Calculating I for a Circular Hoop with Tangential Axis

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SUMMARY

The value of the moment of inertia (I) for a circular hoop of mass M and radius R about an axis tangent to the hoop and lying in the hoop's plane is calculated using the integral I = 2θ ∫ from -R to R of z²√(R² - z²)dz. The correct evaluation leads to I = (1/4)MR². However, a participant in the discussion argues that the integral must account for the circular nature of the hoop, suggesting that the mass density should be integrated around the circle from 0 to 2π, leading to a larger moment of inertia than the initial calculation.

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Homework Statement

What is the value of I for a circular hoop of mass M and Radius R about an axis that is tangent to the hoop and lies in the hoops plane?



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The Attempt at a Solution

I=2 theta intergral from -R to R, z^2 square root of (R^2 - z^2)dz = theta pi R^4/4 = 1/4 MR^2
 
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I'm picturing a circle with a vertical line one left side, just touching the circle. If so, I don't think your integral is correct. I don't know what your Z stands for, but I believe you must integrate around the circle (zero to 2 pi) because the linear mass density is m/(2πR) per unit distance all the way round, whereas integrating linearly the mass density is much higher as you approach the sides of the circle. I get a much larger answer than you have. My intuition is telling me the answer should be larger than mR² because a good part of the mass is distance 2R away from the axis. My answer agrees; yours does not.
 

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