Moment of Inertia on Semi-Hollow Cylinder

  • Thread starter trautlein
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  • #1
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Homework Statement



A hollow cylinder has mass m, and outside radius [tex]r_{2}[/tex], and an inside radius [tex]r_{1}[/tex]. Use intergration to show that the moment of inertia about its axis is give by [tex]I=\frac{1}{2}m(r^2_2+r_1^2).[/tex]

Uniform composition is assumed.

Homework Equations



[tex]I=\int r^2dm[/tex]
[tex]dm=\sigma dA[/tex]

and

[tex]A=\pi r^2[/tex]
(note, later on I constrict the lower and upper bounds of integration to account for the now entire area of the circular side, which sounds as though it is assumed here)

The Attempt at a Solution



Note: Using equations above.

[tex]I=\int r^2dm[/tex]

[tex]I=\int r^2\sigma dA[/tex]

[tex]dA=2\pi rdr[/tex]

[tex]I=\sigma 2\pi\int r^3dr[/tex]

(here I insert my upper and lower bounds)
[tex]I=\sigma 2\pi\int_{r_1}^{r_2} r^3dr[/tex]

[tex]I=\sigma 2\pi \left(\frac{r^4_2}{4}-\frac{r^4_1}{4} \right )[/tex]

[tex]I=\sigma \pi \left(\frac{r^4_2}{2}-\frac{r^4_1}{2} \right )[/tex]

[tex]I=\frac {1}{2}\sigma \pi \left(r^4_2-r^4_1 \right )[/tex]

[tex]\sigma=\frac{m}{A}[/tex]

and since

[tex]A=\pi r^2_1-\pi r^2_2[/tex]
then

[tex]I=\frac{m}{2(\pi r^2_1-\pi r^2_2)} \pi \left(r^4_2-r^4_1 \right )[/tex]


[tex]I=\frac{\left(r^4_2-r^4_1 \right )m}{2 (r^2_1-r^2_2)}[/tex]


This is as far as I venture into the problem before I realize that I've done something wrong if I'm supposed to get the answer they assume at the beginning of the problem. Where did I go wrong?

Thank you very much for the help, it's greatly appreciated.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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That is correct so far. Remember a2-b2=(a+b)(a-b). So if you factorize the numerator what do you get?
 
  • #3
15
1
Given:

[tex]I=\frac{\left(r^4_2-r^4_1 \right )m}{2 (r^2_1-r^2_2)}[/tex]

Then:

[tex]I=\frac{1}{2}m(r^2_1+r^2_2)[/tex]

Great! Thank you so much! I actually forgot how those exponents would go into each other, that was my issue. Thanks for leading me toward the right answer there!
 

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