Answer to Question #328909 in Physics for Liam

Question #328909

A 9 kg solid cylinder has a radius 0.10m and thickness 5 cm. the axis of rotation is

located at the center of the cylinder. Compute for the moment of inertia.

Expert's answer

"J=\\frac{M}{2}(R_1^2+R_2^2)"

"J=\\rm\\frac{9\\: kg}{2}((0.1\\: m)^2+(0.05 \\: m)^2)=0.056\\: kg\\cdot m^2"

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