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Perpendicular-Axis Theorem. Consider a rigid body that is

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 95P Chapter 9

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 95P

Perpendicular-Axis Theorem?. Consider a rigid body that is a thin, plane sheet of arbitrary shape. Take the body to lie in the ?xy?-plane and let the origin ?O? of coordinates be located at any point within or outside the body. Let ?Ix? and ?Iy? be the moments of inertia about the ?x?- and ?y?-axes, and let ?IO? be the moment of inertia about an axis through ?O? perpendicular to the plane. (a) By considering mass elements ?mi?, with coordinates (?xi, yi?), show that ?Ix?+ ?Iy? = ?IO?. This is called the perpendicular-axis theorem. Note that point ?O? does not have to be the center of mass. (b) For a thin washer with mass ?M? and with inner and outer radii ?R?1 and R?2, use the perpendicular-axis theorem to find the moment of inertia about an axis that is in the plane of the washer and that passes through its center. You may use the information in Table 9.2. (c) Use the perpendicular-axis theorem to show that for a thin, square sheet with mass ?M? and side ?L?, the moment of inertia about ?any axis in the plane of the sheet that passes through the center of the sheet is . You may use the information in Table 9.2.

Step-by-Step Solution:

Solution 95P Introduction We have to prove the perpendicular axis theorem using an arbitrary thin sheet of object and then we have to use this theorem to calculate the moment of inertia of given object about given axis. Step 1 The above figure represents an arbitrary thin shit. Let us consider an elementary surface area dA i dx dy in ai position (x ,y ). Letius ilso consider that the density of the object is given by (x iy i. So the mass of the elementary area is dm =i(x ,y idx iy i i So the moment of inertia of the elementary about x-axis is given by dI = x dm = x (x ,y )dx dy xi i i i i i Hence the total moment of inertia of the object with respect to the x-axis is given by 2 Ix= dI= xx(x y )dx dyi i i i Similarly the moment of inertia about the y-axis is given by Iy= y (x ,y )di dyi i i Now the moment of inertia of the object about the z-axis is given by 2 I = r (x ,y )dx dy == x + y 2 (x ,y )dx dy = x (x ,y )dx dy + y (x ,y )dx dy = I + I z i i i i i ( i i ) i i i i i i i i i i i i x y Hence the perpendicular axis theorem is proved.

Step 2 of 3

Chapter 9, Problem 95P is Solved
Step 3 of 3

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Perpendicular-Axis Theorem. Consider a rigid body that is