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An object consists of a rod (of length 3.0 m and negligible moment of inertia) to which four small 2.0-kg masses are attached, one at each end and one at each point on the rod 1.0 m from each end. (The masses are one meter apart.) The moment of inertia of this object about an axis perpendicular to the rod and through one of the inner masses:

Respuesta :

Answer:

I = 12 Kg.m²

Explanation:

given,

mass of the small masses = 2 Kg

distance between the masses = 1 m

moment of inertia of object through one of the inner masses.

moment of inertia

taking second block from the left as the reference point

so,

I = m r₁² + m r₂² + m r₃² + m r₄²

r₁ = -1 m , r₂ = 0 m , r₃ = 1 m , r₄ = 2 m

I = m( r₁² +  r₂² +  r₃² +  r₄² )

I = 2 x ( (-1)² +  (0)² +  (1)² +  (2)² )

I = 2 x 6

I = 12 Kg.m²

Hence, the moment of inertia of the object is equal to 12 Kg.m²

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