What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpendicular) to and through the center of a face of the cube? Watch your units. *Note that moment of inertia is a purely geometric property of a rigid body, and in lab we will call this the static moment of inertia. We will also experimentally determine a dynamic moment of inertia which will be measured from the rigid body's angular acceleration.

The moment of inertia I is

I = 2.205x10^-4 kg/m^2

Explanation:

Given mass m = 0.5 kg

And side lenght = 0.03 m

Moment of inertia I = mass x radius of rotation squared

I = mr^2

In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.

Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m

Therefore,

I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow