Question:
Three point masses 3Kg, 2kg, 1kg are placed at A, B, C of a triangle. The distance AB = 0.3m, AC = 0.5m and BC = 0.4m. the M.I of the system about an axis through A and normal to the plane of triangle is? (Ans : 0.43km2)
Answer:
The diagram, indicates the 3 points A, B, C with masses of 3 Kg, 2Kg and 1 Kg at points A, B and C.
AB = 0.3 m, AC = 0.5 m, BC = 0.4 m
We need to calculate the moment of inertia of the system about the axis passing through A and perpendicular to the plane of the triangle. This is indicated by the dotted diagram, passing through A.
Moment of inertia of the system I = I1 + I2 + I3 = = m1r12 + m2r22 + m3r32
From the diagram, m1 = 3Kg, m2= 2Kg, m3 = 1Kg, r1 = 0 (As the MI about the point A is taken) , r2= 0.3 m (AB), r3 = 0.5 m (AC)
Hence, I = 3 (0)(0) + 2(0.3)(0.3) + 1(0.5)(0.5) = 0.18 + 0.25 = 0.43 kgm2 .
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