Calculating I for a Circular Hoop with Tangential Axis

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The discussion focuses on calculating the moment of inertia (I) for a circular hoop with mass M and radius R about a tangential axis in the hoop's plane. One participant suggests an integral approach that leads to a result of I = 1/4 MR², while another challenges this calculation, arguing that the integration should account for the entire circular path and the distribution of mass. They emphasize that the mass further from the axis contributes significantly, implying that the moment of inertia should be larger than mR². The disagreement highlights the importance of correctly setting up the integral for accurate results. The conversation underscores the complexity of calculating moments of inertia for non-central axes.
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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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