How Do You Calculate Moment of Inertia for Connected Masses?

[Jaklynn429]

Homework Statement

http://i241.photobucket.com/albums/ff4/alg5045/p13-17.gif
The four masses shown in the figure below are connected by massless, rigid rods.
(a) Find the coordinates of the center of mass if MA = 130 g and MB = MC = MD = 260 g.
(b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.

Homework Equations

I=mA(rA^2)+mB(rB^2)...

The Attempt at a Solution

I got part A as being .057 m for both x and y.
now wouldn't i use:
(.130*0)+(.260*.05^2)+(.260*.0707^2)+(.260*.05^2)
?? Its telling me my answer is wrong?!

 

[alphysicist]

Hi Jaklynn429,

Homework Statement

http://i241.photobucket.com/albums/ff4/alg5045/p13-17.gif
The four masses shown in the figure below are connected by massless, rigid rods.
(a) Find the coordinates of the center of mass if MA = 130 g and MB = MC = MD = 260 g.
(b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.

Homework Equations

I=mA(rA^2)+mB(rB^2)...

The Attempt at a Solution

I got part A as being .057 m for both x and y.
now wouldn't i use:
(.130*0)+(.260*.05^2)+(.260*.0707^2)+(.260*.05^2)

In the formula, r is the distance from each particle to the axis. The first term here is correct, because the 0.130kg particle is 0m from the axis. The other three terms seem to be saying that there are two 0.260kg particles that are 5cm from the axis, and one 0.26kg particle is is 7.07cm from the axis. Do you see what these last three terms need to be?