University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2022
Views:

230

Pages:

2
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 180. (i) Using Equation (9.30), we have ( ) ( ) ( ) I x + I y + I z = ∫ y 2 + z 2 dm + ∫ z + x dm + ∫ x + y dm ( ) = 2∫ x 2 + y 2 + z 2 dm = 2∫ r 2dm where r is the distance from the origin O to the element of mass dm. Now assume that the given body can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m and radius a which is centered at O; it then follows that m1 = m2 mbody = msphere = m . ( ) Then ( I x + I y + I z )body = ( I x + I y + I z )sphere + ( I x + I y + I z )V 1 ( − Ix + I y + Iz )V 2 or ( I x + I y + I z )body = ( I x + I y + I z )sphere + 2∫ m r 2dm − 2∫ m r 2dm 1 2 Now, m1 = m2 and r1 ≥ r2 for all elements of mass dm in volumes 1 and 2. ∴ ∫ m r 2dm − ∫ m r 2dm ≥ 0 1 2 so that ( I x + I y + I z )body ≥ ( I x + I y + I z )sphere Q.E.D. continued COSMOS: Complete Online Solutions Manual Organization System (ii) First note from Figure 9.28 that for a sphere Ix = I y = Iz = Thus, 2 2 ma 5 ( I x + I y + I z )sphere = 65 ma2 For a solid of revolution, where the x axis is the axis of revolution, have I y = Iz Then, using the results of part i ( I x + 2I y )body ≥ 65 ma2 Iy ≥ From Problem 9.178 have or 1 Ix 2 ( 2I y − I x )body ≥ 0 Adding the last two inequalities yields ( 4I y )body ≥ 65 ma 2 or ( I y )body ≥ 103 ma 2 Q.E.D.

COSMOS Chapter 9 Solution 180

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