Moment of Inertia of Compound Disk

[capst]

Homework Statement

A compound disk of outside diameter 152 cm is made up of a uniform solid disk of radius 41.0 cm and area density 3.30 g/cm² surrounded by a concentric ring of inner radius 41.0 cm , outer radius 76.0 cm , and area density 2.10 g/cm² .

Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center (in kg*m²).

Homework Equations

Moment of inertia of solid cylinder (a thin cylinder is a disk) = I = .5mr²

The Attempt at a Solution

This object is basically one inner disk with mass m_i surrounded by an outer disk with mass m_o. Finding the moment of inertia of each of these and adding them together should give the solution.

m_i in kg = (area * density)/1000 = (pi*41² *3.3)/1000 = 17.427

m_o in kg = [(area - inner area) * density]/1000 = [(pi*71² - pi*41²) * 2.1]/1000 = 22.1671

Using the moment of inertia for solid cylinder and adding yields:
.5m_i*.41²+.5m_o*.76²= 7.87 kg*m²

The answer given is 11.5 kg*m². What am I doing wrong?

 

[haruspex]

I assume the 8.76 is a typo for .76. But you have effectively taken the outer disk's mass as being spread over a disk of radius .76. In fact, it is concentrated between .41 and .76 radius, increasing the MI.
The simplest approach would be to treat it as a solid disk radius .76 and density 2.1, plus a solid disk radius .41 and density 3.3-2.1=1.2.

 

[capst]

Oh, thanks! And yup, fixed the typo.