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Formulas/physics/Rotational Motion/Moment of Inertia of a Ring

Moment of Inertia of a Ring

All mass at distance R from axis.
Class 11Class JEE
Derivation

Result

I=MR2(about axis through centre, perpendicular to plane)I = MR^2 \quad \text{(about axis through centre, perpendicular to plane)}

Derivation

Every mass element of the ring is at the same distance RR from the central axis.

I=∫r2 dm=R2∫dm=MR2I = \int r^2 \, dm = R^2 \int dm = MR^2

The integral is trivial — r=Rr = R for all elements.

Other axes via theorems

About a diameter (using perpendicular axis theorem):

Iz=Ix+IyI_z = I_x + I_y, by symmetry Ix=IyI_x = I_y:

MR2=2Idiameter  ⟹  Idiameter=MR22MR^2 = 2I_{diameter} \implies I_{diameter} = \frac{MR^2}{2}

About a tangent perpendicular to plane (parallel axis theorem, d=Rd = R):

I=MR2+MR2=2MR2I = MR^2 + MR^2 = 2MR^2

About a tangent in the plane (parallel axis theorem applied to diameter, d=Rd = R):

I=MR22+MR2=3MR22I = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}

Summary for ring

AxisII
Central perpendicularMR2MR^2
DiameterMR22\frac{MR^2}{2}
Tangent perpendicular to plane2MR22MR^2
Tangent in plane3MR22\frac{3MR^2}{2}
Remember
The ring has the largest MI among standard shapes for its size — all its mass is at the maximum distance $R$ from the axis. This makes it the best shape for storing rotational energy (flywheels are ring-like for this reason).