Moment of Inertia of a Rod

About the centre

$I_{cm} = \frac{1}{12}ML^2$

Derivation: Place the rod along the $x$-axis, centre at origin. Linear mass density $\lambda = M/L$.

$I_{cm} = \int_{-L/2}^{L/2} x^2 \cdot \frac{M}{L} \, dx = \frac{M}{L} \cdot \frac{x^3}{3}\Big|_{-L/2}^{L/2} = \frac{M}{L} \cdot \frac{2(L/2)^3}{3} = \frac{M}{L} \cdot \frac{L^3}{12} = \frac{ML^2}{12}$

About one end

$I_{end} = \frac{1}{3}ML^2$

Via parallel axis theorem: $d = L/2$ (distance from centre to end):

$I_{end} = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3}$

Direct derivation: Place rod from $x = 0$ to $x = L$:

$I_{end} = \int_0^L x^2 \cdot \frac{M}{L} \, dx = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3}$

The factor of 4

$I_{end} = 4 \times I_{cm}$. Rotating a rod about its end requires 4 times the torque (for the same angular acceleration) as rotating it about its centre. This is why doors are easier to push near the hinged end — wait, no: they are harder to push at the hinge and easier at the far edge because of the torque formula $\tau = Fr$. But the MI is higher about the hinged end, requiring more torque for the same angular acceleration.

About an axis at distance $d$ from centre

By parallel axis theorem:

$I = \frac{ML^2}{12} + Md^2$

Summary for rod