### Question Description

Four objects are held in position at the corners of a rectangle by light rods as shown in the figure below. (The mass values are given in the table.)

m_{1} (kg) | m_{2} (kg) | m_{3} (kg) | m_{4} (kg) |

2.70 | 2.10 | 4.50 | 2.30 |

*x*-axis.

kg · m

^{2}

(b) Find the moment of inertia of the system about the

*y*-axis.

kg · m

^{2}

(c) Find the moment of inertia of the system about an axis through

*O*and perpendicular to the page.

## Final Answer

Thank you for the opportunity to help you with your question!

moment of inertia =Sum[ mass x distance^2]

where the distance is the distance from the chosen rotation axis

a) here, moment of inertia =

(2.7+2.1+4.5+2.3)3^2 = 104.4kgm^2

because each mass is 3 m from the x axis

b) now we consider distances from the y axis; each mass is 2m from the y axis:

I=(2.7+2.1+4.5+2.3)2^2=46.4 kgm^2

c) the distance from the rotation axis is now found from the pythagorean theorem

distance^2=2^2+3^2=13

I=(2.7+2.1+4.5+2.3)*13=150.8 kgm^2