Question

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.

Answer 1

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

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