Moment of inertia...?

Question

Can you please help me with the following problem?
I have four masses connected by massless rigid rods so that they form the square in xy plane (each rod has length 10cm). I need to to find the moment of inertia about an axis that passes through the mass positioned at the lower left vertex of the square and perpendicular to the page. The mass at that point is 100g and all other masses are 200g.
I have a feeling that this should be very simple but all I've got is the center of mass - (5.7cm, 5.7cm) using the coordinate system with the origin at that lower left corner, and (0,10), (10, 0) and (10, 10) as the coordinates of other masses.I have the final solution but I really need the procedure how to get there.
Thanks for any help!

Answer 1

I = ∑mr^2 = 100*200 + 100*200 + 200*200 + 0*100
I = 80,000 gm-cm^2

Answer 2

The moment of inertia of a point mass(m) at a distance (r) from the Axis of rotation is (1/2)mr^2,

Just add the moments of inertia of the various point Masses.

The 100g is at 0cm its MI = 0.
The 200g at the opposite vertex is at r^2 = 200, so its MI = (1/2)200*200=20000
The other two are at 10 cm, each has an MI=(1/2)200*100 = 10000

Total MI of all 4 = 0 + 20000+ 2*10000 =40000 gm-cm^2

Answer 3

You seem to have issues returned to front. something heavy includes greater kinetic skill than something lighter shifting on the comparable velocity. the bigger the radius the quicker the load is shifting for a given rpm. So a large heavy flywheel is greater effective than a small mild one.