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

The moment of inertia of the disk for rotation about the center is 0.01 kg•m^2, and for rotation about the edge is 0.02 kg•m^2.

Step-by-step explanation:

The moment of inertia of a solid disk about an axis through its center is given by the formula 1/2 * MR^2. In this case, the mass of the disk is 2.0 kg and the radius is 10 cm (half of the diameter). So, substituting the values into the formula, the moment of inertia about the center axis is:

1/2 * 2.0 kg * (0.1 m)^2 = 0.01 kg•m^2.

For rotation about an axis through the edge of the disk, the moment of inertia is:

MR^2 = 2.0 kg * (0.1 m)^2 = 0.02 kg•m^2.

Answer 2

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Step-by-step 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=(1)/(2)MR^(2)\\ I=(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)`