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A uniform thin rod of mass M- 4.27 kg pivots about an axis through its center and perpendicular to its L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I -0.983 kg m2? Number rm H7I 7i

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Answer #1

MOI of rod about center axis = ML^2 / 12 = 4.27 * (L^2)/12

MOI of objects = MR^2 + MR^2 = > 2 * (0.593)* (L/2)^2

2 * (0.593)* (L/2)^2 + 4.27 * (L^2)/12 = 0.983

2 * (0.593)* (1/4) * (L)^2 + 4.27 * 1/12 * (L^2) = 0.983

L = 1.22m

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