Question

Four identical particles of mass 0.514 kg each are placed at the vertices of a 4.70 m x 4.70 m square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square (c) lies in the plane of the square and passes through two diagonally opposite particles?

Answer 1

In order to solve the problem it is necessary to take into account the concepts related to the moment of inertia, and the center of mass of the object.

Our values are given by:

m = 0.514kg

I = 4.7m (each side)

A) For the case when the axis passes through the midpoints of opposite sidesand lies in the plane of the square. The formula is given by,

$I = 4m (\frac{l}{2})^2\\I = 4*(0.514)(\frac{4.7}{2})^2\\I = 11.35kgm^2$

B) For the case when the axis passes through the midpoint of one of the sides and is perpendicular to the plane of the square. The formula is given by,

$I = 2m(\frac{l}{2})^2+2m(\frac{l}{2}^2+l^2)\\I = 2*(0.514)*(\frac{4.7}{2})^2+2*(0.514)*((\frac{4.7}{2})^2+4.7^2)\\I = 34.06Kgm^2$

C) For the case when the axis lies in the plane of square f passes through two diagonally opposite particles,

$I = \frac{1}{2}*l^2$

$I = \frac{1}{2}*(4.7)^2$

$I = 11.045kgm^2$