Physics Technology Update (4th Edition)

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

Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 371: 81

Answer

$116W$

Work Step by Step

We know that $\tau=Fr$ $\tau=3.68oz.in\times\frac{1lb}{16oz}\times \frac{4.45}{1lb}\times \frac{1m}{39.4in}$ $\tau=0.0260N.m$ we convert the angular velocity into rad/s as $\omega=42500\frac{rev}{min}\times \frac{2\pi rad}{rev}\times \frac{1min}{60s}=4450\frac{rad}{s}$ Now we can calculate the power output as $P=\tau \omega$ We plug in the known values to obtain: $P=(0.0260)(4450)$ $P=116W$
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