implies I{\omega}+I{\omega}=I{\omega}+mr^2{\omega}$ We plug in the known values to obtain: $820(0.95)+0=820{\omega}+75(3)^2\omega$ $779=1495\omega$ $\omega=\frac{779}{1495}=0.52\frac{rad}{s}$ (b) Initial Rotational Kinetic energy of the system of platform plus person: $\frac{1}{2}I\omega^2=\frac{1}{2}(820)(0.95)^2=370J$ Final Rotational Kinetic energy of the system of platform plus person: $\frac{1}{2}(820)(0.52)^2+\frac{1}{2}(75)(3)^2(0.52)^2=2.0\times 10^2J$
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.