What is the Moment of Inertia of a Square System with Aligned Masses?

[veronicak5678]

Homework Statement

Four 3 kg masses are aligned in a square, 8 meters on a side.
a- find the moment of inertia at the center, along an axis running perpendicular to the paper.
b- Find the moment of inertia along a line that passes through the center in the plane of the paper.
c- Find the moment of inertia of the system at mass 'c' ( mass in the bottom right corner) along an axis running perpendicular to the page.

Homework Equations

I = summation mr^2

The Attempt at a Solution

a- 4 ( 3 kg)(8m)^2 /2 = 384 kg m^2

b- 3 kg ( 4 ( 8.00 m ^2/2) = 48 kg m^2

c- 3kg ( 4 ( 8*8* 11.3 m ) ^2 = 6276218 kg m^2? I am obviously not setting this up correctly...

 

[Office_Shredder]

You have to SUM each individual moment of inertia, not just multiply the radii together. So it will be

3*0² + 3*8² + 3*8² + 3*11.3²

Incidentally, it doesn't look you did (b) right either

 

[veronicak5678]

So the third question should come to 767 kg m^2.

On the second, what will the radii be? I don't understand a moment of inertia along a line rather than a point.

 

[Office_Shredder]

It's the same concept... in parts (a) and (c) you weren't measuring to 'a point', you were measuring to an axis. It just so happens that the shortest distance from a point to an axis perpendicular through plane the points are on happens to be where the axis intersects the plane (which is where the point comes from). Draw the square, draw the axis it describes, and find what the shortest distance from any vertex to the axis is

 

[veronicak5678]

OK... Sorry I'm not getting this. We didn't go over it and he just threw it at us in our homework set.

So for part b the shortest distance will always be 4 meters, so I sum up the 4 moments as

3kg * (4 m)^2 * 4 = 192 kg m^2?

 

[Office_Shredder]

That looks right to me.

 

[veronicak5678]

Cool. Thanks for your help!