Moment of Inertia of a Thin Rod about one End: ML²/3

Derivation

Integrate from 0 to L: I = (M/L) ∫₀ᴸ x² dx = (M/L)(L³/3) = ML²/3.
Check with the parallel axis theorem: ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3.

Body and axis	I
Thin ring, through the centre, perpendicular to its plane	MR²
Thin ring, about a diameter	MR²/2
Thin ring, about a tangent in its plane	3MR²/2
Thin ring, about a tangent perpendicular to its plane	2MR²
Uniform disc, through the centre, perpendicular to its plane	MR²/2
Uniform disc, about a diameter	MR²/4
Uniform disc, about a tangent in its plane	5MR²/4
Uniform disc, about a tangent perpendicular to its plane	3MR²/2
Thin rod, through the centre, perpendicular to its length	ML²/12
Thin rod, through one end, perpendicular to its length	ML²/3
Solid sphere, about a diameter	2MR²/5
Thin hollow sphere (spherical shell), about a diameter	2MR²/3
Solid sphere, about a tangent	7MR²/5
Thin hollow sphere, about a tangent	5MR²/3
Solid cylinder, about its own axis	MR²/2
Thin hollow cylinder, about its own axis	MR²
Solid cylinder, through the centre, perpendicular to its axis	M(R²/4 + L²/12)
Rectangular plate (sides a and b), through the centre, perpendicular to the plate	M(a² + b²)/12
Annular disc (radii R₁ and R₂), through the centre, perpendicular to its plane	M(R₁² + R₂²)/2
Solid cone, about its own axis	3MR²/10

What is the moment of inertia of a thin rod through one end, perpendicular to its length?

I = ML²/3, so the radius of gyration is k = L/√3.

How is the moment of inertia of a thin rod through one end, perpendicular to its length derived?

Integrate from 0 to L: I = (M/L) ∫₀ᴸ x² dx = (M/L)(L³/3) = ML²/3. Check with the parallel axis theorem: ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3.