October 31, 2007 - Moment of Inertia - Solutions

# October 31, 2007 - Moment of Inertia - Solutions - 9.63...

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9.63. IDENTIFY: Apply Eq.(9.20). SET UP: For this case, . dm dx EXECUTE: (a) 22 0 0 L L xL M dm x dx   (b) 44 0 0 ( ) . 2 L L M I x x dx L This is larger than the moment of inertia of a uniform rod of the same mass and length, since the mass density is greater further away from the axis than nearer the axis. (c) 2 3 4 4 2 2 2 3 2 2 00 0 ( ) ( 2 ) 2 2 3 4 12 6 L LL x x x L M I L x xdx L x Lx x dx L L L    . This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end. EVALUATE: For a uniform rod with an axis at one end, 2 1 3 I ML . The result in (b) is larger than this and the result in (a) is smaller than this. 9.59. IDENTIFY: Use the equations in Table 9.2. I for the rod is the sum of I for each segment. The parallel-axis theorem says 2 p cm I I Md  . SET UP: The bent rod and axes a and b are shown in Figure 9.59. Each segment has length /2 L and mass M .

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## This note was uploaded on 04/03/2008 for the course PHYSICS 140 taught by Professor Evrard during the Fall '07 term at University of Michigan.

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October 31, 2007 - Moment of Inertia - Solutions - 9.63...

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