A 6.00 kg mass is located at (2.00 m, 2.00 m, 2.00 m) and a 5.00 kg mass is located at (−1.0 m, 3.00 m, −2.00 m). The rotational inertia of this system of masses about the Y-axis, perpendicular to the Z-X plane, is

The rotational inertia, also known as the moment of inertia, of a system of point masses about an axis is given by the sum of the products of each mass and the square of its perpendicular distance from the axis.

The formula for the moment of inertia (I) is:

I = Σ mi * ri^2

where mi is the mass of the ith particle and ri is the perpendicular distance of the ith particle from the axis of rotation.

The y-axis is perpendicular to the x-z plane, so the distance of each mass from the y-axis is given by the Pythagorean theorem in the x-z plane:

ri = sqrt(xi^2 + zi^2)

For the 6.00 kg mass:

r1 = sqrt((2.00 m)^2 + (2.00 m)^2) = sqrt(8.00 m^2) = 2.83 m

For the 5.00 kg mass:

r2 = sqrt((-1.00 m)^2 + (-2.00 m)^2) = sqrt(5.00 m^2) = 2.24 m

Now we can calculate the moment of inertia:

I = (6.00 kg * (2.83 m)^2) + (5.00 kg * (2.24 m)^2)

I = 47.88 kgm^2 + 25.12 kgm^2

I = 73.00 kg*m^2

So, the rotational inertia of this system of masses about the Y-axis, perpendicular to the Z-X plane, is 73.00 kg*m^2.