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3 point particles of masses 2kg each placed on 3 corners pf an equilateral triangle of perimeter 12m . This triangular structure suppose to have niglible mass. Find moment of inertia of this system about an axis passing through one of its side.plz help me guys.... |
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Answer» perimeter = 12 side = 12/3 = 4 let axis pass through BD.its perpendicular BISECTOR and its equilateral triangle so it will pasd through A As PARTICLE is at A also so its moment of inertia along axis is 0 BC= CD = 4/2 = 2 SO Moment of inertia = 2 M( BC)^2 = 2 (2)( 2)^2 = 16 |
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