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

2024
Views:

368

Pages:

2
Author:

omiliar74be

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 30. From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is ( J C )cir = π 2 r4 The area of the circle is Acir = π r 2 So that A2  J C ( A)  = cir 2π Two methods of solution will be presented. However, both methods depend upon the observation that as a given element of area dA is moved closer to some point C, The value of J C will be decreased ( J C = ∫ r 2dA ; as r decreases, so must J C ). Solution 1 Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its area is equal to A. To minimize the value of ( J C ) A , the area would have to be distributed as closely as possible about C. This is accomplished by winding the strip into a tightly wound roll with C as its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus increasing the value of ( J C ) A . (The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll is circular, with the centroid of its area at C, it follows that ( JC ) A ≥ A2 Q.E.D. 2π where the equality applies when the original area is circular. continued COSMOS: Complete Online Solutions Manual Organization System Solution 2 Consider an area A, with its centroid at point C, and a circular area of area A, with its center (and centroid) at point C. Without loss of generality, assume that A1 = A2 A3 = A4 It then follows that ( J C ) A = ( J C )cir +  J C ( A1 ) − J C ( A2 ) + J C ( A3 ) − J C ( A4 )  Now observe that J C ( A1 ) − J C ( A2 ) ≥ 0 J C ( A3 ) − J C ( A4 ) ≥ 0 since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase. ∴ ( J C ) A ≥ ( J C )cir or ( JC ) A ≥ A2 Q.E.D. 2π