Mass Moment of Inertia - Thin - Cylinder

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The discussion centers on calculating the mass moment of inertia for a thin cylinder, represented by the integral I = 4wρ ∫₀ᴿ x²√(R²-x²) dx. The integral is noted to be complex, but one participant found a substitution method using x = Rcos(θ) to simplify the calculation. This substitution led to a final value of (πR²)/16 for the moment of inertia. The original formula for the mass moment of inertia about the specified axis is given as I = (1/4)MR². The conversation highlights the challenges and solutions in deriving the moment of inertia for a cylinder.
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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