Moment of Inertia

Four small spheres, each of which you can regard as a point of mass (0.2)Kg are arranged in a square (0.4)m on a side and connected by extremely light rods. Find the moment of inertia of the system about an axis through the centre of the square.
Discussion:

The length of each side of a square is (0.4)m. Thus, the perpendicular bisector is at (0.2)m. The length of the bisector dropped from the point of intersection of the diagonals is (0.2)m. Therefore, the distance of the vertices from the point of intersection of diagonals is

$r=\sqrt{(0.2)^2+(0.2^2)}=0.2828m$

The moment of inertia

$I=MR^2= 4(0.2)(0.2828)$

=

$0.0640kg m^2$

Four small spheres, each of which you can regard as a point of mass (0.2)Kg are arranged in a square (0.4)m on a side and connected by extremely light rods. Find the moment of inertia of the system about an axis through the centre of the square.
Discussion:

The length of each side of a square is (0.4)m. Thus, the perpendicular bisector is at (0.2)m. The length of the bisector dropped from the point of intersection of the diagonals is (0.2)m. Therefore, the distance of the vertices from the point of intersection of diagonals is

$r=\sqrt{(0.2)^2+(0.2^2)}=0.2828m$

The moment of inertia

$I=MR^2= 4(0.2)(0.2828)$

=

$0.0640kg m^2$