University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 334: 10.64

Answer

$a = \frac{F}{2M}$ $F_f = \frac{F}{2}$

Work Step by Step

We can set up a torque equation for the cylinder. $\tau = I\alpha$ $F_f~R = (MR^2)(\frac{a}{R})$ $F_f = Ma$ We can use this expression in the force equation for the cylinder. $\sum F = Ma$ $F-F_f = Ma$ $F-Ma= Ma$ $F = 2Ma$ $a = \frac{F}{2M}$ We can find the friction force. $F_f = Ma$ $F_f = (M)(\frac{F}{2M})$ $F_f = \frac{F}{2}$
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