Question

What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) through the edge of the disk?

Answer 1

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

$I=\frac{1}{2}MR^{2}\\ I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\ I=0.01 kg m^{2}$

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

$I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\ I=0.03kgm^{2}$