Three particles of mass m is situated at the ends of an equilateral triangle ABC of length L.Find the moment of inertia of about line Ax and perpendicular to AB.
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see diagram.
The position of the axis Ax is not clear. But I hope this is what is intended.
CD = perpendicular from C onto Ax = L Sin 30° = L/2
Moment of Inertia of the system of the three masses situated at A, B and C about the line Ax :
= Sum of MOI due to A , B and C respectively
= 0 + m L² + m (L/2)²
MOI due to mass at A is 0 as it is on the axis Ax.
MOI = 5 m L² /4
The position of the axis Ax is not clear. But I hope this is what is intended.
CD = perpendicular from C onto Ax = L Sin 30° = L/2
Moment of Inertia of the system of the three masses situated at A, B and C about the line Ax :
= Sum of MOI due to A , B and C respectively
= 0 + m L² + m (L/2)²
MOI due to mass at A is 0 as it is on the axis Ax.
MOI = 5 m L² /4
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kvnmurty:
clik on red heart thanks above pls
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