Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises - Page 326: 43

Answer

$a_t=0.742\frac{m}{s^2}$ $a_{cp}=1300\frac{m}{s^2}$

Work Step by Step

We know that $a_t=r\alpha$ We plug in the known values to obtain: $a_t=(0.00320)(232)=0.742\frac{m}{s^2}$ similarly, $a_{cp}=r\omega^2$ We plug in the known values to obtain: $a_{cp}=(0.00320)(640)$ $a_{cp}=1300\frac{m}{s^2}$
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