Moment of Inertia: Solid Disk

A solid disk has its mass spread evenly from the axis out to the rim, and integrating $r^2$ over that distribution gives exactly half of $mr^2$: $I = \tfrac{1}{2}mr^2$. The coefficient is a pure statement about geometry. Compare the two extremes and it makes sense: a thin hoop with all its mass at the rim has $I = mr^2$, and a mass concentrated at the very centre would have $I = 0$. A uniform disk lands halfway between, though not for the reason the word "halfway" suggests — most of a disk's area is in its outer half, and the two effects happen to cancel to a clean $\tfrac{1}{2}$.

A steel grinding wheel 300 mm across and weighing 8 kg has $r = 0.15$ m, so $I = 0.5 \times 8 \times 0.15^2 = 0.09$ kg·m². At 3600 rpm, $\omega = 377$ rad/s, it stores $\tfrac{1}{2} \times 0.09 \times 377^2 \approx 6.4$ kJ — enough that it will keep turning for a long time after the power is cut, and enough to do serious harm if it lets go.

Nothing in the formula mentions thickness, and that is not an omission: a solid cylinder of any length has the same $\tfrac{1}{2}mr^2$ about its central axis, because stacking disks along the axis does not move any mass closer to or further from it. So this one expression covers a coin, a flywheel, a roller and a shaft alike. It is the value behind the stored energy in flywheels, the spin-up time of hard-disk platters, and the rolling behaviour of any wheel treated as a uniform disk.

This is the moment of inertia about the central axis — the one the disk naturally spins about — and only that one. Flip the disk so it turns about a diameter instead, like a coin rolled on edge and spun about a horizontal line through its centre, and the correct value is $\tfrac{1}{4}mr^2$, half as much. The two get confused because the same disk is involved. The second error is applying $\tfrac{1}{2}mr^2$ to something that is not solid. A tube, a pipe, or a rim-heavy flywheel with a light web has more of its mass out near the radius, and its moment of inertia is $\tfrac{1}{2}m(r_1^2 + r_2^2)$ using both the inner and outer radii — for a thin-walled tube that approaches $mr^2$, twice the solid-disk figure. Treating a fabricated flywheel as a solid disk understates its inertia badly, and since flywheels are deliberately built rim-heavy, it understates exactly the case where you were relying on the number. And, as ever, $r$ is a radius: a "300 mm wheel" gives 0.15, and using 0.3 overstates $I$ fourfold.