Mass Moment of Inertia - Thin - Cylinder

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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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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