Question

A solid circular disk has a mass of 1.2 kg and a radius of 0.16m. Each of three identical thin rods has a mass of 0.16kg. 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

Answer:

0.027648 kgm²

Explanation:

M = Mass of disc = 1.2 kg

r = Radius of disc = 0.16 m

m = Mass of rod = 0.16 kg

R = Rod distance = 0.16 m

Moment of inertia of disk is given by

$I_1=\dfrac{1}{2}Mr^2\\\Rightarrow I_1=\dfrac{1}{2}1.2\times 0.16^2\\\Rightarrow I_1=0.01536\ kgm^2$

Moment of inertia of the three rods

$I_2=3mr^2\\\Rightarrow I_2=3\times 0.16\times 0.16^2\\\Rightarrow I_2=0.012288\ kgm^2$

The total moment of inertia is given by

$I=I_1+I_2=0.01536+0.012288\\\Rightarrow I=0.027648\ kgm^2$

The moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center is 0.027648 kgm²