AIM: Determine, by calculations, i) the location of the mass center, o, in units of in,...

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AIM: Determine, by calculations, i) the location of the mass center, o, in units of in, ft, and m, and ii) the moment of inertia at G* about an axis perpendicular to the paper of the given beam in units of slug-in, slug ft, Ibm-ft, and kg-m The beam has 7 holes as shown and the diameter of each hole is 5/8". With the beam in the upright position (holes on top), each hole serves as a pivot. Weight of beam: 29.4 oz Length of beam: Thickness of beam: 36" (tip to tip) 1/5" Width of beam: 1" Cross-section of beam 1/5" REPORT DUE: Submit next class. Missing the deadline means your report will not be graded, therefore, it is a zero. So, do not miss the deadline! Moment of inertia of a body at its mass center, G, is frdm. body For a uniform rectangular parallelepiped of length, (, width, w, and mass, m, the moment of inertia at the center 1 of mass is m(l + w). 12 For a uniform disc of radius, R, and mass, m, the moment of inertia at the mass center for the axis perpendicular to the disc is mR2 2 For a uniform semi-circular disc of radius, r, and mass, m, the mass center 4r is located at for the axis 1 16 G 2 972 The moment of inertia at the mass center, l perpendicular to mr2 the disc is 187 (9 - 32). 36 1/2" Body AIM: Determine, by calculations, i) the location of the mass center, o, in units of in, ft, and m, and ii) the moment of inertia at G* about an axis perpendicular to the paper of the given beam in units of slug-in, slug ft, Ibm-ft, and kg-m The beam has 7 holes as shown and the diameter of each hole is 5/8". With the beam in the upright position (holes on top), each hole serves as a pivot. Weight of beam: 29.4 oz Length of beam: Thickness of beam: 36" (tip to tip) 1/5" Width of beam: 1" Cross-section of beam 1/5" REPORT DUE: Submit next class. Missing the deadline means your report will not be graded, therefore, it is a zero. So, do not miss the deadline! Moment of inertia of a body at its mass center, G, is frdm. body For a uniform rectangular parallelepiped of length, (, width, w, and mass, m, the moment of inertia at the center 1 of mass is m(l + w). 12 For a uniform disc of radius, R, and mass, m, the moment of inertia at the mass center for the axis perpendicular to the disc is mR2 2 For a uniform semi-circular disc of radius, r, and mass, m, the mass center 4r is located at for the axis 1 16 G 2 972 The moment of inertia at the mass center, l perpendicular to mr2 the disc is 187 (9 - 32). 36 1/2" Body