MOMENT OF INERTIA ($kg \cdot m^2$)
$I_{cm}$ (Center)0.00
$+ Md^2$ (Shift)0.00
Total $I$0.00
Theorem Mechanics:
■ $I_{cm}$ (Uniform Thin Rod):
$I_{cm} = \frac{1}{12} M L^2$
The minimum possible moment of inertia for any axis parallel to the chosen orientation occurs when the axis passes exactly through the Center of Mass (CM).
■ The $Md^2$ Penalty:
As you shift the pivot axis away from the CM by distance $d$, the rotational resistance increases quadratically. When $d = L/2$ (end of rod), $I = \frac{1}{3} M L^2$.
■ $I_{cm}$ (Uniform Thin Rod):
$I_{cm} = \frac{1}{12} M L^2$
The minimum possible moment of inertia for any axis parallel to the chosen orientation occurs when the axis passes exactly through the Center of Mass (CM).
■ The $Md^2$ Penalty:
As you shift the pivot axis away from the CM by distance $d$, the rotational resistance increases quadratically. When $d = L/2$ (end of rod), $I = \frac{1}{3} M L^2$.