Rotational Inertia, Need Help With Integrals

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SUMMARY

The discussion focuses on calculating the rotational inertia of a two-disk assembly consisting of a small disk (radius r = 2.00 cm) and a larger disk (radius R = 4.00 cm), both with a uniform density of 1.50 x 103 kg/m3 and thickness of 4.60 mm. The correct approach involves using the parallel axis theorem to find the rotational inertia of the smaller disk about point O, the center of the larger disk. The total rotational inertia is obtained by summing the individual inertias of both disks. The user seeks assistance in understanding the integration process necessary for these calculations.

PREREQUISITES
  • Understanding of rotational inertia and its formulas
  • Familiarity with the parallel axis theorem
  • Basic knowledge of calculus, specifically integration
  • Concept of mass density and volume calculations
NEXT STEPS
  • Study the application of the parallel axis theorem in rotational dynamics
  • Learn how to perform integration for calculating mass distributions
  • Explore the derivation of rotational inertia formulas for various shapes
  • Review examples of composite bodies and their rotational inertia calculations
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This discussion is beneficial for physics students, educators, and anyone involved in mechanical engineering or dynamics, particularly those focusing on rotational motion and inertia calculations.

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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

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 I am 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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No integration is needed. Find the rotational inertia of each disk about point O, then add them for the total rotational inertia of the two-disk assembly. Hint: To find the rotational inertia of the smaller disk about point O, use the parallel axis theorem.
 


Got it, Thank you!
 

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