- #1

- 30

- 0

**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=[tex]\frac{MR\stackrel{2}{}}{2}[/tex]

I=[tex]\int[/tex]r[tex]\stackrel{2}{}[/tex]dm

[tex]\rho[/tex]= [tex]\frac{m}{v}[/tex]

com= [tex]\frac{m1x1 + m2x2 + m3x3...m(n)x(n)}{M}[/tex]

I=Icom + Mh[tex]\stackrel{2}{}[/tex]

## The Attempt at a Solution

My initial attempt basically included the following steps:

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

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

=> I(com)= [tex]\frac{1}{2}[/tex]MR[tex]\stackrel{2}{}[/tex] 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!

Last edited by a moderator: