$F = ma$ has a rotational twin with the same three sockets: $\tau = I\alpha$, read aloud tau equals I alpha. $\tau$ — tau — is the net torque, the turning effect, in newton-metres. $\alpha$ is the angular acceleration it produces, in $\mathrm{rad/s^2}$. And $I$ is the moment of inertia, in $\mathrm{kg \cdot m^2}$ — the rotational stand-in for mass. Whichever of the three the question withholds is the one you solve for.

But $I$ is not simply the mass, and that is the interesting part. Rotation cares where the metal sits, because metal far from the axis has further to travel for the same angle. So the distance enters squared. A compact mass $m$ in kilograms parked at radius $r$ in metres gives $I = m r^{2}$. A uniform solid disk of mass $m$ and outer radius $r$, turning about its own centre, gives $I = \tfrac{1}{2} m r^{2}$ — the ½ is the discount for all the metal that sits nearer the axis than the rim does.

That squared radius is why a flywheel designer would always rather add a centimetre of diameter than a kilogram of steel, and why a figure skater speeds up by pulling her arms in. Push the units through every rearrangement: $\mathrm{kg \cdot m^2}$ times $\mathrm{rad/s^2}$ is a newton-metre. If your rearrangement's units refuse to land there, the rearrangement is wrong — no appeal. Units that DO work out never prove you right; the check only ever runs one way.