Point masses `1, 2, 3 "and" 4 "kg` are lying at the point `(0, 0, 0), (2, 0, 0), (0, 3, 0)` and `(–2, –2, 0)` respectively. The moment of inertia of this system about x-axis will be

Options

A. `43 kg- m^(2)`
B. `34 kg- m^(2)`
C. `27 kg- m^(2)`
D. `72 kg- m^(2)`

Answer

A

Explanation

`I= 3 xx (3)^(2) + 4(-2)^(2)`
`=27+ 16= 43 kg- m^(2)`