Question

A solid circular disk has a mass of 1.2 kg and a radius of 0.19 m. Each of three identical thin rods has a mass of 0.13 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to form a three-legged stool (see the drawing). Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center. (Hint: When considering the moment of inertia of each rod, note that all of the mass of each rod is located at the same perpendicular distance from the axis.)

Answer 1

The moment of inertia of the stool is the sum of the individual moments of inertia of its parts

a circular disk of radius R has a moment of inertia of:

I (disk)= 1/2 M (disk) R^2

Each thin rod is attached perpendicular to the disk at its outer edge. Therefore, each particle in a rod is located at a perpendicular distance from the axis that is equal to the radius of the disk. This means that each of the rods has a moment of inertia of

I (rod)= M (rod) R^2

Remembering that the stool has three legs, we find that the its moment of inertia is:

I (stool)= 0.5*(1.2) (0.19 ^2) + 3(0.13)(0.19^2)

= 0.035739 kg-m^2