Ask Me Help Desk - I tried this question but doesn't seem right help!

1. The center of mass $(\bar x, \bar y, \bar z)$ can be found using:

$ \bar x = \frac {\Sigma M_i \times x_i} {\Sigma M_i} \\ \bar y = \frac {\Sigma M_i \times y_i} {\Sigma M_i} \\ \bar z = \frac {\Sigma M_i \times z_i} {\Sigma M_i} $

I'll do the x coordinate for you:

$ \bar x = \frac {( 2.5KG \times 2m + 5 Kg \times 0m + 2.5 Kg \times 0 m + 5 Kg \times 2m)} {2.5 Kg + 5 Kg + 2.5 Kg + 5 Kg)} = \frac {15 Kg-m}{15 Kg} = 1 m. $

Repeat this process for the y and z coordinates.

2. The rotational intertia of each mass is equal to the mass times the square of its distance from the center of rotation. So taking the center of rotation as the x-axis, for masses 2 and 3 the distance is zero, and for masses 1 and for the distance is 2m:

$ I = 2.5 Kg \times (2m)^2 + 5 Kg \times (0m)^2 + 2.5 Kg \times (0m)^2 + 5 Kg \times (2m)^2 $

3. The moment of inertia about a line distance d from the center of gravity (cg) is equal to:

$ I_d = I_{cg} + Md^2 $

where M is the total mass of the system. You already have $I_d$ from (2), so you can find $I_{cg}$ .

Post back your results if you want confirmation.