Find the center of mass of the object shown in the figure below?

LINK TO FIGURE:
Untitled.jpg picture by amy_parker1 - Photobucket (http://s1168.photobucket.com/albums/r497/amy_parker1/?action=view&current=Untitled.jpg)

1. Find the center of mass of the object shown in the figure below.
2. Calculate the rotational inertia of the object about the x-axis.
3. From this value, deduce the rotational inertia of the object about an axis parallel to the x-axis, and going through the center of mass.

m1=2.5kg , m2=5kg, m3=2.5kg , m4=5kg

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.