Rotational inertia of particle system

[snoggerT]

The masses and coordinates of four particles are indicated in the following table.
40 g x = 1.0 cm y = -4.0 cm
25 g x = 2.0 cm y = 4.0 cm
30 g x = -3.0 cm y = 5.0 cm
35 g x = 4.0 cm y = 4.0 cm

(a) What is the rotational inertia of this collection about the x axis?
g·cm2

(b) What is the rotational inertia of this collection about the y axis?
g·cm2

The Attempt at a Solution

I honestly don't really know where to start with this problem. I know it's not a continuous body, so I wouldn't solve it through integration, but there's not really anything in the chapter that discusses multiple points in the xy plane.

 

[Chi Meson]

Rotation inertia is simply the resistance to rotational (or "angular" acceleration). Every object will have a rotational inertia with respect to any point in the universe. The simple equation for rotational inertia is I=mr^2.

Careful: when x=1 cm, that means it is 1 cm from the y-axis.

For a group of individual points, you just sum all the individual inertias. The next step will be integration for solid objects.

 

[snoggerT]

Rotation inertia is simply the resistance to rotational (or "angular" acceleration). Every object will have a rotational inertia with respect to any point in the universe. The simple equation for rotational inertia is I=mr^2.

Careful: when x=1 cm, that means it is 1 cm from the y-axis.

For a group of individual points, you just sum all the individual inertias. The next step will be integration for solid objects.

- That's what I was thinking about doing, but my problem was that I wasn't sure of where to take my origin at so I could get my r. Would I just take the point (0,0) as the origin and get the r from that?

 

[Chi Meson]

No, it's simpler. The system of points will be rotated about the x-axis first, so the "r's" are the straight line distances to the axes.

If you were to rotate the system about the z-axis, then you would find the distance to the origin