Newton's Second Law for Rotation (τ = Iα)

This is $F = ma$ for things that spin. Torque takes the role of force, angular acceleration takes the role of linear acceleration, and moment of inertia takes the role of mass: $\tau = I\alpha$. The correspondence is exact, and it is the reason rotational dynamics feels familiar once you accept that $I$ is doing a job that mass alone cannot.

What makes $I$ different from mass is that it depends not only on how much material a body contains but on where that material sits relative to the axis. Mass far from the axis resists being spun up far more than the same mass close in, because the far-out mass has to be accelerated to a much higher linear speed to achieve the same angular one. Apply 10 N·m to a flywheel with $I = 2$ kg·m² and it accelerates at $\alpha = 10/2 = 5$ rad/s². Build a flywheel of the same mass but twice the radius and $I$ becomes four times as large, so the same torque manages only 1.25 rad/s².

Run backwards, this is how drives get sized. Decide the angular acceleration a duty cycle requires, look up or compute the moment of inertia of everything the motor has to turn, and the product is the torque the motor must produce over and above whatever the load already demands. It is also the equation behind the reflected-inertia calculation in a geared drive, where a gearbox of ratio $n$ reduces the load inertia seen at the motor by $n^2$.

The mistake that invalidates the whole calculation is using a moment of inertia quoted about the wrong axis. Unlike mass, $I$ is not a property of a body — it is a property of a body and a chosen axis. A rod's moment of inertia about its centre is $\tfrac{1}{12}mL^2$ and about one end it is $\tfrac{1}{3}mL^2$, four times larger, and the parallel-axis theorem $I = I_{cm} + md^2$ is what converts between them. Taking a handbook value without checking which axis it refers to is the standard way to get a plausible answer that is wrong by a factor of several. The second point is the same one that $F = ma$ has: $\tau$ is the net torque. Bearing friction, windage, and whatever load is attached all subtract from the driving torque, and it is only what remains that produces $\alpha$. A motor rated at 10 N·m driving through a gearbox that absorbs 2 N·m accelerates the load as though 8 N·m were applied — which is why a rig that models beautifully can still fail to reach speed in the time the specification allows.