Mass Moment of Inertia - Thin - Cylinder

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SUMMARY

The discussion focuses on calculating the mass moment of inertia for a thin cylinder, specifically using the integral $$ I = 4 w \rho \int_0^R x^2 \sqrt{R^2-x^2} dx $$ to derive the result. Participants confirm that the mass moment of inertia is equivalent to $$ I = \frac{1}{4}MR^2 $$, with one user suggesting a substitution of x for Rcos to simplify the integral. The final value obtained is $$ \frac{\pi R^2}{16} $$ after applying the limits of integration.

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erobz
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Homework Statement
Apparently the Mass moment of inertia of a cylinder about the axis shown is ## I = \frac{1}{4}MR^2##
Relevant Equations
## \int_V \rho r^2 dm ##
1743632039598.png


The integral is a bit nasty.

$$ I = 4 w \rho \int_0^R x^2 \sqrt{R^2-x^2} dx $$

Just inquiring if I try it, I'm not chasing an approximation, or verify someone gets/has gotten the result.
 
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Never mind- I figured out how to confirm it with symbolic calculator. It does appear to be equivalent.

1743634759794.png
 
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erobz said:
Homework Statement: Apparently the Mass moment of inertia of a cylinder about the axis shown is ## I = \frac{1}{4}MR^2##
Relevant Equations: ## \int_V \rho r^2 dm ##

View attachment 359392

The integral is a bit nasty.

$$ I = 4 w \rho \int_0^R x^2 \sqrt{R^2-x^2} dx $$

Just inquiring if I try it, I'm not chasing an approximation, or verify someone gets/has gotten the result.

Well you can substitute x for Rcos$ and it would become way easier , i got the final value after putting in the limits (πR^2)/16 . Thanks
 

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