標題:
Applied Maths - Mechanics 53
發問:
Three rectangular areas, 0.6m by 4cm, 0.9m by 4cm, and 0.3m by 4cm, are fitted together to form an I figure, the longest and shortest areas forming the cross-pieces. Find the moment of inertia of the figure about the outer edge of the shortest area. 更新: How about the graph ?
最佳解答:
Suppose that the mass of the whole I figure is M, then, for easy notation, we denote: Piece A: 0.3 m by 4 cm (of mass M/6) Piece B: 0.6 m by 4 cm (of mass M/3) Piece C: 0.9 m by 4 cm (of mass M/2) Hence the axis for taking moment of inertia is the outer edge of piece A. First of all, pieces A and C are placed with their lengths parallel to the axis and piece B is placed with its width parallel to the axis. Then, their respective moments of inertia, taken along the line parallel to the axis and through their respective centres, are given by: IA = (1/3)(M/6)(0.02)2 IB = (1/3)(M/3)(0.3)2 IC = (1/3)(M/2)(0.02)2 Then by parallel axis theorem, we can obtain their respective moment of inertia with respect to the axis as follows: Piece A: 0.02 m from the reference line of IA, giving IA' = (1/3)(M/6)(0.02)2 + (M/6)(0.02)2 Piece B: 0.34 m from the reference line of IB, giving IB' = (1/3)(M/3)(0.3)2 + (M/3)(0.34)2 Piece C: 0.66 m from the reference line of IC, giving IC' = (1/3)(M/2)(0.02)2 + (M/2)(0.66)2 Finally, the total moment of inertia is: IA' + IB' + IC' = 0.266M 2009-08-21 08:55:05 補充： Will append ASAP. Sorry for the inconvenience caused. 2009-08-21 09:49:38 補充： Pls. refer to: http://i388.photobucket.com/albums/oo325/loyitak1990/Aug09/Crazymech15.jpg
其他解答:
Applied Maths - Mechanics 53
發問:
Three rectangular areas, 0.6m by 4cm, 0.9m by 4cm, and 0.3m by 4cm, are fitted together to form an I figure, the longest and shortest areas forming the cross-pieces. Find the moment of inertia of the figure about the outer edge of the shortest area. 更新: How about the graph ?
最佳解答:
Suppose that the mass of the whole I figure is M, then, for easy notation, we denote: Piece A: 0.3 m by 4 cm (of mass M/6) Piece B: 0.6 m by 4 cm (of mass M/3) Piece C: 0.9 m by 4 cm (of mass M/2) Hence the axis for taking moment of inertia is the outer edge of piece A. First of all, pieces A and C are placed with their lengths parallel to the axis and piece B is placed with its width parallel to the axis. Then, their respective moments of inertia, taken along the line parallel to the axis and through their respective centres, are given by: IA = (1/3)(M/6)(0.02)2 IB = (1/3)(M/3)(0.3)2 IC = (1/3)(M/2)(0.02)2 Then by parallel axis theorem, we can obtain their respective moment of inertia with respect to the axis as follows: Piece A: 0.02 m from the reference line of IA, giving IA' = (1/3)(M/6)(0.02)2 + (M/6)(0.02)2 Piece B: 0.34 m from the reference line of IB, giving IB' = (1/3)(M/3)(0.3)2 + (M/3)(0.34)2 Piece C: 0.66 m from the reference line of IC, giving IC' = (1/3)(M/2)(0.02)2 + (M/2)(0.66)2 Finally, the total moment of inertia is: IA' + IB' + IC' = 0.266M 2009-08-21 08:55:05 補充： Will append ASAP. Sorry for the inconvenience caused. 2009-08-21 09:49:38 補充： Pls. refer to: http://i388.photobucket.com/albums/oo325/loyitak1990/Aug09/Crazymech15.jpg
其他解答:
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