What is the rotational inertia of a solid iron disk of mass 45 kg, with a thickness of 3.44 cm and radius of 19.0 cm, about an axis through its center and perpendicular to it?

r = .19 meter

I = (1/2) m r^2 is moment of inertia of disk
I = (1/2)(45)(.19)^2
That is the rotational moment of inertia, I suspect what you want. If you really want the rotational inertia you must multiply that by omega, the angular velocity in radians per second.

To find the rotational inertia of a solid disk, you can use the formula:

I = (1/2) * m * r^2

Where:
I is the rotational inertia,
m is the mass of the disk, and
r is the radius of the disk.

In this case, the mass is given as 45 kg and the radius is given as 19.0 cm.

First, convert the radius to meters:
r = 19.0 cm = 0.19 m

Now, substitute the values into the formula:
I = (1/2) * 45 kg * (0.19 m)^2

Calculating the result:
I = 0.5 * 45 kg * (0.19 m)^2
I = 0.5 * 45 kg * 0.0361 m^2
I = 0.5 * 1.6255 kg * m^2
I ≈ 0.8128 kg * m^2

Therefore, the rotational inertia of the solid iron disk about an axis through its center and perpendicular to it is approximately 0.8128 kg * m^2.