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Formulas/physics/Rotational Motion/Moment of Inertia of a Solid Cylinder

Moment of Inertia of a Solid Cylinder

Uniform solid cylinder about its symmetry axis. Same as disc.
Class 11Class JEE
Derivation

Result

I=12MR2(about symmetry axis)I = \frac{1}{2}MR^2 \quad \text{(about symmetry axis)}

Why same as disc

A solid cylinder is a stack of thin discs. Each disc of mass dmdm contributes 12dm⋅R2\frac{1}{2}dm \cdot R^2 to the MI about the symmetry axis. Summing:

I=∫12R2 dm=12R2∫dm=12MR2I = \int \frac{1}{2}R^2 \, dm = \frac{1}{2}R^2 \int dm = \frac{1}{2}MR^2

The length of the cylinder doesn't matter — only the radius and total mass determine the MI about the symmetry axis.

About an axis through centre, perpendicular to symmetry axis

For a cylinder of length LL and radius RR:

I⊥=M12(3R2+L2)I_\perp = \frac{M}{12}(3R^2 + L^2)

This involves both RR and LL — unlike the symmetry axis result.

Summary for solid cylinder

AxisII
Symmetry axis12MR2\frac{1}{2}MR^2
Perpendicular through centreM(3R2+L2)12\frac{M(3R^2+L^2)}{12}
Tangent parallel to symmetry axis32MR2\frac{3}{2}MR^2
Remember
Solid cylinder and disc have the same $k^2/R^2 = 1/2$. This means they roll and accelerate identically on an incline (same $a = \frac{g\sin\theta}{1+k^2/R^2}$). All solid cylinders, regardless of radius or length, roll at the same speed on the same incline.