Question

Four identical particles of mass 0.60 kg each are placed at the vertices of a 2.5 m ✕ 2.5 m square and held there by four massless rods, which form the sides of the square. (

a) What is the rotational inertia of this rigid body about an axis that passes through the midpoints of opposite sides and lies in the plane of the square?

(b) What is the rotational inertia of this rigid body about an axis that passes through the midpoint of one of the sides and is perpendicular to the plane of the square?

Answer 1

given,

mass of the four identical particle , m= 0.60 kg

dimension of the square = 2.5 m x 2.5 m

a) Rotational inertia when axis is passing through midpoints of opposite sides and lies in the plane of the square.

m₁ = m₂ = m₃ = m₄ = m

r₁ = r₂ = r₃ = r₄ = (a/2) = 2.5/2 = 1.25 m where a is the side of the square

I = m₁r₁² + m₂r₂² + m₃r₃² +m₄r₄²

I = 4 x m (a/2)²

I = 4 x 0.6 x 1.25²

I = 3.75 kg.m²

b) now,

m₁ = m₂ = m₃ = m₄ = m

I = m₁r₁² + m₂r₂² + m₃r₃² +m₄r₄²

I =
`0.6 * ((a)/(2))^2 +0.6 * ((a)/(2))^2 + 0.6 * (((a)/(2))^2+a^2) + 0.6 * (((a)/(2))^2+ a^2)`

I =
`0.6 * ((2.5)/(2))^2 +0.6 * ((2.5)/(2))^2 + 0.6 * (((2.5)/(2))^2+2.5^2) + 0.6 * (((2.5)/(2))^2+ 2.5^2)`

I = 11.25 kg.m²