Moment of Inertia of a Hollow Cylinder

Result

$I = MR^2 \quad \text{(thin cylindrical shell about symmetry axis)}$

For a thick cylindrical shell with inner radius $R_1$ and outer radius $R_2$:

$I = \frac{M(R_1^2 + R_2^2)}{2}$

Derivation — thin shell

A thin cylindrical shell is a stack of rings, each of radius $R$. Each ring contributes $dm \cdot R^2$ to the MI:

$I = \int R^2 \, dm = R^2 M$

Derivation — thick shell

Divide into thin cylindrical shells of radius $r$, thickness $dr$:

Volume of shell: $2\pi r L \, dr$ (length $L$)

$dm = \rho \cdot 2\pi r L \, dr, \quad \rho = \frac{M}{\pi(R_2^2 - R_1^2)L}$

$I = \int_{R_1}^{R_2} r^2 \cdot \frac{2\pi r L M}{\pi(R_2^2-R_1^2)L} \, dr = \frac{2M}{R_2^2-R_1^2}\int_{R_1}^{R_2} r^3 \, dr$

$= \frac{2M}{R_2^2-R_1^2} \cdot \frac{R_2^4-R_1^4}{4} = \frac{M(R_2^2+R_1^2)}{2}$

For thin shell ($R_1 = R_2 = R$): $I = MR^2$ ✓

For solid cylinder ($R_1 = 0$, $R_2 = R$): $I = \frac{MR^2}{2}$ ✓

Hollow vs solid cylinder — rolling race

A hollow cylinder has $k^2/R^2 = 1$ (for thin shell). A solid cylinder has $k^2/R^2 = 1/2$.

Rolling acceleration: $a = \frac{g\sin\theta}{1 + k^2/R^2}$

• Hollow: $a = \frac{g\sin\theta}{2}$
• Solid: $a = \frac{2g\sin\theta}{3}$

The solid cylinder accelerates faster and wins a rolling race down an incline — it has less rotational inertia relative to its mass.