Four identical particles (mass of each = 0.40 kg) are placed at the vertices of a rectangle (2.5 m × 4.0 m) and held in those positions by four light rods which form the sides of the rectangle. What is the moment of inertia of this rigid body about an axis that passes through the mid-points of the shorter sides and is parallel to the longer sides? A) 2.2 kg⋅m² B) 2.8 kg⋅m² C) 2.5 kg⋅m² D) 3.1 kg⋅m² E) 1.6 kg⋅m²

Question

Quizplus · Accepted Answer

The moment of inertia of the rectangle about an axis that passes through the mid-points of the shorter sides and is parallel to the longer sides can be found using the parallel axis theorem. The moment of inertia of the rectangle about an axis passing through its center and perpendicular to its plane is given by:

$I_c = \frac{1}{12}M(2.5^2+4^2) = 1.167$ kg⋅m²

where $M$ is the total mass of the rectangle (which is equal to the sum of the masses of the four particles since they are identical).

If we consider an axis passing through one of the mid-points of the shorter sides and parallel to the longer sides, the distance between this axis and the center of the rectangle is $\frac{4}{2}=2$ m. Therefore, the moment of inertia of the rectangle about this axis is:

$I = I_c + Md^2 = 1.167 + 4(0.4)(2^2) = 2.5$ kg⋅m²

Therefore, the answer is (C).

Four Identical Particles (Mass of Each = 0