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 367: 13

Answer

$2.10N.m$

Work Step by Step

We can find the required torque as follows: $\tau=I\alpha$ $\implies \tau=(\frac{1}{12}mL^2)\alpha$ We plug in the known values to obtain: $\tau=\frac{1}{12}(8.42Kg)(3.15)^2(0.302rad/s^2)$ $\tau=2.10N.m$
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