Cylinder Moment of Inertia

Moment of inertia of a cylinder: I = (1/2)·M·R²

Spin a solid cylinder about its central (longitudinal) axis and you get I = (1/2)·M·R². A thin hollow cylinder (a shell) gives I = M·R². For a thick hollow cylinder with inner radius R₁ and outer R₂, it becomes I = (1/2)·M·(R₁² + R₂²). Switch to a transverse axis through the center, perpendicular to the length L, and you have I = (1/4)·M·R² + (1/12)·M·L². Think of moment of inertia as mass for rotation, with rotational kinetic energy given by E_rot = (1/2)·I·ω². Say you have a flywheel of M = 50 kg and R = 0.3 m: I = (1/2)·50·0.09 = 2.25 kg·m², and at ω = 100 rad/s it holds E_rot = (1/2)·2.25·10000 = 11.25 kJ.

Applications

Engine flywheels lean on it to store rotational energy and smooth out power delivery. So do industrial centrifuges, gears, driveshafts and cardan shafts, rotating machinery on the factory floor, and gyroscope design. Wherever someone has to know how much torque it takes to spin up a cylindrical part, or to brake it, this is the number they want.

FAQ

Why is the hollow cylinder's I larger than the solid one's at the same M and R? Its mass sits farther out from the rotation axis, and since I grows with r², mass that is farther away resists angular acceleration more.

Does the length L matter for the central axis? No. When the rotation is about the longitudinal axis, L never enters the formula. It only comes into play for the transverse axis.

How do I find I about an off-center axis? Reach for the parallel-axis theorem, I = I_cm + M·d², where d is the distance from the center of mass out to the new axis.