Find Moment of Inertia of Horizontal Rod - Help with Integral Setup

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SUMMARY

The discussion focuses on calculating the moment of inertia (I) for a horizontal rod with radius a and length L, using cylindrical coordinates. The user initially set up the integral incorrectly, leading to confusion regarding the dimensions and the expected result. The correct formula for the moment of inertia of a thin rod is I = m((L^2/12) + (a^2/4)), which accounts for both the length and the radius of the rod. The user sought clarification on the integral setup and dimensional analysis to arrive at the correct answer.

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fredrick08
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Homework Statement


find I of a horizontal rod, with radius a and length L, with axis through the centre in the z direction

The Attempt at a Solution


ok can someone please help me set up the integral for this?

i got rho=m/v=m/2pi*a^2*L, I=int(rho*r^2)dV, i used r=rho for cylindrical=L/2

I=(m/2pi)int from -a to a, 0 to 2pi, 0 to L/2(rho^2/a^2*L)drho dtheta dz, that gave me (mL^2)/12, which apparently is the answer for a thin rod, of infinite thinness... according to wiki... but my answer is supposed to be m((L^2/12)+(a^2/4))... can anyone help me understand this better... because I am getting confused with my dimensions.
 
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sorry, what i did was wrong in the first place...
 

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