Respuesta :

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

[tex]I=\int\limits dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2[/tex]

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

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