Moment of Inertia: Find 0.25-kg 4-Point Mass Square Rigid Body

[endeavor]

Four 0.25-kg point masses connected by rods of negligible mass form a square rigid body with 0.10-m sides Find the moment of inertia of the body about each of the follwoing axes: (a) an axis perpendicular to the plane of the squuare and through its center, (b) an axis perpendicular to the plane and through one of hte masses, (c) an axis in the plane and along one side through two of the masses, and (d) an axis in the plane running diagonally through two fo the masses.

my answers are a. 1.7 * 10^-3, b. 2.2 * 10^-2, c. 3.3 * 10^-3, and I don't know how to solve d. However, when I checked my answers in the back of my textbook, a. is 5.0 * 10^-3, b. 1.0 * 10^-2, c. 5.0 * 10^-3.

What am I doing wrong...?

 

[assyrian_77]

Remember that you can get the moment of inertia through

$$I=\sum m_{i}r_{i}^2$$

where r is the distance from the mass to the axis. So in part a, the distances from the masses to the axis are equal and by simple geometry $$0.1/\sqrt{2}$$. The rest should be done in a similar way.

 

[endeavor]

okay, i got a,b, c, and d. thanks