Moment of Inertia (Solid Cylinder)

Moment of inertia of a cylinder

Spin a solid cylinder of mass M and radius R about its central axis and you get I = (1/2)·M·R². Push all the mass out to the rim and you have a thin hollow cylinder, where I = M·R². For a thick-walled hollow cylinder with inner radius R₁ and outer radius R₂, the value lands in between: I = (1/2)·M·(R₁² + R₂²). Plug in M = 5 kg and R = 0.2 m for the solid case and you get I = 0.5·5·0.04 = 0.1 kg·m².

Think of moment of inertia as mass for rotation. It tells you how much torque you need to spin a body up, through τ = I·α, and how much rotational energy it holds once it is turning at angular velocity ω, through E = (1/2)·I·ω².

Applications: flywheels, centrifuges and driveshafts

In an engine, the flywheel banks kinetic energy between combustion pulses so the rotation runs smoothly. Modern flywheel batteries take this further, spinning composite rotors at tens of thousands of RPM to store grid energy. For a centrifuge, lab or industrial, you size it from I and ω to work out spin-up time and braking energy. And drive shafts, paper-mill rollers and rolling-mill cylinders all lean on I to predict their torsional behaviour and the startup torque their industrial motors have to deliver.

FAQ

Why is the hollow cylinder's I larger than the solid one? Mass that sits farther from the axis counts for more, since the contribution scales with r². A thin shell keeps every bit at r = R, while a solid cylinder buries half of its mass closer in, which drags the average r² down.

Does length matter? Not when you rotate about the central long axis, where only M and R show up. It starts to matter once you rotate about a transverse axis instead.

Units of I? kg·m² in SI. Multiply it by ω² (rad²/s²) and you land on joules of rotational kinetic energy.

Why are flywheels heavy at the rim? It is the way to wring the most I out of a fixed mass. Sitting matter at large R buys you the most energy storage per kg.