# Rotational Inertia, Need Help With Integrals

Rotational Inertia, Need Help With Integrals!!!

## Homework Statement

In the figure below, a small disk of radius r = 2.00 cm has been glued to the edge of a larger disk of radius R = 4.00 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.50 103 kg/m3 and a uniform thickness of 4.60 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O?

http://www.webassign.net/halliday8e/art/images/halliday8019c10/image_n/nfg051.gif [Broken]

## Homework Equations

I=$$\frac{MR\stackrel{2}{}}{2}$$

I=$$\int$$r$$\stackrel{2}{}$$dm

$$\rho$$= $$\frac{m}{v}$$

com= $$\frac{m1x1 + m2x2 + m3x3...m(n)x(n)}{M}$$

I=Icom + Mh$$\stackrel{2}{}$$

## The Attempt at a Solution

My initial attempt basically included the following steps:

m(circle1)= $$\pi$$r1$$\stackrel{2}{}$$ * width [.0046m] * density [1.5*10$$\stackrel{3}{}$$kg/m$$\stackrel{3}{}$$]

m(circle2)= $$\pi$$r2$$\stackrel{2}{}$$ * width [.0046m] * density [1.5*10$$\stackrel{3}{}$$kg/m$$\stackrel{3}{}$$]

=> I(com)= $$\frac{1}{2}$$MR$$\stackrel{2}{}$$ Where R is the radius' of both circles added together

This clearly was the wrong approach and I'm fairly certain that in order to get the correct answer for this, i do not know how to take the integral (im only somewhat familiar with the process).

That integral is supposed to give me the I (com) and then im fairly certain all i have to do is find Mh^2 and add the two together to get the I (sys).

Would anybody mind helping me with this process, by explaining how to take a simple integral (preferably using these exact same terms), and what exactly this is doing for the calculations. This kind of assistance would be much appreciated!

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