Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 330: 17


(a) $I = 6.9~kg~m^2$ (b) $I = 4.1~kg~m^2$

Work Step by Step

(a) We can find the moment of inertia of the door when the axis is on its hinges. Let $L$ be the width of the door. $I = \frac{1}{3}ML^2$ $I = \frac{1}{3}(25~kg)(0.91~m)^2$ $I = 6.9~kg~m^2$ (b) The moment of inertia about an axis through the midpoint of the door's width is $I_{cm} = \frac{1}{12}ML^2$. If the vertical axis is 15 cm from one edge of the door, we can find the distance $x$ from the midpoint to to the axis of rotation. $x = \frac{0.91~m}{2}-0.15~m$ $x = 0.305~m$ We can use the parallel axis theorem to find the moment of inertia about the axis that is 15 cm from the edge of the door. $I = I_{cm}+Mx^2$ $I = \frac{1}{12}ML^2+Mx^2$ $I = \frac{1}{12}(25~kg)(0.91~m)^2+(25~kg)(0.305~m)^2$ $I = 4.1~kg~m^2$
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