195_Problem CHAPTER 9

195_Problem CHAPTER 9 - To the Instructor: The following...

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To the Instructor: The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.144–9.147. From Figure 9.28 Cylinder: () 2 cyl cyl 1 2 x I ma = () () ( ) 22 cyl cyl cyl 1 3 12 yz I Im a L == + Symmetry and the definition of the mass moment of inertia 2 I rdm =∫ imply semicylinder cylinder 1 2 II = 2 cyl sc 11 x I  ∴=   and cyl sc sc 3 21 2 I a L + However, sc cyl 1 2 mm =
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