Calculating Moment of Inertia of Particles & Cubes

[joemama69]

Homework Statement

a) Four point particles, each of mass m, are arranged in a square and attached by massless rods that run along the sides of the square, and along the diagonals. The length of each side of the square is a. The construction is then pivoted about an axis perpendicular to the square, and through its center. Calculate moment of inertia Ia about this axis.

b) pose a sphere of radius a/4 is cut out of a cube. The center of the excised sphere is at the center of the cube. What is the moment of inertia Id of the resulting object, pivoted about an axis perpendicular to one of the sides and through the center of the cube?

Homework Equations

The Attempt at a Solution

part a

I = $$\sum$$mr² where r = $$\sqrt{2a^2}$$

I = 4(mr²) = 8ma²

part b

I_cube = 1/12 m(2l²)...I_sphere = 2/5 mr²

I = I_c - I_s = 1/12 m(2a²) - 2/5 mr² = 17/120 ma²

 

[ideasrule]

Do you just want confirmation, or are you unsure about what you did? Both answers are correct.

 

[joemama69]

both i suppose, thanks